Area of a Triangle By Heron's formula
 Books Name
 ABCD CLASSES Mathematics Book
 Publication
 ABCD CLASSES
 Course
 CBSE Class 9
 Subject
 Mathmatics
Heron's Formula
Heron’s formula is used to find the area of a triangle when we know the length of all its sides. It is also termed Hero’s Formula. We can Heron’s formula to find different types of triangles, such as scalene, isosceles and equilateral triangles. We don’t have to need to know the angle measurement of a triangle to calculate its area, by using Heron’s formula.
What is Heron’s Formula?
Heron’s formula is a formula to calculate the area of triangles, given the three sides of the triangle. This formula is also used to find the area of the quadrilateral, by dividing the quadrilateral into two triangles, along its diagonal.
If a, b and c are the three sides of a triangle, respectively, then Heron’s formula is given by:
Semiperimeter, s= Perimeter of triangle/2 = (a+b+c)/2 
History of Heron’s Formula
Hero of Alexandria was a great mathematician who derived the formula for the calculation of the area of a triangle using the length of all three sides. He also extended this idea to find the area of quadrilateral and also higherorder polygons. This formula has its huge applications in trigonometry such as proving the law of cosines or the law of cotangents, etc.
Heron’s Formula for Triangles
According to Heron, we can find the area of any given triangle, whether it is a scalene, isosceles or equilateral, by using the formula, provided by the sides of the triangle.
Suppose, a triangle ABC, whose sides are a, b and c, respectively. Thus, the area of a triangle can be given by;
Where “s” is semiperimeter = (a+b+c) / 2
And a, b, c are the three sides of the triangle.
Proof of Heron’s Formula
There are two methods by which we can derive Heron’s formula.
 First, by using trigonometric identities and cosine rule.
 Secondly, solving algebraic expressions using the Pythagoras theorem.
Let us see one by one both the proofs and derivation.
Using Cosine Rule
Let us prove the result using the law of cosines:
Let a, b, and c be the sides of the triangle and α, β, and γ are opposite angles to the sides.
We know that the law of cosines is
Again, using trig identity, we have
=
Here, Base of triangle = a
Altitude = b sinγ
Now,
Using Pythagoras Theorem
Area of a Triangle with 3 Sides
Area of ∆ABC is given by
A = 1/2 bh _ _ _ _ (i)
Draw a perpendicular BD on AC
Consider a ∆ADB
x^{2} + h^{2} = c^{2}
x^{2} = c^{2} − h^{2}—(ii)
⇒x = √(c^{2}−h^{2})−−−−−−—(iii)
Consider a ∆CDB,
(b−x)^{2} + h^{2} = a^{2}
(b−x)^{2} = a^{2} − h^{2}
b^{2} − 2bx + x^{2} = a^{2}–h^{2}
Substituting the value of x and x^{2} from equation (ii) and (iii), we get
b^{2} – 2b√(c^{2}−h^{2})+ c^{2}−h^{2} = a^{2} − h^{2}
b^{2} + c^{2} − a^{2} = 2b√(c^{2} − h^{2})
Squaring on both sides, we get;
(b^{2}+c^{2}–a^{2})^{2} = 4b^{2}(c^{2}−h^{2})
The perimeter of a ∆ABC is
P= a+b+c
Substituting the value of h in equation (i), we get;
Semi perimeter(s) =
Note: Heron’s formula is applicable to all types of triangles and the formula can also be derived using the law of cosines and the law of Cotangents.
How to Find the Area Using Heron’s Formula?
To find the area of a triangle using Heron’s formula, we have to follow two steps:
 Find the perimeter of the given triangle
 Then, find the value of the semiperimeter of the given triangle; S = (a+b+c)/2
 Now use Heron’s formula to find the area of a triangle (√(s(s – a)(s – b)(s – c)))
 Finally, represent the area with the accurate square units (such as m^{2}, cm^{2}, in^{2}, etc.)
Heron’s Formula for Equilateral Triangle
As we know the equilateral triangle has all its sides equal. To find the area of the equilateral triangle let us first find the semi perimeter of the equilateral triangle will be:
s = (a+a+a)/2
s=3a/2
where a is the length of the side.
Now, as per the heron’s formula, we know;
Since, a = b = c
Therefore,
A = √[s(sa)^{3}]
which is the required formula.
Heron’s Formula for Isosceles Triangle
An isosceles triangle has two of its sides equal and the angles corresponding to these sides are congruent. To find the area of an isosceles triangle, we can derive the heron’s formula as given below:
Let a be the length of the congruent sides and b be the length of the base.
Semiperimeter (s) = (a + a + b)/2
s = (2a + b)/2
Using the heron’s formula of a triangle,
Area = √[s(s – a)(s – b)(s – c)]
By substituting the sides of an isosceles triangle,
Area = √[s(s – a)(s – a)(s – b)]
= √[s(s – a)^{2}(s – b)]
Or
= (s – a)√[s(s – b)]
which is the required formula to find the area for the given isosceles triangle.
Heron’s Formula for Scalene Triangle
As we know, a scalene triangle has all three unequal sides, thus, we can directly use the original Heron’s formula to find the area.
Area of scalene triangle = √s(sa)(sb)(sc),
where s = (a + b + c)/2
Heron’s Formula For Quadrilateral
Let us learn how to find the area of quadrilateral using Heron’s formula here.
If ABCD is a quadrilateral, where ABCD and AC & BD are the diagonals.
AC divides the quad.ABCD into two triangles ADC and ABC.
Now we have two triangles here.
Area of quad.ABCD = Area of ∆ADC + Area of ∆ABC
So, if we know the lengths of all sides of a quadrilateral and the length of diagonal AC, then we can use Heron’s formula to find the total area.
Hence, we will first find the area of ∆ADC and area of ∆ABC using Heron’s formula and at last, will add them to get the final value.
Applications of Heron’s Formula
The two major applications of Heron’s formula are:
 To find the area of different types of a triangle (when the length of three sides is given)
 To find the area of a quadrilateral (when the length of all three sides is given)
Solved Examples
Let us now look into some examples to have a brief insight into the topic:
Example 1: Find the area of a trapezium, the length of whose parallel sides is given as 22 cm and 12 cm and the length of the other sides is 14 cm each.
Solution: Let PQRS be the given trapezium in which PQ = 22 cm, SR = 12 cm,
PS=QR=14cm.
Constructions: Draw ORPS
Now, PORS is a parallelogram in which PSOR and POSR
Therefore, PO=SR=12cm
⇒OQ = PQPO = 22 12 = 10cm
In ∆OQR , we have
Area of ∆OQR =
=
=
=
………(i)
We know that Area =
…………..(ii)
Area of trapezium =
=
=
Example 2: Find the area of the triangle whose sides measure 10 cm, 17 cm and 21 cm. Also, determine the length of the altitude on the side which measures 17 cm.
Solution: s =
=
= 24
Area of Triangle=
= 84 square cm
Taking 17 cm as the base length we need to find the height
Area, A = 1/2 x base x height
1/2 x 17 x h = 84 or h = 168/17 = 9.88 cm (Rounded to the nearest hundredth).
Example 3: A triangle PQR has sides 4 cm, 13 cm and 15 cm. Find the area of the triangle.
Solution:
Semiperimeter of triangle PQR, s = (4+13+15)/2 = 32/2 = 16
By heron’s formula, we know;
A = √[s(sa)(sb)(sc)]
Hence, A = √[16(164)(1613)(1615)] = √(16 x 12 x 3 x 1) = √576 = 24 sq.cm
Heron’s Formula – Practice Questions
A few practice questions are given below that will help you in improving your knowledge of how to find the area of a given triangle.
 Find the area of a triangle whose perimeter is 54 cm and two of its sides measure 12 cm and 25 cm. [Answer: 90 sq.cm]
 If the length of equal sides of an isosceles triangle is 5 cm and the base is 6 cm, then find its area using heron’s formula. [Answer: 12 sq.cm]
 The sides of a quadrilateral field, taken in order are 26 cm, 27 cm, 7 cm, and 24 cm respectively. The angle contained by the last two sides is a right angle. Find its area. [Answer: 375.85 sq.cm.]
Area of a Triangle  Heron's Formula
The basic formula for the area of a triangle is only helpful if you know the base and the height.
So what do you do if you only know the three side lengths?
There are just two steps.
Step 1: Calculate half the perimeter of the triangle and call it s.
Step 2: Use the s in the following formula:
Let's use the formula to determine the area of the triangle above.
Step 1: The perimeter of the triangle is equal to 12 + 22 + 16 = 50
Therefore, the s value is half of 50 or 25.
Step 2: Replace the s in the area formula with 25 and solve.
A =
A =
A =
A = 93.7 m^{2}
Here is a couple more example to try.
Example 1:
Step 1: Determine half the perimeter.
s =
Step 2: Use the s in Heron's formula.
A =
A =
A =
A = 24 units^{2}
Now, this triangle does give the base and height. So let's compare our work to the formula you might be more familiar with using to solve.
A =
A =
A = 24 units^{2}
Here we have the same answer as we got from using Heron's method.
Example 2:
Step1: Determine half the perimeter.
s =
s =
s = 15.5
Step 2: Use the s in Heron's formula.
A =
A =
A =
A = 33.7 mm^{2}
When the solution is not rational, the answer can be rounded. In this example, we rounded to the nearest tenth.
Let's Review
If you are given the three sides of a triangle, you can use the perimeter and Heron's formula to determine the area. There are just two steps.
Step 1: Determine half the perimeter.
Step 2: Use the three side lengths and the half perimeter in Heron's formula.
Some of the areas will be irrational numbers. That means that they cannot be represented as a fraction. Instead, they are decimals that never end and never repeat. When this happens, the area can be rounded to whatever place value you may have been asked for.
Area of a Triangle By Heron's formula
Chapter 12  Heron's Formula
Heron's Formula
Heron’s formula is used to find the area of a triangle when we know the length of all its sides. It is also termed as Hero’s Formula. We can Heron’s formula to find different types of triangles, such as scalene, isosceles and equilateral triangles. We don’t have to need to know the angle measurement of a triangle to calculate its area, by using Heron’s formula.
What is Heron’s Formula?
Heron’s formula is a formula to calculate the area of triangles, given the three sides of the triangle. This formula is also used to find the area of the quadrilateral, by dividing the quadrilateral into two triangles, along its diagonal.
If a, b and c are the three sides of a triangle, respectively, then Heron’s formula is given by:
Semiperimeter, s= Perimeter of triangle/2 = (a+b+c)/2 
History of Heron’s Formula
Hero of Alexandria was a great mathematician who derived the formula for the calculation of the area of a triangle using the length of all three sides. He also extended this idea to find the area of quadrilateral and also higherorder polygons. This formula has its huge applications in trigonometry such as proving the law of cosines or the law of cotangents, etc.
Heron’s Formula for Triangles
According to Heron, we can find the area of any given triangle, whether it is a scalene, isosceles or equilateral, by using the formula, provided the sides of the triangle.
Suppose, a triangle ABC, whose sides are a, b and c, respectively. Thus, the area of a triangle can be given by;
Where “s” is semiperimeter = (a+b+c) / 2
And a, b, c are the three sides of the triangle.
Proof of Heron’s Formula
There are two methods by which we can derive Heron’s formula.
 First, by using trigonometric identities and cosine rule.
 Secondly, solving algebraic expressions using the Pythagoras theorem.
Let us see one by one both the proofs or derivation.
Using Cosine Rule
Let us prove the result using the law of cosines:
Let a, b, c be the sides of the triangle and α, β, γ are opposite angles to the sides.
We know that, the law of cosines is
Again, using trig identity, we have
=
Here, Base of triangle = a
Altitude = b sinγ
Now,
Using Pythagoras Theorem
Area of a Triangle with 3 Sides
Area of ∆ABC is given by
A = 1/2 bh _ _ _ _ (i)
Draw a perpendicular BD on AC
Consider a ∆ADB
x^{2} + h^{2} = c^{2}
x^{2} = c^{2} − h^{2}—(ii)
⇒x = √(c^{2}−h^{2})−−−−−−—(iii)
Consider a ∆CDB,
(b−x)^{2} + h^{2} = a^{2}
(b−x)^{2} = a^{2} − h^{2}
b^{2} − 2bx + x^{2} = a^{2}–h^{2}
Substituting the value of x and x^{2} from equation (ii) and (iii), we get
b^{2} – 2b√(c^{2}−h^{2})+ c^{2}−h^{2} = a^{2} − h^{2}
b^{2} + c^{2} − a^{2} = 2b√(c^{2} − h^{2})
Squaring on both sides, we get;
(b^{2}+c^{2}–a^{2})^{2} = 4b^{2}(c^{2}−h^{2})
The perimeter of a ∆ABC is
P= a+b+c
Substituting the value of h in equation (i), we get;
Semi perimeter(s) =
Note: Heron’s formula is applicable to all types of triangles and the formula can also be derived using the law of cosines and the law of Cotangents.
How to Find the Area Using Heron’s Formula?
To find the area of a triangle using Heron’s formula, we have to follow two steps:
 Find the perimeter of the given triangle
 Then, find the value of the semiperimeter of the given triangle; S = (a+b+c)/2
 Now use Heron’s formula to find the area of a triangle (√(s(s – a)(s – b)(s – c)))
 Finally, represent the area with the accurate square units (such as m^{2}, cm^{2}, in^{2}, etc.)
Heron’s Formula for Equilateral Triangle
As we know the equilateral triangle has all its sides equal. To find the area of the equilateral triangle let us first find the semi perimeter of the equilateral triangle will be:
s = (a+a+a)/2
s=3a/2
where a is the length of the side.
Now, as per the heron’s formula, we know;
Since, a = b = c
Therefore,
A = √[s(sa)^{3}]
which is the required formula.
Heron’s Formula for Isosceles Triangle
An isosceles triangle has two of its sides equal and the angles corresponding to these sides are congruent. To find the area of an isosceles triangle, we can derive the heron’s formula as given below:
Let a be the length of the congruent sides and b be the length of the base.
Semiperimeter (s) = (a + a + b)/2
s = (2a + b)/2
Using the heron’s formula of a triangle,
Area = √[s(s – a)(s – b)(s – c)]
By substituting the sides of an isosceles triangle,
Area = √[s(s – a)(s – a)(s – b)]
= √[s(s – a)^{2}(s – b)]
Or
= (s – a)√[s(s – b)]
which is the required formula to find the area for the given isosceles triangle.
Heron’s Formula for Scalene Triangle
As we know, a scalene triangle has all three unequal sides, thus, we can directly use the original Heron’s formula to find the area.
Area of scalene triangle = √s(sa)(sb)(sc),
where s = (a + b + c)/2
Heron’s Formula For Quadrilateral
Let us learn how to find the area of quadrilateral using Heron’s formula here.
If ABCD is a quadrilateral, where ABCD and AC & BD are the diagonals.
AC divides the quad.ABCD into two triangles ADC and ABC.
Now we have two triangles here.
Area of quad.ABCD = Area of ∆ADC + Area of ∆ABC
So, if we know the lengths of all sides of a quadrilateral and the length of diagonal AC, then we can use Heron’s formula to find the total area.
Hence, we will first find the area of ∆ADC and area of ∆ABC using Heron’s formula and at last, will add them to get the final value.
Applications of Heron’s Formula
The two major applications of Heron’s formula are:
 To find the area of different types of a triangle (when the length of three sides are given)
 To find the area of a quadrilateral (when the length of all three sides are given)
Solved Examples
Let us now look into some examples to have a brief insight into the topic:
Example 1: Find the area of a trapezium, the length of whose parallel sides is given as 22 cm and 12 cm and the length of other sides is 14 cm each.
Solution: Let PQRS be the given trapezium in which PQ = 22 cm, SR = 12 cm,
PS=QR=14cm.
Constructions: Draw ORPS
Now, PORS is a parallelogram in which PSOR and POSR
Therefore, PO=SR=12cm
⇒OQ = PQPO = 22 12 = 10cm
In ∆OQR , we have
Area of ∆OQR =
=
=
=
………(i)
We know that Area =
…………..(ii)
Area of trapezium =
=
=
Example 2: Find the area of the triangle whose sides measure 10 cm, 17 cm and 21 cm. Also, determine the length of the altitude on the side which measures 17 cm.
Solution: s =
=
= 24
Area of Triangle=
= 84 square cm
Taking 17 cm as the base length we need to find the height
Area, A = 1/2 x base x height
1/2 x 17 x h = 84 or h = 168/17 = 9.88 cm (Rounded to the nearest hundredth).
Example 3: A triangle PQR has sides 4 cm, 13 cm and 15 cm. Find the area of the triangle.
Solution:
Semiperimeter of triangle PQR, s = (4+13+15)/2 = 32/2 = 16
By heron’s formula, we know;
A = √[s(sa)(sb)(sc)]
Hence, A = √[16(164)(1613)(1615)] = √(16 x 12 x 3 x 1) = √576 = 24 sq.cm
Heron’s Formula – Practice Questions
A few practice questions are given below that will help you in improving your knowledge of how to find the area of a given triangle.
 Find the area of a triangle whose perimeter is 54 cm and two of its sides measure 12 cm and 25 cm. [Answer: 90 sq.cm]
 If the length of equal sides of an isosceles triangle is 5 cm and base is 6 cm, then find its area using heron’s formula. [Answer: 12 sq.cm]
 The sides of a quadrilateral field, taken in order are 26 cm, 27 cm, 7 cm, 24 cm respectively. The angle contained by the last two sides is a right angle. Find its area. [Answer: 375.85 sq.cm.]
Area of a Triangle  Heron's Formula
The basic formula for the area of a triangle is only helpful if you know the base and the height.
So what do you do if you only know the three side lengths?
There are just two steps.
Step 1: Calculate half the perimeter of the triangle and call it s.
Step 2: Use the s in the following formula:
Let's use the formula to determine the area of the triangle above.
Step 1: The perimeter of the triangle is equal to 12 + 22 + 16 = 50
Therefore, the s value is half of 50 or 25.
Step 2: Replace the s in the area formula with 25 and solve.
A =
A =
A =
A = 93.7 m^{2}
Here are a couple more example to try.
Example 1:
Step 1: Determine half the perimeter.
s =
Step 2: Use the s in Heron's formula.
A =
A =
A =
A = 24 units^{2}
Now, this triangle does give the base and height. So let's compare our work to the formula you might be more familiar with using to solve.
A =
A =
A = 24 units^{2}
Here we have the same answer as we got from using Heron's method.
Example 2:
Step1: Determine half the perimeter.
s =
s =
s = 15.5
Step 2: Use the s in Heron's formula.
A =
A =
A =
A = 33.7 mm^{2}
When the solution is not rational, the answer can be rounded. In this example, we rounded to the nearest tenth.
Let's Review
If you are given the three sides of a triangle, you can use the perimeter and Heron's formula to determine the area. There are just two steps.
Step 1: Determine half the perimeter.
Step 2: Use the three side lengths and the half perimeter in Heron's formula.
Some of the areas will be irrational numbers. That means that they cannot be represented as a fraction. Instead they are decimals that never end and never repeat. When this happens, the area can be rounded to whatever place value you may have been asked for.
Surface Area of Cuboid and a Cube
 Books Name
 ABCD CLASSES Mathematics Book
 Publication
 ABCD CLASSES
 Course
 CBSE Class 9
 Subject
 Mathmatics
Surface Areas and Volume
Surface area and volume are calculated for any threedimensional geometrical shape. The surface area of any given object is the area or region occupied by the surface of the object. Whereas volume is the amount of space available in an object.
In geometry, there are different shapes and sizes such as spheres, cubes, cuboids, cones, cylinders, etc. Each shape has its surface area as well as volume. But in the case of twodimensional figures like squares, circles, rectangles, triangles, etc., we can measure only the area covered by these figures and there is no volume available. Now, let us see the formulas of surface areas and volumes for different 3dshapes.
What is Surface Area?
The space occupied by a twodimensional flat surface is called the area. It is measured in square units. The area occupied by a threedimensional object by its outer surface is called a surface area. It is also measured in square units.
Generally, Area can be of two types:
(i) Total Surface Area
(ii) Curved Surface Area/Lateral Surface Area
Total surface area
Total surface area refers to the area including the base(s) and the curved part. It is the total of the area covered by the surface of the object. If the shape has a curved surface and base, then the total area will be the sum of the two areas.
Curved surface area/Lateral surface area
Curved surface area refers to the area of only the curved part of the shape excluding its base(s). It is also referred to as lateral surface area for shapes such as a cylinder.
What is Volume?
The amount of space, measured in cubic units, that an object or substance occupies is called volume. Twodimensional doesn’t have volume but has area only. For example, the Volume of the Circle cannot be found, though the Volume of the sphere can be. It is so because a sphere is a threedimensional shape.
Surface Area and Volume Formulas
Below given is the table for the calculating Surface area and Volume for the basic geometrical figures:
Name: Square
Perimeter: 4a
Total Surface Area: a^{2}
Curved Surface Area/Lateral Surface Area: —
Volume: 
Figure: 
Name: Rectangle
Perimeter: 2(w+h)
Total Surface Area: w.h
Curved Surface Area/Lateral Surface Area: —
Volume: 
Figure: 
Name: Parallelogram
Perimeter: 2(a+b)
Total Surface Area: b.h
Curved Surface Area/Lateral Surface Area: —
Volume: 
Figure: 
Name: Trapezoid
Perimeter: a+b+c+d
Total Surface Area: 1/2(a+b).h
Curved Surface Area/Lateral Surface Area: —
Volume: 
Figure: 
Name: Circle
Perimeter: 2 π r
Total Surface Area: π r^{2}
Curved Surface Area/Lateral Surface Area: —
Volume: 
Figure: 
Name: Ellipse
Perimeter: 2π√(a^{2} + b^{2})/2
Total Surface Area: π a.b
Curved Surface Area/Lateral Surface Area: —
Volume: 
Figure: 
Name: Triangle
Perimeter: a+b+c
Total Surface Area: 1/2 * b * h
Curved Surface Area/Lateral Surface Area: —
Volume: 
Figure: 
Name: Cuboid
Perimeter: 4(l+b+h)
Total Surface Area: 2(lb+bh+hl)
Curved Surface Area/Lateral Surface Area: 2h(l+b)
Volume: l * b * h
Figure: 
Name: Cube
Perimeter: 6a
Total Surface Area: 6a^{2}
Curved Surface Area/Lateral Surface Area: 4a^{2}
Volume: a^{3}
Figure: 
Name: Cylinder
Perimeter: 
Total Surface Area: 2 π r(r+h)
Curved Surface Area/Lateral Surface Area: 2πrh
Volume: π r^{2} h
Figure: 
Name: Cone
Perimeter: 
Total Surface Area: π r(r+l)
Curved Surface Area/Lateral Surface Area: π r l
Volume: 1/3π r^{2} h
Figure: 
Name: Sphere
Perimeter: 
Total Surface Area: 4 π r^{2}
Curved Surface Area/Lateral Surface Area: 4π r^{2}
Volume: 4/3π r^{3}
Figure: 
Name: Hemisphere
Perimeter: 
Total Surface Area: 3 π r^{2}
Curved Surface Area/Lateral Surface Area: 2 π r^{2}
Volume: 2/3π r^{3}
Figure: 
Surface Area of Cube and Cuboid
Cube and cuboid are threedimensional shapes that consist of six faces, eight vertices and twelve edges. The primary difference between them is a cube has all its sides equal whereas the length, width and height of a cuboid are different. Both shapes look almost the same but have different properties. The area and volume of a cube, cuboid and also cylinder differ from each other.
In everyday life, we have seen many objects like wooden boxes, a matchbox, a tea packets, a chalk box, a dice, a book, etc. All these objects have a similar shape. All these objects are made of six rectangular planes or square planes. In mathematics, the shape of these objects is either a cuboid or a cube. Here, in this article, we will learn the difference between cube and cuboid shapes with the help of their properties and formulas of surface area and volume.
Definition of Cube and Cuboid Shape
The cube and cuboid shapes in Maths are threedimensional shapes. The cube and cuboid are obtained from the rotation of the twodimensional shapes called square and rectangle respectively.
Cube: A cube is a threedimensional shape that is defined XYZ plane. It has six faces, eight vertices and twelve edges. All the faces of the cube are in square shape and have equal dimensions.
Cuboid: A cuboid is also a polyhedron having six faces, eight vertices and twelve edges. The faces of the cuboid are parallel. But not all the faces of a cuboid are equal in dimensions.
Hence, cube and cuboid shapes have six faces, eight vertices and twelve edges.
Difference Between Cube and Cuboid
The difference between the cube and cuboid shapes are as follows:
 The sides of the cube are equal but for cuboids they are different.
 The sides of the cube are square in shape but for cuboid, they are in a rectangular shape.
 All the diagonals of the cube are equal but a cuboid has equal diagonals for only parallel sides.
Cube and Cuboid Shape
As we already know, both cube and cuboid are in 3D shape, whose axes go along the xaxis, yaxis and zaxis. Now, let us learn in detail.
A cuboid is a closed 3dimensional geometrical figure bounded by six rectangular plane regions.
Properties of a Cuboid
Below are the properties of the cuboid, its faces, base and lateral faces, edges and vertices.
Faces of Cuboid
A Cuboid is made up of six rectangles, each of the rectangles is called the face. In the figure above, ABFE, DAEH, DCGH, CBFG, ABCD and EFGH are the 6 faces of the cuboid.
The top face ABCD and bottom face EFGH form a pair of opposite faces. Similarly, ABFE, DCGH, and DAEH, CBFG are pairs of opposite faces. Any two faces other than the opposite faces are called adjacent faces.
Consider a face ABCD, the adjacent face to this are ABFE, BCGF, CDHG, and ADHE.
Base and lateral faces
Any face of a cuboid may be called the base of the cuboid. The four faces which are adjacent to the base are called the lateral faces of the cuboid. Usually, the surface on which a solid rests is known to be the base of the solid.
In Figure (1) above, EFGH represents the base of a cuboid.
Edges
The edge of the cuboid is a line segment between any two adjacent vertices.
There are 12 edges, they are AB, AD, AE, HD, HE, HG, GF, GC, FE, FB, EF and CD and the opposite sides of a rectangle are equal.
Hence, AB = CD = GH = EF, AE = DH = BF = CG and EH = FG = AD = BC.
Vertices of Cuboid
The point of intersection of the 3 edges of a cuboid is called the vertex of a cuboid.
A cuboid has 8 vertices. A, B, C, D, E, F, G and H represent vertices of the cuboid in fig 1.
By observation, the twelve edges of a cuboid can be grouped into three groups, such that all edges in one group are equal in length, so there are three distinct groups and the groups are named as length, breadth and height.
A solid having its length, breadth, and height all to be equal in measurement is called a cube. A cube is a solid bounded by six square plane regions, where the side of the cube is called the edge.
Properties of Cube
 A cube has six faces and twelve edges of equal length.
 It has squareshaped faces.
 The angles of the cube in the plane are at a right angle.
 Each face of the cube meets four other faces.
 Each vertex of the cube meets three faces and three edges.
 The opposite edges of the cube are parallel to each other.
Cube and Cuboid Formulas
The formulas for cube and cuboid shapes are defined based on their surface areas, lateral surface areas and volume.
Cube 
Cuboid 
Total Surface Area = 6(side)^{2} 
Total Surface area = 2 (length × breadth + breadth × height + length × height) 
Lateral Surface Area = 4 (Side)^{2} 
Lateral Surface area = 2 height(length + breadth) 
Volume of cube = (Side)^{3} 
Volume of the cuboid = (length × breadth × height) 
Diagonal of a cube = √3l 
Diagonal of the cuboid =√( l^{2} + b^{2} +h^{2}) 
Perimeter of cube = 12 × side 
Perimeter of cuboid = 4 (length + breadth + height) 
Surface Area of Cube and Cuboid
The surface area of a cuboid is equal to the sum of the areas of its six rectangular faces.
Consider a cuboid having the length to be ‘l’ cm, breadth be ‘b’ cm and height be ‘h’ cm.
 Area of face EFGH = Area of Face ABCD = (l × b) cm^{2}
 Area of face BFGC = Area of face AEHD = (b × h) cm^{2}
 Area of face DHGC = Area of face ABFE = (l × h) cm^{2}
Total surface area of a cuboid = Sum of the areas of all its 6 rectangular faces
Total Surface Area of Cuboid= 2(lb + bh +lh)
Example: If the length, breadth and height of a cuboid are 5 cm, 3 cm and 4 cm, then find its total surface area.
Given, Length, l = 5 cm, Breadth, b = 3 cm and Height, h = 4 cm.
Total surface area (TSA) = 2(lb + bh + lh)
= 2(5 × 3 + 3 × 4 + 5 × 4)
= 2(15 + 12 + 20)
= 2(47)
= 94 sq.cm.
Lateral surface area of a Cuboid
The sum of surface areas of all faces except the top and bottom face of a solid is defined as the lateral surface area of a solid.
Consider a Cuboid of length, breadth and height to be l, b and h respectively.
Lateral surface area of the cuboid= Area of face ADHE + Area of face BCGF + Area of face ABFE + Area of face DCGH
=2(b × h) + 2(l × h)
=2h(l + b)
LSA of Cuboid = 2h(l +b)
Example: If the length, breadth and height of a cuboid are 5 cm, 3 cm and 4 cm, then find its lateral surface area.
Given, Length = 5 cm, Breadth = 3 cm and Height = 4 cm
LSA = 2h(l + b)
LSA = 2 × 4(5 + 3)
LSA = 2 × 4(8)
LSA = 2 × 32 = 64 cm^{2}
Surface Area of a Cube
For cube, length = breadth = height
Suppose the length of an edge = l
Hence, surface area of the cube = 2(l × l +l × l + l × l) = 2 x 3l^{2 } = 6l^{2}
Total Surface Area of Cube= 6l^{2}
Example: If the length of the side of the cube is 6 cm, then find its total surface area.
Given, side length = 6 cm
TSA of cube = 6l^{2}
TSA = 6 (6)^{2}
TSA = 6 × 36
TSA = 216 sq.cm
Lateral surface area of a Cube
Formula to find the Lateral surface area of the cube is:
2(l × l + l × l) = 4l^{2}
LSA of Cube = 4l2
Example: If the length of the side of the cube is 6 cm, then find its lateral surface area.
Given,
Side length, l = 6 cm
LSA of cube = 4l^{2}
LSA = 4 (6)^{2}
LSA = 4 x 36 = 144 sq.cm
Volume of the Cube and Cuboid
Volume of Cuboid:
The volume of the cuboid is equal to the product of the area of one surface and height.
Volume of the cuboid = (length × breadth × height) cubic units
Volume of the cuboid = ( l × b × h) cubic units
Example: If the length, breadth and height of a cuboid are 5 cm, 3 cm and 4 cm, then find its volume.
Given, Length (l) = 5 cm, Breadth (b) = 3 cm and Height (h) = 4 cm
Volume of cuboid = l × b × h
V = 5 × 3 × 4
V = 60 cubic cm
Volume of the Cube:
The volume of the cube is equal to the product of the area of the base of a cube and its height. As we know already, all the edges of the cube are of the same length. Hence,
Volume of the cube = l^{2} × h
Since, l = h
Therefore,
Volume of the cube = l^{2} × l
Volume of the cube = l3 cubic units
Example: If the length of the side of the cube is 6 cm, then find its volume.
Given, side length = 6 cm
Volume of cube = side^{3}
V = 6^{3}
V = 216 cubic cm
Diagonal of Cube and Cuboid
The length of the diagonal of the cuboid is given by:
Diagonal of the cuboid =√( l^{2 }+ b^{2} +h^{2})
The length of the diagonal of a cube is given by:
Diagonal of a cube = √3l
The perimeter of Cube and Cuboid
The perimeter of the cuboid is based on its length, width and height. Since the cuboid has 12 edges and the value of its edges are different from each other, therefore, the perimeter is given by:
Perimeter of a cuboid = 4 (l + b + h)
where l is the length
b is the breadth
h is the height
Example: If the length, width and height of a cuboid are 5 cm, 3 cm and 4 cm, find its Perimeter.
Given, Length = 5 cm, Width = 3 cm and Height = 4 cm
Perimeter = 4 (l + b + h) = 4 (5 + 3 + 4)
P = 4 (12)
P = 48 cm
The perimeter of the cube also depends upon the number of edges it has and the length of the edges. Since the cube has 12 edges and all the edges have equal length, the perimeter of the cube is given by:
The perimeter of a cube = 12l
where l is the length of the edge of the cube
Example: If the side length of the cube is 6 cm, then find its perimeter.
Given , l = 6 cm
The perimeter of the cube = 12l
P = 12 × 6
P = 72 cm
Example Problems on Cube and Cuboid Shape
Example 1:
Find the total surface area of the cuboid with dimensions 2 inches × 3 inches × 7 inches.
Solution:
Given,
Length (l) = 2 inches
Breadth (b) = 3 inches
Height (h) = 7 inches
Total Surface Area(TSA) = 2 (lb + bh + hl )
TSA = 2 ( 2 × 3 + 3 × 7 + 7 × 2)
= 2 ( 6 + 21 + 14 )
= 2 × 41
= 82
So, the total surface area of this cuboid is 82 inches^{2}
Example 2:
The length, width and height of a cuboid are 12 cm, 13 cm and 15 cm, respectively. Find the lateral surface area of a cuboid.
Solution:
Given,
Length (l) = 12 cm
Width (w) = 13 cm
Height (h) = 15 cm
Lateral surface area of a cuboid is given by:
LSA = 2h ( l + w )
LSA = 2 × 15 ( 12 + 13 )
= 30 × 25
= 750 cm^{2}
Example 3:
Find the surface area of a cube having its sides equal to 8 cm.
Solution: Given,
Length of the side ‘a’= 8 cm
Surface area = 6a^{2}
= 6 × 8^{2}
= 6 × 64
= 384 cm^{2}
Example 4:
If the side length of the cube shape object is 3 cm and the dimensions of the cuboidshaped object are 2 cm × 4 cm × 6 cm. Find the volume of cube and cuboidshaped objects.
Solution:
Given: Side length of cube, l = 3 cm.
We know that the volume of cube = l^{3} cubic units.
The volume of cubeshaped object = 3^{3} = 27 cm^{3}.
Given: Dimension of the cuboidshaped object = 2 cm × 4 cm × 6 cm.
We know that the volume of cuboid = lbh cubic units
The volume of cuboidshaped object = 2 cm × 4 cm × 6 cm = 48 cm^{3}.
Therefore, the volume of the cube and cuboidshaped object are 27 cm^{3} and 48 cm^{3} respectively.
Surface Area of Cuboid and a Cube
Chapter 13  Surface areas and Volumes
Surface Areas and Volume
Surface area and volume are calculated for any threedimensional geometrical shape. The surface area of any given object is the area or region occupied by the surface of the object. Whereas volume is the amount of space available in an object.
In geometry, there are different shapes and sizes such as sphere, cube, cuboid, cone, cylinder, etc. Each shape has its surface area as well as volume. But in the case of twodimensional figures like square, circle, rectangle, triangle, etc., we can measure only the area covered by these figures and there is no volume available. Now, let us see the formulas of surface areas and volumes for different 3dshapes.
What is Surface Area?
The space occupied by a twodimensional flat surface is called the area. It is measured in square units. The area occupied by a threedimensional object by its outer surface is called surface area. It is also measured in square units.
Generally, Area can be of two types:
(i) Total Surface Area
(ii) Curved Surface Area/Lateral Surface Area
Total surface area
Total surface area refers to the area including the base(s) and the curved part. It is total of the area covered by the surface of the object. If the shape has curved surface and base, then total area will be the sum of the two areas.
Curved surface area/Lateral surface area
Curved surface area refers to the area of only the curved part of the shape excluding its base(s). It is also referred to as lateral surface area for shapes such as a cylinder.
What is Volume?
The amount of space, measured in cubic units, that an object or substance occupies is called volume. Twodimensional doesn’t have volume but has area only. For example, Volume of Circle cannot be found, though Volume of the sphere can be. It is so because a sphere is a threedimensional shape.
Surface Area and Volume Formulas
Below given is the table for calculating Surface area and Volume for the basic geometrical figures:
Name 
Perimeter 
Total Surface Area 
Curved Surface Area/Lateral Surface Area 
Volume 
Figure 
Square 
4a 
a^{2} 
— 
— 

Rectangle 
2(w+h) 
w.h 
— 
— 

Parallelogram 
2(a+b) 
b.h 
— 
— 

Trapezoid 
a+b+c+d 
1/2(a+b).h 
— 
— 

Circle 
2 π r 
π r^{2} 
— 
— 

Ellipse 
2π√(a^{2} + b^{2})/2 
π a.b 
— 
— 

Triangle 
a+b+c 
1/2 * b * h 
— 
— 

Cuboid 
4(l+b+h) 
2(lb+bh+hl) 
2h(l+b) 
l * b * h 

Cube 
6a 
6a^{2} 
4a^{2} 
a^{3} 

Cylinder 
— 
2 π r(r+h) 
2πrh 
π r^{2} h 

Cone 
— 
π r(r+l) 
π r l 
1/3π r^{2} h 

Sphere 
— 
4 π r^{2} 
4π r^{2} 
4/3π r^{3} 

Hemisphere 
— 
3 π r^{2} 
2 π r^{2} 
2/3π r^{3} 

Surface Area of Cube and Cuboid
Cube and cuboid are threedimensional shapes that consist of six faces, eight vertices and twelve edges. The primary difference between them is a cube has all its sides equal whereas the length, width and height of a cuboid are different. Both shapes look almost the same but have different properties. The area and volume of cube, cuboid and also cylinder differ from each other.
In everyday life, we have seen many objects like a wooden box, a matchbox, a tea packet, a chalk box, a dice, a book, etc. All these objects have a similar shape. All these objects are made of six rectangular planes or square planes. In mathematics, the shape of these objects is either a cuboid or a cube. Here, in this article, we will learn the difference between cube and cuboid shapes with the help of their properties and formulas of surface area and volume.
Definition of Cube and Cuboid Shape
The cube and cuboid shapes in Maths are threedimensional shapes. The cube and cuboid are obtained from the rotation of the twodimensional shapes called square and rectangle respectively.
Cube: A cube is a threedimensional shape that is defined XYZ plane. It has six faces, eight vertices and twelve edges. All the faces of the cube are in square shape and have equal dimensions.
Cuboid: A cuboid is also a polyhedron having six faces, eight vertices and twelve edges. The faces of the cuboid are parallel. But not all the faces of a cuboid are equal in dimensions.
Hence, cube and cuboid shapes have six faces, eight vertices and twelve edges.
Difference Between Cube and Cuboid
The difference between the cube and cuboid shapes are as follows:
 The sides of the cube are equal but for cuboids they are different.
 The sides of the cube are square in shape but for cuboid, they are in a rectangular shape.
 All the diagonals of the cube are equal but a cuboid has equal diagonals for only parallel sides.
Cube and Cuboid Shape
As we already know, both cube and cuboid are in 3D shape, whose axes go along the xaxis, yaxis and zaxis. Now, let us learn in detail.
A cuboid is a closed 3dimensional geometrical figure bounded by six rectangular plane regions.
Properties of a Cuboid
Below are the properties of the cuboid, its faces, base and lateral faces, edges and vertices.
Faces of Cuboid
A Cuboid is made up of six rectangles, each of the rectangles is called the face. In the figure above, ABFE, DAEH, DCGH, CBFG, ABCD and EFGH are the 6 faces of the cuboid.
The top face ABCD and bottom face EFGH form a pair of opposite faces. Similarly, ABFE, DCGH, and DAEH, CBFG are pairs of opposite faces. Any two faces other than the opposite faces are called adjacent faces.
Consider a face ABCD, the adjacent face to this are ABFE, BCGF, CDHG, and ADHE.
Base and lateral faces
Any face of a cuboid may be called the base of the cuboid. The four faces which are adjacent to the base are called the lateral faces of the cuboid. Usually, the surface on which a solid rests is known to be the base of the solid.
In Figure (1) above, EFGH represents the base of a cuboid.
Edges
The edge of the cuboid is a line segment between any two adjacent vertices.
There are 12 edges, they are AB, AD, AE, HD, HE, HG, GF, GC, FE, FB, EF and CD and the opposite sides of a rectangle are equal.
Hence, AB = CD = GH = EF, AE = DH = BF = CG and EH = FG = AD = BC.
Vertices of Cuboid
The point of intersection of the 3 edges of a cuboid is called the vertex of a cuboid.
A cuboid has 8 vertices. A, B, C, D, E, F, G and H represent vertices of the cuboid in fig 1.
By observation, the twelve edges of a cuboid can be grouped into three groups, such that all edges in one group are equal in length, so there are three distinct groups and the groups are named as length, breadth and height.
A solid having its length, breadth, and height all to be equal in measurement is called a cube. A cube is a solid bounded by six square plane regions, where the side of the cube is called the edge.
Properties of Cube
 A cube has six faces and twelve edges of equal length.
 It has squareshaped faces.
 The angles of the cube in the plane are at a right angle.
 Each face of the cube meets four other faces.
 Each vertex of the cube meets three faces and three edges.
 The opposite edges of the cube are parallel to each other.
Cube and Cuboid Formulas
The formulas for cube and cuboid shapes are defined based on their surface areas, lateral surface areas and volume.
Surface Area of Cube and Cuboid
The surface area of a cuboid is equal to the sum of the areas of its six rectangular faces.
Consider a cuboid having the length to be ‘l’ cm, breadth be ‘b’ cm and height be ‘h’ cm.
 Area of face EFGH = Area of Face ABCD = (l × b) cm^{2}
 Area of face BFGC = Area of face AEHD = (b × h) cm^{2}
 Area of face DHGC = Area of face ABFE = (l × h) cm^{2}
Total surface area of a cuboid = Sum of the areas of all its 6 rectangular faces
Total Surface Area of Cuboid= 2(lb + bh +lh)
Example: If the length, breadth and height of a cuboid are 5 cm, 3 cm and 4 cm, then find its total surface area.
Given, Length, l = 5 cm, Breadth, b = 3 cm and Height, h = 4 cm.
Total surface area (TSA) = 2(lb + bh + lh)
= 2(5 × 3 + 3 × 4 + 5 × 4)
= 2(15 + 12 + 20)
= 2(47)
= 94 sq.cm.
Lateral surface area of a Cuboid
The sum of surface areas of all faces except the top and bottom face of a solid is defined as the lateral surface area of a solid.
Consider a Cuboid of length, breadth and height to be l, b and h respectively.
Lateral surface area of the cuboid= Area of face ADHE + Area of face BCGF + Area of face ABFE + Area of face DCGH
=2(b × h) + 2(l × h)
=2h(l + b)
LSA of Cuboid = 2h(l +b)
Example: If the length, breadth and height of a cuboid are 5 cm, 3 cm and 4 cm, then find its lateral surface area.
Given, Length = 5 cm, Breadth = 3 cm and Height = 4 cm
LSA = 2h(l + b)
LSA = 2 × 4(5 + 3)
LSA = 2 × 4(8)
LSA = 2 × 32 = 64 cm^{2}
Surface Area of a Cube
For cube, length = breadth = height
Suppose the length of an edge = l
Hence, surface area of the cube = 2(l × l +l × l + l × l) = 2 x 3l^{2 } = 6l^{2}
Total Surface Area of Cube= 6l^{2}
Example: If the length of the side of the cube is 6 cm, then find its total surface area.
Given, side length = 6 cm
TSA of cube = 6l^{2}
TSA = 6 (6)^{2}
TSA = 6 × 36
TSA = 216 sq.cm
Lateral surface area of a Cube
Formula to find the Lateral surface area of the cube is:
2(l × l + l × l) = 4l^{2}
LSA of Cube = 4l^{2}
Example: If the length of the side of the cube is 6 cm, then find its lateral surface area.
Given,
Side length, l = 6 cm
LSA of cube = 4l^{2}
LSA = 4 (6)^{2}
LSA = 4 x 36 = 144 sq.cm
Volume of the Cube and Cuboid
Volume of Cuboid:
The volume of the cuboid is equal to the product of the area of one surface and height.
Volume of the cuboid = (length × breadth × height) cubic units
Volume of the cuboid = ( l × b × h) cubic units
Example: If the length, breadth and height of a cuboid are 5 cm, 3 cm and 4 cm, then find its volume.
Given, Length (l) = 5 cm, Breadth (b) = 3 cm and Height (h) = 4 cm
Volume of cuboid = l × b × h
V = 5 × 3 × 4
V = 60 cubic cm
Volume of the Cube:
The volume of the cube is equal to the product of the area of the base of a cube and its height. As we know already, all the edges of the cube are of the same length. Hence,
Volume of the cube = l^{2} × h
Since, l = h
Therefore,
Volume of the cube = l^{2} × l
Volume of the cube = l^{3 }cubic units
Example: If the length of the side of the cube is 6 cm, then find its volume.
Given, side length = 6 cm
Volume of cube = side^{3}
V = 6^{3}
V = 216 cubic cm
For More Information On Volumes of Cubes and Cuboid, Watch The Below Video:
29,804
Diagonal of Cube and Cuboid
The length of diagonal of the cuboid is given by:
Diagonal of the cuboid =√( l^{2 }+ b^{2} +h^{2})
The length of diagonal of a cube is given by:
Diagonal of a cube = √3l
Perimeter of Cube and Cuboid
The perimeter of the cuboid is based on its length, width and height. Since the cuboid has 12 edges and the value of its edges are different from each other, therefore, the perimeter is given by:
Perimeter of a cuboid = 4 (l + b + h)
where l is the length
b is the breadth
h is the height
Example: If the length, width and height of a cuboid are 5 cm, 3 cm and 4 cm, find its Perimeter.
Given, Length = 5 cm, Width = 3 cm and Height = 4 cm
Perimeter = 4 (l + b + h) = 4 (5 + 3 + 4)
P = 4 (12)
P = 48 cm
The perimeter of the cube also depends upon the number of edges it has and the length of the edges. Since the cube has 12 edges and all the edges have equal length, the perimeter of the cube is given by:
Perimeter of a cube = 12l
where l is the length of the edge of the cube
Example: If the side length of the cube is 6 cm, then find its perimeter.
Given , l = 6 cm
The perimeter of cube = 12l
P = 12 × 6
P = 72 cm
Example Problems on Cube and Cuboid Shape
Example 1:
Find the total surface area of the cuboid with dimensions 2 inches × 3 inches × 7 inches.
Solution:
Given,
Length (l) = 2 inches
Breadth (b) = 3 inches
Height (h) = 7 inches
Total Surface Area(TSA) = 2 (lb + bh + hl )
TSA = 2 ( 2 × 3 + 3 × 7 + 7 × 2)
= 2 ( 6 + 21 + 14 )
= 2 × 41
= 82
So, the total surface area of this cuboid is 82 inches^{2}
Example 2:
The length, width and height of a cuboid are 12 cm, 13 cm and 15 cm, respectively. Find the lateral surface area of a cuboid.
Solution:
Given,
Length (l) = 12 cm
Width (w) = 13 cm
Height (h) = 15 cm
Lateral surface area of a cuboid is given by:
LSA = 2h ( l + w )
LSA = 2 × 15 ( 12 + 13 )
= 30 × 25
= 750 cm^{2}
Example 3:
Find the surface area of a cube having its sides equal to 8 cm.
Solution: Given,
Length of the side ‘a’= 8 cm
Surface area = 6a^{2}
= 6 × 8^{2}
= 6 × 64
= 384 cm^{2}
Example 4:
If the side length of the cube shape object is 3 cm and the dimensions of the cuboidshaped object are 2 cm × 4 cm × 6 cm. Find the volume of cube and cuboid shaped objects.
Solution:
Given: Side length of cube, l = 3 cm.
We know that the volume of cube = l^{3} cubic units.
The volume of cubeshaped object = 3^{3} = 27 cm^{3}.
Given: Dimension of the cuboidshaped object = 2 cm × 4 cm × 6 cm.
We know that the volume of cuboid = lbh cubic units
The volume of cuboidshaped object = 2 cm × 4 cm × 6 cm = 48 cm^{3}.
Therefore, the volume of the cube and cuboid shaped object are 27 cm^{3} and 48 cm^{3} respectively.
Application of Heron's formula in finding areas of quadrilaterals
 Books Name
 ABCD CLASSES Mathematics Book
 Publication
 ABCD CLASSES
 Course
 CBSE Class 9
 Subject
 Mathmatics
Application of Heron’s Formula in Finding Areas of Quadrilaterals
In geometry, the shapes can be classified into twodimensional shapes and threedimensional shapes. 2D shapes are the plane figures, which are made up of two dimensions such as length and breadth, whereas 3D shapes are the solid figures, which are made up of three dimensions such as length, breadth and height. We know that a triangle is a 2D shape, whose area can be determined by using the formula ½ × b × h if the base and height of the triangle are given. What if only the sides of a triangle are given? How to find the area of a triangle if the side lengths are given?
The solution to this is by applying Heron’s formula. One can find the area of a triangle using Heron’s formula, only if the side measurements are given. Now, let us have a quick recall about Heron’s formula.
Heron’s Formula – Finding Area of a Triangle
If a, b and c are the sides of a triangle, and s is the semi perimeter of a triangle, then the formula to find the area of the triangle using Heron’s formula is:
Area of Triangle = √[s(sa)(sb)(sc)] Square units.
Note: s is the semi perimeter, which means half the perimeter of a triangle, and it is found using the formula:
s = (a+b+c)/2.
Application of Heron’s Formula in Finding Areas of Quadrilaterals Examples
Heron’s formula is also applicable to find the area of a quadrilateral if the quadrilaterals are divided into triangular parts. Once the quadrilaterals are divided into triangular forms, apply Heron’s formula to find the area of individual triangular parts. After finding the areas of each triangular part, add all the areas to find the area of the quadrilateral. Now, let us understand the application of Heron’s formula in finding the areas of quadrilaterals with the help of solved examples.
Example 1:
Masha has a piece of land that is shaped like a rhombus. She wants her one son and one daughter to work on the land to produce different crops. So, she has divided the land into two equal parts. Determine how much area each of them will get for their crop production if the perimeter of the land is 400 m and one of its diagonals is 160 m.
Solution: Given: ABCD is a rhombusshaped land.
Given that, perimeter = 400 m
Therefore, each side length of rhombus = 400/4 = 100 cm.
Thus, we can take that AD = AB = 100 m
Diagonal, BD = 160 m.
Now, consider the triangle ABD,
The semiperimeter of ∆ ABD, s = (100+ 100+ 160)/2
s = 360/2 m
s = 180 m.
Therefore, the semi perimeter of a triangle = 180m.
Thus, the area of triangle ABD = √[s(sa)(sb)(sc)] square units.
Now, substitute s= 180 m, a= 100, b= 100 and c= 160 in the Heron’s formula, we get
A = √[180 (180 – 100)(180 – 100)(180 – 160)] m^{2}
A = √[180(80)(80)(20)] m^{2}
A = √23040000 m^{2}
A = 4800 m^{2}
Hence, each of them will get land for their crop production of 4800 m^{2}
Example 2:
The school students staged a rally for a cleanliness campaign. They walked through the lane into two groups. One group walked through the lanes AB, BC and CA, and the other group walked through the lanes AC, CD and DA as shown in the figure. They cleaned the area enclosed within the lanes. Find the total area cleaned by the students (Neglect the width of the lane), if AB = 9m, BC = 40m, CD = 15m, DA = 28m and ∠B = 90º. Also, find which group cleaned more area and how much.
Solution:
Given that, AB = 9m, BC = 40m, and ∠B = 90º
Therefore, AC = √(9^{2}+40^{2}) = √(81+1600)
AC = √1681 = 41m.
Thus, the first group of students cleaned the area of triangle ABC (which is rightangled)
Therefore, the area of triangle = (½)(40)(9) m^{2}
A = 180m^{2}
Hence, the first group of students to clean the lane ABC is 180m^{2}.
Now, the second group of students cleans the lane ACD, which is a scalene triangle having side lengths 41m, 28m and 15m.
Now, use Heron’s formula to find the area of ACD.
So, s = (41+28+15)/2
s = 42m
By, using Heron’s formula, A =√[s(sa)(sb)(sc)], we get
A = √[42(4241)(4228)(4215)] m^{2}
A = √[42(1)(14)(27)] m^{2}
A = √15876 m^{2}
A = 126 m^{2}
Hence, the second group of students cleaned the lane ACD is 126m^{2}.
The total area cleaned by all the students = (180+126)m^{2} = 306m^{2}.
Therefore, the first group of students cleaned more lane areas than the area cleaned by the second group of students.
Practice Problems
Solve the following problems on the application of Heron’s formula in finding areas of the quadrilateral.
 A field is in the shape of a rhombus and it has green grass for 18 cows to graze. Determine how much area of grass field each cow will be getting, if each side of the rhombus is 30m and its longest diagonal is 48m.
 A kite is in a square shape. Its diagonal is 32 cm and its isosceles triangle of base 8 cm and side 6 cm each is made of three different shades as shown in the figure. Find how much paper of each shade has been used in it to make a kite.
Application of Heron's formula in finding areas of quadrilaterals
Application of Heron’s Formula in Finding Areas of Quadrilaterals
In geometry, the shapes can be classified into twodimensional shapes and threedimensional shapes. 2D shapes are the plane figures, which are made up of two dimensions such as length and breadth, whereas 3D shapes are the solid figures, which are made up of threedimensions such as length, breadth and height. We know that a triangle is a 2D shape, whose area can be determined by using the formula ½ × b × h, if base and height of the triangle are given. What if only the sides of a triangle are given? How to find the area of a triangle if the side lengths are given?
The solution to this is by applying Heron’s formula. One can find the area of a triangle using Heron’s formula, only if the side measurements are given. Now, let us have a quick recall about Heron’s formula.
Heron’s Formula – Finding Area of a Triangle
If a, b and c are the sides of a triangle, and s is the semiperimeter of a triangle, then the formula to find the area of triangle using Heron’s formula is:
Area of Triangle = √[s(sa)(sb)(sc)] Square units.
Note: s is the semiperimeter, which means half the perimeter of a triangle, and it is found using the formula:
s = (a+b+c)/2.
Application of Heron’s Formula in Finding Areas of Quadrilaterals Examples
Heron’s formula is also applicable to find the area of a quadrilateral if the quadrilaterals are divided into triangular parts. Once the quadrilaterals are divided into triangular forms, apply Heron’s formula to find the area of individual triangular parts. After finding the areas of each triangular part, add all the areas to find the area of quadrilateral. Now, let us understand the application of Heron’s formula in finding the areas of quadrilaterals with the help of solved examples.
Example 1:
Masha has a piece of land which is shaped like a rhombus. She wants her one son and one daughter to work on the land to produce different crops. So, she has divided the land into two equal parts. Determine how much area each of them will get for their crop production if the perimeter of the land is 400 m and one of its diagonals is 160 m.
Solution:
Given: ABCD is a rhombusshaped land.
Given that, perimeter = 400 m
Therefore, each side length of rhombus = 400/4 = 100 cm.
Thus, we can take that AD = AB = 100 m
Diagonal, BD = 160 m.
Now, consider the triangle ABD,
The semiperimeter of ∆ ABD, s = (100+ 100+ 160)/2
s = 360/2 m
s = 180 m.
Therefore, the semi perimeter of a triangle = 180m.
Thus, the area of triangle ABD = √[s(sa)(sb)(sc)] square units.
Now, substitute s= 180 m, a= 100, b= 100 and c= 160 in the Heron’s formula, we get
A = √[180 (180 – 100)(180 – 100)(180 – 160)] m^{2}
A = √[180(80)(80)(20)] m^{2}
A = √23040000 m^{2}
A = 4800 m^{2}
Hence, each of them will get land for their crop production of 4800 m^{2}
Example 2:
The school students staged a rally for a cleanliness campaign. They walked through the lane into two groups. One group walked through the lanes AB, BC and CA, and the other group walked through the lanes AC, CD and DA as shown in the figure. They cleaned the area enclosed within the lanes. Find the total area cleaned by the students (Neglect the width of the lane), if AB = 9m, BC = 40m, CD = 15m, DA = 28m and ∠B = 90º. Also, find which group cleaned more area and how much.
Solution:
Given that, AB = 9m, BC = 40m, and ∠B = 90º
Therefore, AC = √(9^{2}+40^{2}) = √(81+1600)
AC = √1681 = 41m.
Thus, the first group of students cleaned the area of triangle ABC (which is rightangled)
Therefore, the area of triangle = (½)(40)(9) m^{2}
A = 180m^{2}
Hence, the first group of students to clean the lane ABC is 180m^{2}.
Now, the second group of students clean the lane ACD, which is a scalene triangle having side lengths 41m, 28m and 15m.
Now, use Heron’s formula to find the area of ACD.
So, s = (41+28+15)/2
s = 42m
By, using Heron’s formula, A =√[s(sa)(sb)(sc)], we get
A = √[42(4241)(4228)(4215)] m^{2}
A = √[42(1)(14)(27)] m^{2}
A = √15876 m^{2}
A = 126 m^{2}
Hence, the second group of students cleaned the lane ACD is 126m^{2}.
The total area cleaned by all the students = (180+126)m^{2} = 306m^{2}.
Therefore, the first group of students cleaned more lane areas than the area cleaned by the second group of students.
Practice Problems
Solve the following problems on the application of Heron’s formula in finding areas of the quadrilateral.
 A field is in the shape of a rhombus and it has green grass for 18 cows to graze. Determine how much area of grass field each cow will be getting, if each side of the rhombus is 30m and its longest diagonal is 48m.
 A kite is in a square shape. Its diagonal is 32 cm and its isosceles triangle of base 8 cm and side 6 cm each is made of three different shades as shown in the figure. Find how much paper of each shade has been used in it to make a kite.
Surface Area of a Right Circular Cylinder
 Books Name
 ABCD CLASSES Mathematics Book
 Publication
 ABCD CLASSES
 Course
 CBSE Class 9
 Subject
 Mathmatics
Surface Area of a Right Circular Cylinder
A right circular cylinder is a cylinder that has a closed circular surface having two parallel bases on both the ends and whose elements are perpendicular to its base. It is also called a right cylinder. All the points, in a right circular cylinder, lying on the closed circular surface is at a fixed distance from a straight line known as the axis of the cylinder. The two circular bases of the right circular cylinder have the same radius and are parallel to each other. It is one such geometric shape that is used frequently in real life. Basically, to derive the formulas for the surface area and volume of the cylinder, the right cylinder is considered. If the bases of the cylinder are not parallel to each other, then such cylinder is known as an oblique cylinder in 3D geometry.
What is a Right Circular Cylinder?
A cylinder whose bases are circular in shape and parallel to each other is called the right circular cylinder. It is a threedimensional shape. The axis of the cylinder joins the center of the two bases of the cylinder. This is the most common type of cylinder used in daytoday life. Whereas the oblique cylinder is another type of cylinder, which does not have parallel bases and resembles a tilted structure.
In the above figure, r is the radius of the circular bases and h is the height of the right cylinder.
Parts of Right Circular Cylinder
The three parts of the right circular cylinder are:
 Top circular base
 Curved lateral face
 Bottom circular face
Properties of Right Circular Cylinder
You must have learned about the properties of the cylinder before. Here, let us discuss the right circular cylinder properties.
 The line joining the centers of the circle is called the axis.
 When we revolve a rectangle about one side as the axis of revolution, a right cylinder is formed.
 The section obtained on cutting a right circular cylinder by a plane, which contains two elements and parallels to the axis of the cylinder is the rectangle.
 If a plane cuts the right cylinder horizontally parallel to the bases, then it’s a circle.
Right Circular Cylinder Formulas
A surface, which is generated by a line that intersects a fixed circle and is perpendicular to the plane of the circle is said to be a right circular cylinder. A right cylinder has two circular bases which are of the same radius and are parallel to each other. The formulas for surface area, curved or lateral surface area and volume of the right circular cylinder are discussed here.
Curved Surface Area of right circular cylinder
The surface area of a closed right circular cylinder is the sum of the area of the curved surface and the area of the two bases. The curved surface that joins the two circular bases is said to be the lateral surface of the right circular cylinder.
Lateral or Curved Area = 2 π r h square units
where r is the radius of circular bases and h is the height of the cylinder.
Total Surface Area of a right circular cylinder
The sum of the lateral surface area and the base area of both the circles will give the total surface area of the right circular cylinder.
TSA = 2 π r (h + r) square units
where r is the radius of circular bases and h is the height of the cylinder.
Volume of right circular cylinder
The volume of a right circular cylinder is given by the product of the area of the top or bottom circle and the height of the cylinder. The volume of a right cylinder is measured in terms of cubic units.
Volume = Area of the circular bases x Height of the Right Cylinder
Volume = πr^{2} h
where r is the radius of circular bases and h is the height of the cylinder.
Solved Examples on Right Circular Cylinder
Let us solve some problems based on the formulas of the right circular cylinder.
Q.1: Find the volume of a right cylinder, if the radius and height of the cylinder are 20 cm and 30 cm respectively.
Solution: We know,
Volume of a right cylinder = πr^{2} h cubic units
Given, r = 20 cm h = 30 cm
Therefore, using the formula, we get;
Volume = 3.14 × 20^{2} × 30
= 3.14 × 20 × 20 × 30
= 37680
Hence, the volume of the given right cylinder is 37680 cm^{3}.
Q.2: The radius and height of a right cylinder are given as 5 m and 6.5 m respectively. Find the volume and total surface area of the right cylinder.
Solution: Given that, r = 5 m h = 6.5 m
We know, by the formula,
Volume of a right cylinder = πr^{2} h cubic units
Therefore,
Volume = 3.14 × 5^{2} × 6.5
= 3.14 × 25 × 6.5
= 510.25
Hence, the volume of the given right cylinder is 510.25 cubic m.
Now we know again, the total surface area of the right cylinder is given by;
TSA = Area of circular base + Curved Surface Area
TSA = 2 π r (h + r) square units
By putting the values of radius and height, we get;
TSA = 2 x π x 5 (6.5 + 5)
TSA = 2 x 3.14 x 5 x 11.5
TSA = 361.1 sq.m
Hence, the total surface area of the given right cylinder is 361.1 m^{2}.
Practice Questions
If the radius of circular bases and height of the right circular cylinder is given, then find the surface areas and volumes.
 Radius = 3 cm, Height = 5 cm
 Radius = 7 cm, Height = 8 cm
 Radius = 0.5 m, Height = 2 m
 Radius = 14 cm, Height = 22 cm
Surface Area of a Right Circular Cylinder
Surface Area of a Right Circular Cylinder
A right circular cylinder is a cylinder that has a closed circular surface having two parallel bases on both the ends and whose elements are perpendicular to its base. It is also called a right cylinder. All the points, in a right circular cylinder, lying on the closed circular surface is at a fixed distance from a straight line known as the axis of the cylinder. The two circular bases of the right circular cylinder have the same radius and are parallel to each other. It is one such geometric shape that is used frequently in real life. Basically, to derive the formulas for the surface area and volume of the cylinder, the right cylinder is considered. If the bases of the cylinder are not parallel to each other, then such cylinder is known as an oblique cylinder in 3D geometry.
What is a Right Circular Cylinder?
A cylinder whose bases are circular in shape and parallel to each other is called the right circular cylinder. It is a threedimensional shape. The axis of the cylinder joins the center of the two bases of the cylinder. This is the most common type of cylinder used in day to day life. Whereas the oblique cylinder is another type of cylinder, which does not have parallel bases and resembles a tilted structure.
In the above figure, r is the radius of the circular bases and h is the height of the right cylinder.
Parts of Right Circular Cylinder
The three parts of the right circular cylinder are:
 Top circular base
 Curved lateral face
 Bottom circular face
Properties of Right Circular Cylinder
You must have learned about the properties of the cylinder before. Here, let us discuss the right circular cylinder properties.
 The line joining the centers of the circle is called the axis.
 When we revolve a rectangle about one side as the axis of revolution, a right cylinder is formed.
 The section obtained on cutting a right circular cylinder by a plane, which contains two elements and parallels to the axis of the cylinder is the rectangle.
 If a plane cuts the right cylinder horizontally parallel to the bases, then it’s a circle.
Right Circular Cylinder Formulas
A surface, which is generated by a line that intersects a fixed circle and is perpendicular to the plane of the circle is said to be a right circular cylinder. A right cylinder has two circular bases which are of the same radius and are parallel to each other. The formulas for surface area, curved or lateral surface area and volume of the right circular cylinder are discussed here.
Curved Surface Area of right circular cylinder
The surface area of a closed right circular cylinder is the sum of the area of the curved surface and the area of the two bases. The curved surface that joins the two circular bases is said to be the lateral surface of the right circular cylinder.
Lateral or Curved Area = 2 π r h square units 
where r is the radius of circular bases and h is the height of cylinder.
Total Surface Area of right circular cylinder
The sum of the lateral surface area and the base area of both the circles will give the total surface area of the right circular cylinder.
TSA = 2 π r (h + r) square units 
where r is the radius of circular bases and h is the height of cylinder.
Volume of right circular cylinder
The volume of a right circular cylinder is given by the product of the area of the top or bottom circle and the height of the cylinder. The volume of a right cylinder is measured in terms of cubic units.
Volume = Area of the circular bases x Height of the Right Cylinder
Volume = πr^{2} h 
where r is the radius of circular bases and h is the height of the cylinder.
Solved Examples on Right Circular Cylinder
Let us solve some problems based on the formulas of the right circular cylinder.
Q.1: Find the volume of a right cylinder, if the radius and height of the cylinder are 20 cm and 30 cm respectively.
Solution: We know,
Volume of a right cylinder = πr^{2} h cubic units
Given, r = 20 cm h = 30 cm
Therefore, using the formula, we get;
Volume = 3.14 × 20^{2} × 30
= 3.14 × 20 × 20 × 30
= 37680
Hence, the volume of the given right cylinder is 37680 cm^{3}.
Q.2: The radius and height of a right cylinder are given as 5 m and 6.5 m respectively. Find the volume and total surface area of the right cylinder.
Solution: Given that, r = 5 m h = 6.5 m
We know, by the formula,
Volume of a right cylinder = πr^{2} h cubic units
Therefore,
Volume = 3.14 × 5^{2} × 6.5
= 3.14 × 25 × 6.5
= 510.25
Hence, the volume of the given right cylinder is 510.25 cubic m.
Now we know again, the total surface area of the right cylinder is given by;
TSA = Area of circular base + Curved Surface Area
TSA = 2 π r (h + r) square units
By putting the values of radius and height, we get;
TSA = 2 x π x 5 (6.5 + 5)
TSA = 2 x 3.14 x 5 x 11.5
TSA = 361.1 sq.m
Hence, the total surface area of the given right cylinder is 361.1 m^{2}.
Practice Questions
If the radius of circular bases and height of the right circular cylinder is given, then find the surface areas and volumes.
 Radius = 3 cm, Height = 5 cm
 Radius = 7 cm, Height = 8 cm
 Radius = 0.5 m, Height = 2 m
 Radius = 14 cm, Height = 22 cm
Surface Area of a Right Circular Cone
 Books Name
 ABCD CLASSES Mathematics Book
 Publication
 ABCD CLASSES
 Course
 CBSE Class 9
 Subject
 Mathmatics
Surface Area of a Right Circular Cone
A right circular cone is a cone where the axis of the cone is the line meeting the vertex to the midpoint of the circular base. That is, the center point of the circular base is joined with the apex of the cone and it forms a right angle. A cone is a threedimensional shape having a circular base and narrowing smoothly to a point above the base. This point is known as apex.
We come across many geometrical shapes while dealing with geometry. We usually study twodimensional and threedimensional figures in school. Twodimensional shapes have length and breadth. They can be drawn on paper, for example – circles, rectangles, squares, triangles, polygons, parallelograms etc. The threedimensional figures cannot be drawn on paper since they have an additional third dimension as height or depth. Examples of 3D shapes are a sphere, hemisphere, cylinder, cone, pyramid, prism etc.
Right Circular Cone Definition
A right circular cone is one whose axis is perpendicular to the plane of the base. We can generate a right cone by revolving a right triangle about one of its legs.
In the figure, you can see a right circular cone, which has a circular base of radius r and whose axis is perpendicular to the base. The line which connects the vertex of the cone to the center of the base is the height of the cone. The length at the outer edge of the cone, which connects a vertex to the end of the circular base is the slant height.
Right Circular Cone Formula
For a right circular cone of radius ‘r’, height ‘h’ and slant height ‘l’, we have;
 Curved surface area of right circular cone = π r l
 Total surface area of a right circular cone = π(r + l) r
 Volume of a right circular cone = 1/3π r^{2} h
Surface Area of a Right Circular Cone
The surface area of any right circular cone is the sum of the area of the base and lateral surface area of a cone. The surface area is measured in terms of square units.
Surface area of a cone = Base Area + Curved Surface Area of a cone
= π r^{2} + π r l
= πr(r + l)
Here, l = √(r^{2}+h^{2})
Where ‘r’ is the radius, ‘l’ is the slant height and ‘h’ is the height of the cone.
Volume of a Right Circular Cone
The volume of a cone is onethird of the product of the area of the base and the height of the cone. The volume is measured in terms of cubic units.
Volume of a right circular cone can be calculated by the following formula,
Volume of a right circular cone = ⅓ (Base area × Height)
Where Base Area = π r^{2}
Hence, Volume = ⅓ π r^{2}h
Properties of Right Circular Cone
 It has a circular base whose center joins its vertex, showing the axis of the respective cone.
 The slant height of this cone is the length of the sides of the cone taken from the vertex to the outer line of the circular base. It is denoted by ‘l’.
 The altitude of a right cone is the perpendicular line from the vertex to the center of the base. It coincides with the axis of the cone and is represented by ‘h’.
 If a right triangle is rotated about its perpendicular, considering the perpendicular as the axis of rotation, the solid constructed here is the required cone. The surface area generated by the hypotenuse of the triangle is the lateral surface area.
 Any section of the right circular cone parallel to the base forms a circle that lies on the axis of the cone.
 A section that contains the vertex and two points of the base of a right circular cone is an isosceles triangle.
Frustum of a Right Circular Cone
A frustum is a portion of the cone between the base and the parallel plane when a right circular cone is cut off by a plane parallel to its base.
Equation of Right Circular Cone
The equation of the right circular cone with vertex origin is:
(x^{2}+y^{2}+z^{2})cos^{2}θ=(lx+my+nz)^{2}
Where θ is the semivertical angle and (l, m, n) are the direction cosines of the axis.
Let us find the equation of the right circular cone whose vertex is the origin, the axis is the line x = y/3 =z/2 and which makes a semivertical angle of 60 degrees.
The direction cosines of the axis are:
[(1/√(1^{2}+3^{2}+2^{2}) , (3/√(1^{2}+3^{2}+2^{2}), (2/√(1^{2}+3^{2}+2^{2})] = (1/√14, 3/√14, 2/√14)
The semivertical angle is, θ = 60°
Therefore, the equation of the right circular cone with vertex (0, 0, 0) is:
(x^{2}+y^{2}+z^{2}) ¼ = 1/14 (x + 3y + 2z)^{2}
7(x^{2}+y^{2}+z^{2}) = 2(x^{2}+9y^{2}+4z^{2}+6xy+12yz+4xz)
5x^{2}11y^{2}z^{2}12xy24yz8xz = 0; which is the required equation.
Examples
Q. 1: Find the surface area of the right cone if the given radius is 6 cm and slant height is 10 cm.
Solution: Given,
Radius (r) = 6 cm
Slant height (l) = 10 cm
Surface area of a cone = π r(r + l)
Solving surface area,
SA = 3.14 × 6(6 + 10)
SA = 3.14 × 6 × 16
SA = 301.44
Therefore, Surface area of the right cone is 301.4 sq. cm.
Question 2: Calculate the Volume of the right cone for the given radius 6 cm and height 10 cm.
Solution: Given,
Radius (r) = 6 cm
Height (h) = 10 cm
Volume of right cone = 1/3π r^{2} h
Volume, V = ⅓ × π × (6)^{2} × 10
V = 3.14 × 12 × 10
V = 376.8
Therefore, Volume of a right cone is 376.8 cubic cm.
Surface Area of a Right Circular Cone
Surface Area of a Right Circular Cone
A right circular cone is a cone where the axis of the cone is the line meeting the vertex to the midpoint of the circular base. That is, the centre point of the circular base is joined with the apex of the cone and it forms a right angle. A cone is a threedimensional shape having a circular base and narrowing smoothly to a point above the base. This point is known as apex.
We come across many geometrical shapes while dealing with geometry. We usually study about two dimensional and threedimensional figures in school. Twodimensional shapes have length and breadth. They can be drawn on paper, for example – circle, rectangle, square, triangle, polygon, parallelogram etc. The threedimensional figures cannot be drawn on paper since they have an additional third dimension as height or depth. The examples of 3D shapes are a sphere, hemisphere, cylinder, cone, pyramid, prism etc.
Right Circular Cone Definition
A right circular cone is one whose axis is perpendicular to the plane of the base. We can generate a right cone by revolving a right triangle about one of its legs.
In the figure, you can see a right circular cone, which has a circular base of radius r and whose axis is perpendicular to the base. The line which connects the vertex of the cone to the centre of the base is the height of the cone. The length at the outer edge of the cone, which connects a vertex to the end of the circular base is the slant height.
Right Circular Cone Formula
For a right circular cone of radius ‘r’, height ‘h’ and slant height ‘l’, we have;

Surface Area of a Right Circular Cone
The surface area of any right circular cone is the sum of the area of the base and lateral surface area of a cone. The surface area is measured in terms of square units.
Surface area of a cone = Base Area + Curved Surface Area of a cone
= π r^{2} + π r l
= πr(r + l)
Here, l = √(r^{2}+h^{2})
Where ‘r’ is the radius, ‘l’ is the slant height and ‘h’ is the height of the cone.
Volume of a Right Circular Cone
The volume of a cone is onethird of the product of the area of the base and the height of the cone. The volume is measured in terms of cubic units.
Volume of a right circular cone can be calculated by the following formula,
Volume of a right circular cone = ⅓ (Base area × Height)
Where Base Area = π r^{2}
Hence, Volume = ⅓ π r^{2}h
Properties of Right Circular Cone
 It has a circular base whose centre joins its vertex, showing the axis of the respective cone.
 The slant height of this cone is the length of the sides of the cone taken from vertex to the outer line of the circular base. It is denoted by ‘l’.
 The altitude of a right cone is the perpendicular line from the vertex to the centre of the base. It coincides with the axis of the cone and is represented by ‘h’.
 If a right triangle is rotated about its perpendicular, considering the perpendicular as the axis of rotation, the solid constructed here is the required cone. The surface area generated by the hypotenuse of the triangle is the lateral surface area.
 Any section of right circular cone parallel to the base forms a circle lies on the axis of the cone.
 A section which contains the vertex and two points of the base of a right circular cone is an isosceles triangle.
Frustum of a Right Circular Cone
A frustum is a portion of the cone between the base and the parallel plane when a right circular cone is cut off by a plane parallel to its base.
Equation of Right Circular Cone
The equation of the right circular cone with vertex origin is:
(x^{2}+y^{2}+z^{2})cos^{2}θ=(lx+my+nz)^{2} 
Where θ is the semivertical angle and (l, m, n) are direction cosines of the axis.
Let us find the equation of the right circular cone whose vertex is the origin, the axis is the line x = y/3 =z/2 and which makes a semivertical angle of 60 degrees.
The direction cosines of the axis are:
[(1/√(1^{2}+3^{2}+2^{2}) , (3/√(1^{2}+3^{2}+2^{2}), (2/√(1^{2}+3^{2}+2^{2})] = (1/√14, 3/√14, 2/√14)
The semivertical angle is, θ = 60°
Therefore, the equation of the right circular cone with vertex (0, 0, 0) is:
(x^{2}+y^{2}+z^{2}) ¼ = 1/14 (x + 3y + 2z)^{2}
7(x^{2}+y^{2}+z^{2}) = 2(x^{2}+9y^{2}+4z^{2}+6xy+12yz+4xz)
5x^{2}11y^{2}z^{2}12xy24yz8xz = 0; which is the required equation.
Examples
Q. 1: Find the surface area of the right cone if the given radius is 6 cm and slant height is 10 cm.
Solution: Given,
Radius (r) = 6 cm
Slant height (l) = 10 cm
Surface area of a cone = π r(r + l)
Solving surface area,
SA = 3.14 × 6(6 + 10)
SA = 3.14 × 6 × 16
SA = 301.44
Therefore, Surface area of the right cone is 301.4 sq. cm.
Question 2: Calculate the Volume of the right cone for the given radius 6 cm and height 10 cm.
Solution: Given,
Radius (r) = 6 cm
Height (h) = 10 cm
Volume of right cone = 1/3π r^{2} h
Volume, V = ⅓ × π × (6)^{2} × 10
V = 3.14 × 12 × 10
V = 376.8
Therefore, Volume of a right cone is 376.8 cubic cm.
Surface Area of a Sphere
 Books Name
 ABCD CLASSES Mathematics Book
 Publication
 ABCD CLASSES
 Course
 CBSE Class 9
 Subject
 Mathmatics
Surface Area of a Sphere
The surface area of a sphere is defined as the region covered by its outer surface in threedimensional space. A Sphere is a threedimensional solid having a round shape, just like a circle. The formula for the total surface area of a sphere in terms of pi (π) is given by:
Surface area = 4 π r^{2 }square units
Area of circle = π r^{2 }The difference between a sphere and a circle is that a circle is a twodimensional figure or a flat shape, whereas, a sphere is a threedimensional shape. Therefore, the area of a circle is different from the area of the sphere.
Definition
From a visual perspective, a sphere has a threedimensional structure that forms by rotating a disc that is circular with one of the diagonals.
Let us consider an instance where spherical ball faces are painted. To paint the whole surface, the paint quantity required has to be known beforehand. Hence, the area of every face has to be known to calculate the paint quantity for painting the same. We define this term as the total surface area.
The surface area of a sphere is equal to the areas of the entire face surrounding it.
Formula
The surface area of a sphere formula is given by,
A = 4 π r^{2 }square units
For any threedimensional shape, the area of the object can be categorized into three types. They are:
 Curved Surface Area
 Lateral Surface Area
 Total Surface Area
Curved Surface Area: The curved surface area is the area of all the curved regions of the solid.
Lateral Surface Area: The lateral surface area is the area of all the regions except bases (i.e., top and bottom).
Total Surface Area: The total surface area is the area of all the sides, top and bottom the solid object.
In the case of a Sphere, it has no flat surface.
Therefore, the Total surface area of a sphere = Curved surface area of a sphere
Solved Examples
Example 1– Calculate the cost required to paint a football which is in the shape of a sphere having a radius of 7 cm. If the painting cost of a football is INR 2.5/square cm. (Take π = 22/7)
Solution
We know,
The total surface area of a sphere = 4 π r^{2 }square units
= 4 × (22/7) × 7 × 7
= 616 cm^{2}
Therefore, total cost of painting the container = 2.5 × 616 = Rs. 1540
Example 2 Calculate the curved surface area of a sphere having a radius equal to 3.5 cm(Take π= 22/7)
Solution–
We know,
Curved surface area = Total surface area = 4 π r^{2 }square units
= 4 × (22/7) × 3.5 × 3.5
Therefore, the curved surface area of a sphere= 154 cm^{2}
Surface Area of a Sphere
Surface Area of a Sphere
The surface area of a sphere is defined as the region covered by its outer surface in threedimensional space. A Sphere is a threedimensional solid having a round shape, just like a circle. The formula of total surface area of a sphere in terms of pi (π) is given by:
Surface area = 4 π r^{2 }square units 
The difference between a sphere and a circle is that a circle is a twodimensional figure or a flat shape, whereas, a sphere is a threedimensional shape. Therefore, the area of circle is different from area of sphere.
Area of circle = π r^{2 }
Definition
From a visual perspective, a sphere has a threedimensional structure that forms by rotating a disc that is circular with one of the diagonals.
Let us consider an instance where spherical ball faces are painted. To paint the whole surface, the paint quantity required has to be known beforehand. Hence, the area of every face has to be known to calculate the paint quantity for painting the same. We define this term as the total surface area.
The surface area of a sphere is equal to the areas of the entire face surrounding it.
Formula
The surface area of a sphere formula is given by,
A = 4 π r^{2 }square units
For any threedimensional shapes, the area of the object can be categorised into three types. They are:
 Curved Surface Area
 Lateral Surface Area
 Total Surface Area
Curved Surface Area: The curved surface area is the area of all the curved regions of the solid.
Lateral Surface Area: The lateral surface area is the area of all the regions except bases (i.e., top and bottom).
Total Surface Area: The total surface area is the area of all the sides, top and bottom the solid object.
In case of a Sphere, it has no flat surface.
Therefore, the Total surface area of a sphere = Curved surface area of a sphere
Solved Examples
Example 1– Calculate the cost required to paint a football which is in the shape of a sphere having a radius of 7 cm. If the painting cost of football is INR 2.5/square cm. (Take π = 22/7) Solution We know, The total surface area of a sphere = 4 π r^{2 }square units = 4 × (22/7) × 7 × 7 = 616 cm^{2} Therefore, total cost of painting the container = 2.5 × 616 = Rs. 1540 Example 2 Calculate the curved surface area of a sphere having radius equals to 3.5 cm(Take π= 22/7) Solution– We know, Curved surface area = Total surface area = 4 π r^{2 }square units = 4 × (22/7) × 3.5 × 3.5 Therefore, the curved surface area of a sphere= 154 cm^{2} 
Volume of a Cube and Cuboid
 Books Name
 ABCD CLASSES Mathematics Book
 Publication
 ABCD CLASSES
 Course
 CBSE Class 9
 Subject
 Mathmatics
Volume of a Cube and Cuboid
Cube, Cuboid, and cylinder form the part of 3D shapes. Generally, in a 3Dshaped figure, you will find the measure of length, width, and height. On the contrary, in the case of a 2D figure, you have just the length and the height. Some of the examples of 2D shapes are triangle, square, and circle. Now, let's look at what’s volume after all? Students generally have a hard time dealing with 3D figures. One of the reasons is their tendency to memorize formulae without actually understanding what it is. So, it would be good to start with understanding what does the volume of any shape means?
Reason for Difference Between Cube and Cylinder Volume
The volume of any 3D figure is the measure of the area enclosed by the area of the figure. In a literal sense, Volume denotes the total capacity or the space occupied by the object which has all the three elements of length, height, and width. The different formulas for each 3D shape are derived by the structure of formation. For instance, the structural difference between cube and cylinder introduces the need for a different formula of volume for each of the shapes in consideration.
Cube
Cube is the first of 3D shapes which is quite tedious to map out. It is a figure enclosed by six identical squares. One of the interesting things about the cube is that a single vertex is formed at the meeting point of three edges.
Properties Of Cube
 Cube is a kind of square prism.
 It has 6 faces, 8 vertices and 12 edges.
 As the faces of the cube are in the shape of a square, the length, breadth and height of the cube are all the same.
 The angle between any two surfaces is 90 degrees.
 In a cube, the opposite faces and edges are parallel to each other.
 Each face of a cube is in contact with the other four faces.
 Each vertice in a cube is in contact with three edges and three faces.
Cuboid
Just a heads up that any object which is in the form of a box is a cuboid. A cuboid essentially is a 3D shape that is characterized by six rectangular faces. Each vertex is formed at right angles.
Properties Of Cuboid
 The surfaces of the cuboid are in the shape of a rectangle.
 It has 8 vertices, 12 edges, 6 faces.
 The two surfaces of the cuboid form an angle of 90 degrees at the vertices.
 The number of diagonals drawn on each face of the cuboid is 2.
 In a cuboid, opposite edges are parallel to each other.
 The number of space diagonals and face diagonals in a cuboid is 4 and 12 respectively.
 The values of length, breadth and height of the cuboid are different.
Cylinder
The basic difference between cube and cylinder structurewise lies in the presence of a third measure i.e height. The diet coke can which you might be holding in your hand while reading this article is an apt example of a cylinder.
A cylinder is a 3D shape, with two parallel sides with a circular or an oval opening at the top and the bottom. Now the openings can be hollow or in the solid state.
Properties Of Cylinder
 A cylinder is a 3D object which has one curved surface and two flat surfaces.
 The two flat surfaces are circular in shape and are identical.
 The size of the cylinder depends on the base radius and height of the curved surface.
 There is no vertex in a cylinder.
 There is the same crosssection everywhere in the cylinder like a prism.
How do you Calculate the Volume of Cube Cuboid and Cylinder
Calculating the volume of Cube, Cuboid, and Cylinder is quite straightforward provided you know the workarounds to arrive at the measurements of the figure at hand.
The difference between cube and cylinder or cuboid and cylinder is quite stark. Also, cylinders do not belong to the polyhedron family of 3D shapes. The reason for its exclusion is mentioned in FAQs. Let us look at the volume of the cube cuboid and cylinder individually.
The Volume of Cube
The total threedimensional space occupied by the cube is known as the volume of the cube. As mentioned earlier, a cube is formed of six identical squares. Thus, each side of the cube will have the same measurement. Now, for calculating the volume of a cube, you just need to find the cube of the given value of the side.
For instance, if the measurement of the side is a then volume of a cube is i.e V= a^{3}
The SI unit of the volume of the cube is m^3 (cubic meter).
Steps for Calculating the Volume of the Cube
The following steps will be helpful in finding the volume of the cube:
 Step 1  Know the length of one edge of the cube as the length of each edge is the same.
 Step 2  Check if the length of the edge is given in the SI unit.
 Step 3  Use the formula in which edge length is used to calculate the volume of the cube. The formula is a^3.
 Step 4  Do the calculations and write the answer in cubic meters.
The Volume of Cuboid
The volume of the cuboid is defined as the space within the cuboid. In other words, The volume of the cuboid is the product of the length, breadth, and height, colloquially dubbed as ‘lbh’. For any cuboid of length L, Breadth B, and height H, the volume ‘V’ is equal to LxBxH.
The SI unit of this volume is a cubic meter (m^3).
Steps for Calculating the Volume of the Cuboid
The procedure for finding the volume of the cuboid is mentioned below:
 Step 1  Measure the length, breadth and height of the cylinder.
 Step 2  Convert all the units of the dimensions into the same unit if the units of the dimensions are different.
 Step 3  Put the values of the dimensions in the given formula (L x B x H) to find the volume.
The Volume of Cylinder
The quantity of material a cylinder can hold or the capacity of the cylinder is termed the volume of the cylinder. The volume of the cylinder involves the use of the area of the circle which is multiplied by the height. It is the easiest way to comprehend the formulae for the volume of a cylinder.
For any cylinder with radius R and height H, the volume ‘V’ is represented as
V= R^{2}H.
It is measured in m^3 (cubic meters).
Steps for Calculating the Volume of the Cylinder
Beneath are the steps for finding the volume of the cylinder:
 Step 1  Find out the height of the cylinder and the base radius.
 Step 2  Substitute all the measured values in the formula of πr^2h.
 Step 3  Perform calculations and write the answer in cubic meters.
Solved Examples
Now, we will see some of the solved examples of volume of Cube Cuboid and Cylinder
1. Find the volume of a cube with a side 9m?
Let the volume of the cube be V, Using the formulae V=a^{3}, we have
V= (9m)^{3} = 729
2. Find the volume of the cuboid with a length 12m, height 8m, and breadth 6m?
Using the formulae for the volume of cuboid i.e V=LBH, we have,
V= (12X8X6)m^{3} = 576m^{3}
3. Find the volume of a cylinder with a radius 14m and height 10m?
Using the formulae for calculating the volume of the cylinder i.e V= R^{2}H.
We have, V= 22/7 x (14m)^{2} X (10m)= 6,160m^{3}.
The volume of a cylinder with Pythagoras theorem application to find the actual height.
4. Find the volume of the cylinder with a slant height 10cm and radius 3cm?
(Note **slant height is the hypotenuse formed by joining the upper end of the oval opening to the opposite bottom end. For more clarity, see the figure below.)
As you can see in the slant height is denoted by d. Now, the formula for calculating the volume of the cylinder involves height. But in the question we do not have the value of height, thus, we have to calculate height using the Pythagorean theorem.
H^{2} (hypotenuse) = L^{2} (length) + B^{2} (breadth square)
Similarly, here we have, d^{2}=h^{2} + D^{2} (diameter)
(10cm)^{2}= h^{2} + (6)^{2}
Solving the above equation, we have h= 8cm
Now, V= R^{2}H
V= 22/7 x (3)^{2} x (8)
Solving the above equation we get V= 226 (approx).
Volume of a Cube and Cuboid
Volume of a Cube and Cuboid
Cube, Cuboid, and cylinder form the part of 3D shapes. Generally, in a 3D shaped figure, you will find the measure of length, width, and height. On the contrary, in the case of a 2D figure, you have just the length and the height. Some of the examples of 2D shapes are triangle, square, and circle. Now, let's look at what’s volume after all? Students generally have a hard time dealing with 3D figures. One of the reasons is their tendency to memorize formulae without actually understanding what it is. So, it would be good to start with understanding what does the volume of any shape means?
Reason for Difference Between Cube and Cylinder Volume
The volume of any 3D figure is the measure of the area enclosed by the area of the figure. In literal sense, Volume denotes the total capacity or the space occupied by the object which has all the three elements length, height, and width. The different formulas for each 3D shape are derived by the structure of formation. For instance, the structural difference between cube and cylinder introduces the need for a different formula of volume for each of the shapes in consideration.
Cube
Cube is the first of 3D shapes which is quite tedious to map out. It is a figure enclosed by six identical squares. One of the interesting things about the cube is that a single vertex is formed at the meeting point of three edges.
Properties Of Cube
 Cube is a kind of square prism.
 It has 6 faces, 8 vertices and 12 edges.
 As the faces of the cube are in the shape of a square, the length, breadth and height of the cube are all the same.
 The angle between any two surfaces is 90 degrees.
 In a cube, the opposite faces and edges are parallel to each other.
 Each face of a cube is in contact with the other four faces.
 Each vertice in a cube is in contact with three edges and three faces.
Cuboid
Just a heads up that any object which is in the form of a box is a cuboid. A cuboid essentially is a 3D shape that is characterized by six rectangular faces. Each vertex is formed at right angles.
Properties Of Cuboid
 The surfaces of the cuboid are in the shape of a rectangle.
 It has 8 vertices, 12 edges, 6 faces.
 The two surfaces of the cuboid form an angle of 90 degrees at the vertices.
 The number of diagonals drawn on each face of the cuboid is 2.
 In a cuboid, opposite edges are parallel to each other.
 The number of space diagonals and face diagonals in a cuboid is 4 and 12 respectively.
 The values of length, breadth and height of the cuboid are different.
Cylinder
The basic difference between cube and cylinder, structurewsie lies in the presence of a third measure i.e height. The diet coke can which you might be holding in your hand while reading this article is an apt example of a cylinder.
A cylinder is a 3D shape, with two parallel sides with a circular or an oval opening at the top and the bottom. Now the openings can be hollow or in the solidstate.
Properties Of Cylinder
 A cylinder is a 3D object which has one curved surface and two flat surfaces.
 The two flat surfaces are circular in shape and are identical.
 The size of the cylinder depends on the base radius and height of the curved surface.
 There is no vertex in a cylinder.
 There is the same crosssection everywhere in the cylinder like a prism.
How do you Calculate the Volume of Cube Cuboid and Cylinder
Calculating the volume of Cube, Cuboid, and Cylinder is quite straightforward provided you know the workarounds to arrive at the measurements of the figure at hand.
The difference between cube and cylinder or cuboid and cylinder is quite stark. Also, cylinders do not belong to the polyhedron family of 3D shapes. The reason for its exclusion is mentioned in FAQs. Let us look at the volume of the cube cuboid and cylinder individually.
The Volume of Cube
The total threedimensional space occupied by the cube is known as the volume of the cube. As mentioned earlier, a cube is formed of six identical squares. Thus, each side of the cube will have the same measurement. Now, for calculating the volume of a cube, you just need to find the cube of the given value of the side.
For instance, if the measurement of the side is a then volume of a cube is i.e V= a^{3}
The SI unit of the volume of the cube is m^3 (cubic meter).
Steps for Calculating the Volume of the Cube
The following steps will be helpful in finding the volume of the cube:
 Step 1  Know the length of one edge of the cube as the length of each edge is the same.
 Step 2  Check if the length of the edge is given in the SI unit.
 Step 3  Use the formula in which edge length is used to calculate the volume of the cube. The formula is a^3.
 Step 4  Do the calculations and write the answer in cubic meters.
The Volume of Cuboid
The volume of the cuboid is defined as the space within the cuboid. In other words, The volume of the cuboid is the product of the length, breadth, and height, colloquially dubbed as ‘lbh’. For any cuboid of length L, Breadth B, and height H, the volume ‘V’ is equal to LxBxH.
The SI unit of this volume is a cubic meter (m^3).
Steps for Calculating the Volume of the Cuboid
The procedure for finding the volume of the cuboid is mentioned below:
 Step 1  Measure the length, breadth and height of the cylinder.
 Step 2  Convert all the units of the dimensions into the same unit if the units of the dimensions are different.
Step 3  Put the values of the dimensions in the given formula (L x B x H) to find the volume.
Volume of a Cylinder
 Books Name
 ABCD CLASSES Mathematics Book
 Publication
 ABCD CLASSES
 Course
 CBSE Class 9
 Subject
 Mathmatics
Volume of a Cylinder
The volume of a cylinder is the density of the cylinder which signifies the amount of material it can carry or how much amount of any material can be immersed in it. Cylinder’s volume is given by the formula, πr^{2}h, where r is the radius of the circular base and h is the height of the cylinder. The material could be a liquid quantity or any substance which can be filled in the cylinder uniformly. Check volume of shapes here.
Volume of cylinder has been explained in this article briefly along with solved examples for better understanding. In Mathematics, geometry is an important branch where we learn the shapes and their properties. Volume and surface area are the two important properties of any 3d shape.
Definition
The cylinder is a threedimensional shape having a circular base. A cylinder can be seen as a set of circular disks that are stacked on one another. Now, think of a scenario where we need to calculate the amount of sugar that can be accommodated in a cylindrical box.
In other words, we mean to calculate the capacity or volume of this box. The capacity of a cylindrical box is basically equal to the volume of the cylinder involved. Thus, the volume of a threedimensional shape is equal to the amount of space occupied by that shape.
Volume of a Cylinder Formula
A cylinder can be seen as a collection of multiple congruent disks, stacked one above the other. In order to calculate the space occupied by a cylinder, we calculate the space occupied by each disk and then add them up. Thus, the volume of the cylinder can be given by the product of the area of base and height.
For any cylinder with base radius ‘r’, and height ‘h’, the volume will be base times the height.
Therefore, the cylinder’s volume of base radius ‘r’, and height ‘h’ = (area of base) × height of the cylinder
Since the base is the circle, it can be written as Volume = πr^{2 }× h
Therefore, the volume of a cylinder = πr^{2}h cubic units.
Volume of Hollow Cylinder
In case of hollow cylinder, we measure two radius, one for inner circle and one for outer circle formed by the base of hollow cylinder. Suppose, r_{1} and r_{2} are the two radii of the given hollow cylinder with ‘h’ as the height, then the volume of this cylinder can be written as;
 V = πh(r_{1}^{2} – r_{2}^{2})
Surface Area of Cylinder
The amount of square units required to cover the surface of the cylinder is the surface area of the cylinder. The formula for the surface area of the cylinder is equal to the total surface area of the bases of the cylinder and surface area of its sides.
 A = 2πr^{2} + 2πrh
Volume of Cylinder in Litres
When we find the volume of the cylinder in cubic centimetres, we can convert the value in litres by knowing the below conversion, i.e.,
1 Litre = 1000 cubic cm or cm^{3}
For example: If a cylindrical tube has a volume of 12 litres, then we can write the volume of the tube as 12 × 1000 cm^{3} = 12,000 cm^{3}
Examples
Question 1: Calculate the volume of a given cylinder having height 20 cm and base radius of 14 cm. (Take pi = 22/7)
Solution:
Given:
Height = 20 cm
radius = 14 cm
we know that;
Volume, V = πr^{2}h cubic units
V=(22/7) × 14 × 14 × 20
V= 12320 cm^{3}
Therefore, the volume of a cylinder = 12320 cm^{3}
Question 2: Calculate the radius of the base of a cylindrical container of volume 440 cm^{3}. Height of the cylindrical container is 35 cm. (Take pi = 22/7)
Solution:
Given:
Volume = 440 cm^{3}
Height = 35 cm
We know from the formula of cylinder;
Volume, V = πr^{2}h cubic units
So, 440 = (22/7) × r^{2} × 35
r^{2 }= (440 × 7)/(22 × 35) = 3080/770 = 4
Therefore, r = 2 cm
Therefore, the radius of a cylinder = 2 cm.
Volume of a Cylinder
Volume of a Cylinder
The volume of a cylinder is the density of the cylinder which signifies the amount of material it can carry or how much amount of any material can be immersed in it. Cylinder’s volume is given by the formula, πr^{2}h, where r is the radius of the circular base and h is the height of the cylinder. The material could be a liquid quantity or any substance which can be filled in the cylinder uniformly. Check volume of shapes here.
Volume of cylinder has been explained in this article briefly along with solved examples for better understanding. In Mathematics, geometry is an important branch where we learn the shapes and their properties. Volume and surface area are the two important properties of any 3d shape.
Definition
The cylinder is a threedimensional shape having a circular base. A cylinder can be seen as a set of circular disks that are stacked on one another. Now, think of a scenario where we need to calculate the amount of sugar that can be accommodated in a cylindrical box.
In other words, we mean to calculate the capacity or volume of this box. The capacity of a cylindrical box is basically equal to the volume of the cylinder involved. Thus, the volume of a threedimensional shape is equal to the amount of space occupied by that shape.
Volume of a Cylinder Formula
A cylinder can be seen as a collection of multiple congruent disks, stacked one above the other. In order to calculate the space occupied by a cylinder, we calculate the space occupied by each disk and then add them up. Thus, the volume of the cylinder can be given by the product of the area of base and height.
For any cylinder with base radius ‘r’, and height ‘h’, the volume will be base times the height.
Therefore, the cylinder’s volume of base radius ‘r’, and height ‘h’ = (area of base) × height of the cylinder
Since the base is the circle, it can be written as
Volume = πr^{2 }× h
Therefore, the volume of a cylinder = πr^{2}h cubic units.
1,05,533
Volume of Hollow Cylinder
In case of hollow cylinder, we measure two radius, one for inner circle and one for outer circle formed by the base of hollow cylinder. Suppose, r_{1} and r_{2} are the two radii of the given hollow cylinder with ‘h’ as the height, then the volume of this cylinder can be written as;
 V = πh(r_{1}^{2} – r_{2}^{2})
Surface Area of Cylinder
The amount of square units required to cover the surface of the cylinder is the surface area of the cylinder. The formula for the surface area of the cylinder is equal to the total surface area of the bases of the cylinder and surface area of its sides.
 A = 2πr^{2} + 2πrh
Volume of Cylinder in Litres
When we find the volume of the cylinder in cubic centimetres, we can convert the value in litres by knowing the below conversion, i.e.,
1 Litre = 1000 cubic cm or cm^{3}
For example: If a cylindrical tube has a volume of 12 litres, then we can write the volume of the tube as 12 × 1000 cm^{3} = 12,000 cm^{3}
Examples
Question 1: Calculate the volume of a given cylinder having height 20 cm and base radius of 14 cm. (Take pi = 22/7)
Solution:
Given:
Height = 20 cm
radius = 14 cm
we know that;
Volume, V = πr^{2}h cubic units
V=(22/7) × 14 × 14 × 20
V= 12320 cm^{3}
Therefore, the volume of a cylinder = 12320 cm^{3}
Question 2: Calculate the radius of the base of a cylindrical container of volume 440 cm^{3}. Height of the cylindrical container is 35 cm. (Take pi = 22/7)
Solution:
Given:
Volume = 440 cm^{3}
Height = 35 cm
We know from the formula of cylinder;
Volume, V = πr^{2}h cubic units
So, 440 = (22/7) × r^{2} × 35
r^{2 }= (440 × 7)/(22 × 35) = 3080/770 = 4
Therefore, r = 2 cm
Therefore, the radius of a cylinder = 2 cm.
Volume of a Right Circular Cone
 Books Name
 ABCD CLASSES Mathematics Book
 Publication
 ABCD CLASSES
 Course
 CBSE Class 9
 Subject
 Mathmatics
Volume of a Cone
The volume of a cone defines the space or the capacity of the cone. A cone is a threedimensional geometric shape having a circular base that tapers from a flat base to a point called apex or vertex. A cone is formed by a set of line segments, halflines or lines connecting a common point, the apex, to all the points on a base that is in a plane that does not contain the apex.
A cone can be seen as a set of noncongruent circular disks that are stacked on one another such that the ratio of the radius of adjacent disks remains constant.
Volume of a Cone Formula
In general, a cone is a pyramid with a circular crosssection. A right cone is a cone with its vertex above the center of the base. It is also called the right circular cone. You can easily find out the volume of a cone if you have the measurements of its height and radius and put it into a formula.
Therefore, the volume of a cone formula is given as
The volume of a cone = (1/3) πr^{2}h cubic units
Where,
‘r’ is the base radius of the cone
‘l’ is the slant height of a cone
‘h’ is the height of the cone
As we can see from the above cone formula, the capacity of a cone is onethird of the capacity of the cylinder. That means if we take 1/3rd of the volume of the cylinder, we get the formula for cone volume.
Note: The formula for the volume of a regular cone or right circular cone and the oblique cone is the same.
Also, read:
Derivation of Cone Volume
You can think of a cone as a triangle that is being rotated about one of its vertices. Now, think of a scenario where we need to calculate the amount of water that can be accommodated in a conical flask. In other words, calculate the capacity of this flask. The capacity of a conical flask is basically equal to the volume of the cone involved. Thus, the volume of a threedimensional shape is equal to the amount of space occupied by that shape. Let us perform an activity to calculate the volume of a cone.
Take a cylindrical container and a conical flask of the same height and same base radius. Add water to the conical flask such that it is filled to the brim. Start adding this water to the cylindrical container you took. You will notice it doesn’t fill up the container fully. Try repeating this experiment once more, you will still observe some vacant space in the container. Repeat this experiment once again; you will notice this time the cylindrical container is completely filled. Thus, the volume of a cone is equal to onethird of the volume of a cylinder having the same base radius and height.
Now let us derive its formula. Suppose a cone has a circular base with a radius ‘r’ and its height is ‘h’. The volume of this cone will be equal to onethird of the product of the area of the base and its height. Therefore,
V = 1/3 x Area of Circular Base x Height of the Cone
Since, we know by the formula of area of the circle, the base of the cone has an area (say B) equal to;
B = πr^{2}
Hence, substituting this value we get;
V = 1/3 x πr^{2} x h
Where V is the volume, r is the radius and h is the height.
Solved Examples
Q.1: Calculate the volume if r= 2 cm and h= 5 cm.
Solution:
Given:
r = 2
h= 5
Using the Volume of Cone formula
The volume of a cone = (1/3) πr^{2}h cubic units
V= (1/3) × 3.14 × 2^{2 }×5
V= (1/3) × 3.14 × 4^{ }×5
V= (1/3) × 3.14 × 20
V = 20.93 cm^{3}
Therefore, the volume of a cone = 20.93 cubic units.
Q.2: If the height of a given cone is 7 cm and the diameter of the circular base is 6 cm. Then find its volume.
Solution: Diameter of the circular base = 6 cm.
So, radius = 6/2 = 3 cm
Height = 7 cm
By the formula of cone volume, we know;
V = 1/3 πr^{2}h
So by putting the values of r and h, we get;
V = 1/3 π 3^{2 }7
Since π = 22/7
Therefore,
V = 1/3 x 22/7 x 3^{2 }x 7
V = 66 cu.cm.
Volume of a Right Circular Cone
Volume of a Cone
The volume of a cone defines the space or the capacity of the cone. A cone is a threedimensional geometric shape having a circular base that tapers from a flat base to a point called apex or vertex. A cone is formed by a set of line segments, halflines or lines connecting a common point, the apex, to all the points on a base that is in a plane that does not contain the apex.
A cone can be seen as a set of noncongruent circular disks that are stacked on one another such that the ratio of the radius of adjacent disks remains constant.
Volume of a Cone Formula
In general, a cone is a pyramid with a circular crosssection. A right cone is a cone with its vertex above the center of the base. It is also called right circular cone. You can easily find out the volume of a cone if you have the measurements of its height and radius and put it into a formula.
Therefore, the volume of a cone formula is given as
The volume of a cone = (1/3) πr^{2}h cubic units
Where,
‘r’ is the base radius of the cone
‘l’ is the slant height of a cone
‘h’ is the height of the cone
As we can see from the above cone formula, the capacity of a cone is onethird of the capacity of the cylinder. That means if we take 1/3rd of the volume of the cylinder, we get the formula for cone volume.
Note: The formula for the volume of a regular cone or right circular cone and the oblique cone is the same.
Also, read:
Derivation of Cone Volume
You can think of a cone as a triangle which is being rotated about one of its vertices. Now, think of a scenario where we need to calculate the amount of water that can be accommodated in a conical flask. In other words, calculate the capacity of this flask. The capacity of a conical flask is basically equal to the volume of the cone involved. Thus, the volume of a threedimensional shape is equal to the amount of space occupied by that shape. Let us perform an activity to calculate the volume of a cone.
Take a cylindrical container and a conical flask of the same height and same base radius. Add water to the conical flask such that it is filled to the brim. Start adding this water to the cylindrical container you took. You will notice it doesn’t fill up the container fully. Try repeating this experiment for once more, you will still observe some vacant space in the container. Repeat this experiment once again; you will notice this time the cylindrical container is completely filled. Thus, the volume of a cone is equal to onethird of the volume of a cylinder having the same base radius and height.
Now let us derive its formula. Suppose a cone has a circular base with radius ‘r’ and its height is ‘h’. The volume of this cone will be equal to onethird of the product of the area of the base and its height. Therefore,
V = 1/3 x Area of Circular Base x Height of the Cone
Since, we know by the formula of area of the circle, the base of the cone has an area (say B) equals to;
B = πr^{2}
Hence, substituting this value we get;
V = 1/3 x πr^{2} x h
Where V is the volume, r is the radius and h is the height.
Solved Examples
Q.1: Calculate the volume if r= 2 cm and h= 5 cm.
Solution:
Given:
r = 2
h= 5
Using the Volume of Cone formula
The volume of a cone = (1/3) πr^{2}h cubic units
V= (1/3) × 3.14 × 2^{2 }×5
V= (1/3) × 3.14 × 4^{ }×5
V= (1/3) × 3.14 × 20
V = 20.93 cm^{3}
Therefore, the volume of a cone = 20.93 cubic units.
Q.2: If the height of a given cone is 7 cm and the diameter of the circular base is 6 cm. Then find its volume.
Solution: Diameter of the circular base = 6 cm.
So, radius = 6/2 = 3 cm
Height = 7 cm
By the formula of cone volume, we know;
V = 1/3 πr^{2}h
So by putting the values of r and h, we get;
V = 1/3 π 3^{2 }7
Since π = 22/7
Therefore,
V = 1/3 x 22/7 x 3^{2 }x 7
V = 66 cu.cm.
Volume of a Sphere
 Books Name
 ABCD CLASSES Mathematics Book
 Publication
 ABCD CLASSES
 Course
 CBSE Class 9
 Subject
 Mathmatics
Volume Of A Sphere Formula
A sphere is a 3D or a solid shape having a completely round structure. If you rotate a circular disc along any of its diameters, the structure thus obtained can be seen as a sphere. You can also define it as a set of points that are located at a fixed distance from a fixed point in a threedimensional space. This fixed point is known as the center of the sphere. And the fixed distance is called its radius.
Volume of a Sphere Formula
In this section, you will learn the formula to compute the volume of a sphere. Volume, as you know, is defined as the capacity of a 3D object. The volume of a sphere is nothing but the space occupied by it. It can be given as:
Where ‘r’ represents the radius of the sphere.
Volume Of A Sphere Derivation
The volume of a sphere can alternatively be viewed as the number of cubic units which is required to fill up the sphere.
Consider a sphere of radius r and divide it into pyramids. In this way, we see that the volume of the sphere is the same as the volume of all the pyramids of height, r and total base area equal to the surface area of the sphere as shown in the figure.
The total volume is calculated by adding the pyramids’ volumes.
Volume of the sphere = Sum of volumes of all pyramids
Volume of the sphere=
Volume of the sphere =
Volume of a Sphere Formula in Real Life:
In our daily life, we come across different types of spheres. Basketball, football, table tennis, etc. are some of the common sports that are played by people all over the world. The balls used in these sports are nothing but spheres of different radii. The volume of sphere formula is useful in designing and calculating the capacity or volume of such spherical objects. You can easily find out the volume of a sphere if you know its radius.
Solved Examples Based on Sphere Volume Formula:
Question 1: A sphere has a radius of 11 feet. Find its volume.
Solution: Given,
r = 11 feet
We know that, volume of a sphere =
=
Question 2: The volume of a spherical ball is
. Find the radius of the ball.
Solution: Given,
Volume of the sphere=
We know that, volume of a sphere =
The radius of the ball is 4.34 cm (approx).
Volume of a Sphere
Volume Of A Sphere Formula
A sphere is 3D or a solid shape having a completely round structure. If you rotate a circular disc along any of its diameters, the structure thus obtained can be seen as a sphere. You can also define it as a set of points which are located at a fixed distance from a fixed point in a threedimensional space. This fixed point is known as the centre of the sphere. And the fixed distance is called its radius.
Volume of a Sphere Formula
In this section, you will learn the formula to compute the volume of a sphere. Volume, as you know, is defined as the capacity of a 3D object. The volume of a sphere is nothing but the space occupied by it. It can be given as:
Where ‘r’ represents the radius of the sphere.
Volume Of A Sphere Derivation
The volume of a sphere can alternatively be viewed as the number of cubic units which is required to fill up the sphere.
Consider a sphere of radius r and divide it into pyramids. In this way, we see that the volume of the sphere is the same as the volume of all the pyramids of height, r and total base area equal to the surface area of the sphere as shown in the figure.
The total volume is calculated by adding the pyramids’ volumes.
Volume of the sphere = Sum of volumes of all pyramids
Volume of the sphere=
Volume of the sphere =
Volume of a Sphere Formula in Real Life:
In our daily life, we come across different types of spheres. Basketball, football, table tennis, etc. are some of the common sports that are played by people all over the world. The balls used in these sports are nothing but spheres of different radii. The volume of sphere formula is useful in designing and calculating the capacity or volume of such spherical objects. You can easily find out the volume of a sphere if you know its radius.
Solved Examples Based on Sphere Volume Formula:
Question 1: A sphere has a radius of 11 feet. Find its volume.
Solution: Given,
r = 11 feet
We know that, volume of a sphere =
=
Question 2: The volume of a spherical ball is
. Find the radius of the ball.
Solution: Given,
Volume of the sphere=
We know that, volume of a sphere =
The radius of the ball is 4.34 cm (approx).