Curved Surface Area Of Combination Of Solids

We know how to find the curved surface areas of three-dimensional figures such as cone, cylinder, sphere, etc. Now, we will learn to find out the curved surface area of a combination of solids.

Let us consider the following figure.

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/468/529/10.1.13.1.1_html_m756200c2.png

The given figure shows a temple. In this figure, the base of the temple is a right circular cylinder and the top is a right circular cone. The height of the conical part is half the height of the cylindrical part and the radius of the base of the temple is 2 ft more than the height of the conical part. The temple has to be plastered from outside at the rate of Rs 5 per square foot.

Can we find the approximate cost of plastering the temple from outside?

In order to find the cost, we have to find the curved surface area of the temple. By adding the curved surface areas of cone and cylinder, we can obtain the curved surface area of the temple.

First of all, let us review the formulae to find the curved surface areas of different solids. The formulae for different solids are given in the following table.

In this way, we can find the curved surface area of a combination of solids. Now, let us solve some more examples to understand the concept better.

Example 1: A toy is in the form of a hemisphere mounted by a cone. The diameter of the hemisphere is 21 cm and height of the whole toy is 24.5 cm. If the surface of the

toy is painted at the rate of Rs 1 per 5 cm2, then find the cost required to paint the entire toy.

Solution:

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/468/529/10.1.13.1.1_html_149e1f57.png

It is given that the diameter of the hemisphere is 21 cm.

∴ Radius of the hemisphere, r = 10.5 cm

Radius of the base of the cone, r = 10.5 cm

From the figure, height of cone, h = 24.5 − 10.5 = 14 cm

Now, l =https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/468/529/10.1.13.1.1_html_7a18fc84.gif =https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/468/529/10.1.13.1.1_html_mcbe597.gif cm

∴ Surface area of the toy = C.S.A. of hemisphere + C.S.A. of cone

= 2πr2 + πrl

= πr (2r + l)

=https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/468/529/10.1.13.1.1_html_m2428d1f3.gif

= 22 × 1.5 × 38.5

= 1270.5 cm2

∴ Cost required for painting = 1270.5 × https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/468/529/10.1.13.1.1_html_3d0b22e5.gif= Rs 254 approximately

Example 2: The given figure shows a post office box. It was painted with red colour and it is observed that the ratio between the cost of painting the hemispherical part and the cylindrical part is 2:7. What is the ratio between the height of the box and the radius of the hemispherical part?

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/468/529/10.1.13.1.1_html_6d21c184.png

Solution:

As the box is painted with a constant cost, irrespective of the hemispherical part and the cylindrical part, we obtain

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/468/529/10.1.13.1.1_html_m78105cdf.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/468/529/10.1.13.1.1_html_2ae12d96.gif, where r is the radius of the hemispherical part and h is height of the

cylindrical part

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/468/529/10.1.13.1.1_html_52d82dfb.gif

Thus, the ratio between the height of the box and the radius of the hemispherical part is 9:2.

Example 3: A tent is in the shape of a cylinder surmounted by a conical top. If the ratio of the height of the cylinder, the radius of cylinder, and the height of the cone is 8:4:3 and the cost of canvas required for making the tent is Rs 10560 at the rate of Rs 10 per m2, then find the height of the tent.

Solution:

Let us consider, height of cylinder = h’, radius of cylinder = r, and height of cone = h

Let h’ = 8x, r = 4x, and h = 3x

Then, l = https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/468/529/10.1.13.1.1_html_7a18fc84.gif= https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/468/529/10.1.13.1.1_html_83293e0.gif= 5x

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/468/529/10.1.13.1.1_html_m2e1735ff.png

Amount of canvas = https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/468/529/10.1.13.1.1_html_2a5501c.gifm2

However, amount of canvas = C.S.A. of cylinder + C.S.A. of cone

1056 = 2πrh’ + πrl

1056 = πr (2h’ + l)

1056 =https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/468/529/10.1.13.1.1_html_m465893fa.gif

1056 =https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/468/529/10.1.13.1.1_html_19c259b7.gif

264x2 = 1056

x2 =https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/468/529/10.1.13.1.1_html_m7245e7ce.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/468/529/10.1.13.1.1_html_38a75f3e.gifx = m

Height of tent = height of cylindrical part + height of conical part

= 8x + 3x

= 11x

= 11× 2

= 22

Thus, the height of the tent is 22 m.

Board question(s) related to this lesson:

Total Surface Areas Of Combination Of Solids

Total surface area of any solid is the sum of its curved surface area and bases. In many situations, we are required to find the total surface area of a combination of solids. Let us consider such a situation.

The largest possible hemisphere is taken out from a cube of side 10 cm. Can we find out the total surface area of the remaining solid?

To find it, consider the following figure.

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/469/531/10.1.13.1.2_GPL_PIY_KSB_SS_html_5627ee4d.png

From the cube, the largest hemisphere is taken out.

That means, diameter of hemisphere = edge of the cube = 10 cm

∴ Radius of the hemisphere =https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/469/531/10.1.13.1.2_GPL_PIY_KSB_SS_html_4a5e1b5c.gif

Total Surface area of the remaining solid = T.S.A. of cube without the circular base + C.S.A. of hemisphere

https://img-nm.mnimgs.com/img/study_content/editlive_lp/56/2012_10_22_16_32_01/mathmlequation8921399014231466133.png

= 678.5 cm2

Thus, the required area is 678.5 cm2.

Let us solve some more examples to understand the concept better.

Example 1: From a cylinder of radius 7 m and height 24 m, a cone of same radius and same height is taken out. Find the surface area of the remaining part. If the inner portion of this remaining part is painted in pink colour at the rate of Rs 23 per 5 m2 and the exterior portion is painted in green colour at the rate of Rs 40 per 11 m2, then find the total cost required to paint such a solid.

Solution:

It is given that the radius and height of the cylinder and cone are the same. Let radius r = 7 m and height h = 24 m

Now, slant height of the cone, l =https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/469/531/10.1.13.1.2_GPL_PIY_KSB_SS_html_7a18fc84.gif= https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/469/531/10.1.13.1.2_GPL_PIY_KSB_SS_html_m6a6e5fee.gifm

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/469/531/10.1.13.1.2_GPL_PIY_KSB_SS_html_404a78e7.png

After taking out the cone from the cylinder,

T.S.A. of remaining outer part = C.S.A. of cylinder + Area of Circle

= 2πrh + πr2

= πr (2h + r)

= https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/469/531/10.1.13.1.2_GPL_PIY_KSB_SS_html_66ad7622.gif

= 22 × 55

= 1210 m2

T.S.A. of remaining inner part = C.S.A. of cone

= πrl

=https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/469/531/10.1.13.1.2_GPL_PIY_KSB_SS_html_7302b390.gif

= 550 m2

Cost of painting the exterior part =https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/469/531/10.1.13.1.2_GPL_PIY_KSB_SS_html_69c05b80.gif

Cost of painting the interior part =https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/469/531/10.1.13.1.2_GPL_PIY_KSB_SS_html_561d89dc.gif

∴ Cost of painting the remaining solid = 4400 + 2530 = Rs 6930

Thus, the required cost is Rs 6930.

Example 2: A right triangle, whose perpendicular sides are 30 cm and 40 cm, is made to revolve about its hypotenuse. Find the surface area of the double cone so obtained.

Solution:

Let ABD be the right-angled triangle such that AB = 30 cm and AD = 40 cm Using Pythagoras theorem, we obtain

BD2 = AB2 + AD2

BD2 =https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/469/531/10.1.13.1.2_GPL_PIY_KSB_SS_html_m20f3e42f.gif

BD = https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/469/531/10.1.13.1.2_GPL_PIY_KSB_SS_html_4cf0b492.gifcm

Let OB = x and OA = y

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/469/531/10.1.13.1.2_GPL_PIY_KSB_SS_html_23aae2f3.png

Applying Pythagoras theorem in ΔAOB, we obtain

OA2 + OB2 = AB2

x2 + y2 = 302

y2 = 900 − x2 … (1)

Applying Pythagoras theorem in ΔAOD, we obtain

OA2 + OD2 = AD2

y2 + (50 − x) 2 = 402

900 − x2 + 2500 + x2 − 100x = 1600 {From equation (1), we have y2 = 900 − x2}

100x = 1800

x = 18

On putting this value in equation (1), we obtain

y2 = 900 − x2 = 900 − 182 = 900 − 324 = 576

y =https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/469/531/10.1.13.1.2_GPL_PIY_KSB_SS_html_17c7565.gif cm

Here, y is the radius of both the cones.

Now, surface area of the solid = C.S.A. of cone ABC + C.S.A. of cone ADC

= π × 24 × 30 + π × 24 × 40

= 1680 π

= 1680 ×https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/469/531/10.1.13.1.2_GPL_PIY_KSB_SS_html_m62ee3405.gif

= 5280 cm2

Thus, the surface area of the double cone so obtained is 5280 cm2.

Example 3: The inner and outer radii of a cylindrical iron container are 10 cm and 11 cm respectively. The thickness of the base is 1.5 cm and the total height of the container is 21.5 cm. Find the cost, if the container has to be thoroughly electroplated with silver metal at the rate of 70 paise per cm2.

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/469/531/10.1.13.1.2_GPL_PIY_KSB_SS_html_7432e057.png

Solution:

To find out the cost, we have to find out the T.S.A. of the container.

Height of the base, h = 1.5 cm

∴Height of the hollow cylinder, H = 21.5 − 1.5 = 20 cm

External radius, R = 11 cm and internal radius, r = 10 cm

T.S.A. of the container = Outer C.S.A. of hollow cylinder + Inner C.S.A. of hollow cylinder + Area of the ring + C.S.A. of solid cylinder + Area of inner base + Area of outer base

= 2π RH + 2πrH + π (R2r2) + 2πRh + πr2 + πR2

= 2π (R + r) H + π (R2r2) + π (2Rh + r2 + R2)

=https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/469/531/10.1.13.1.2_GPL_PIY_KSB_SS_html_5d74c2d2.gif

=https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/469/531/10.1.13.1.2_GPL_PIY_KSB_SS_html_5ad6986f.gif

= 2640 + 66 +https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/469/531/10.1.13.1.2_GPL_PIY_KSB_SS_html_5983fa0e.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/469/531/10.1.13.1.2_GPL_PIY_KSB_SS_html_9cb0bda.gif= cm2

∴ Cost of electroplating =https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/189/469/531/10.1.13.1.2_GPL_PIY_KSB_SS_html_9cb0bda.gif × 70 = 245300 paise = Rs 2453

Thus, the required cost is Rs 2453.

Volume Of Combination Of Solids

We come across many figures in our daily life that are made up of two or more solid figures. Let us consider such an example.

A company produces metallic solid toys that are in the shape of a cylinder with one hemisphere and one cone stuck to their opposite ends. The length of the entire toy is 30 cm; the diameter of the cylinder is 14 cm, while the height of the cone is 10 cm.

Can we find out how much metal should the company order to make 200 toys of this type?

In this way, we can find the volume of any solid figure which is formed by combining two or more basic solids. Let us now look at some more examples.

Example 1: A largest cone is to be taken out from a cube of edge 15 cm. Find the volume of the remaining portion. (Use π = 3.14)

Solution:

The figure can be drawn as follows:

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/470/533/10.1.13.2.1_GPL_PIY_KSB_SS_html_2227167b.png

It is given that length of the cube, l = 15 cm

The base of the cone is a circle whose diameter is equal to the length of the edge of the cube.

∴ Radius of cone, r = https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/470/533/10.1.13.2.1_GPL_PIY_KSB_SS_html_24e2d12b.gifcm

The height of the cone would be equal to the height of cube.

∴ Height of cone, h = 15 cm

Volume of the remaining portion = Volume of Cube − Volume of Cone

=https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/470/533/10.1.13.2.1_GPL_PIY_KSB_SS_html_m7ce5b635.gif

=https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/470/533/10.1.13.2.1_GPL_PIY_KSB_SS_html_17d8b53b.gif

= 3375 − 883.125

= 2491.875 cm3

Thus, the volume of the remaining portion is 2491.875 cm3.

Example 2: A plastic toy is in the following shape. The diameter of the cylindrical shape is 7 cm, but the bottom of the toy has a hemispherical raised portion. The top of the toy is a cone of same base. If the height of the cylinder is 21 cm and cone is 4

cm, find the amount of air inside the toy. (Use π =https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/470/533/10.1.13.2.1_GPL_PIY_KSB_SS_html_m62ee3405.gif )

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/470/533/10.1.13.2.1_GPL_PIY_KSB_SS_html_60432d73.png

Solution:

Diameter of the cylinder = 7 cm

Radius of cylinder, r = https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/470/533/10.1.13.2.1_GPL_PIY_KSB_SS_html_m5dfde585.gifcm

r = https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/470/533/10.1.13.2.1_GPL_PIY_KSB_SS_html_m5dfde585.gif, is also the radius for cone and hemisphere.

Height of the cylinder, H = 21 cm

Height of the cone, h = 4 cm

∴ Volume of the toy = Volume of cylinder + Volume of Cone − Volume of Hemisphere

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/470/533/10.1.13.2.1_GPL_PIY_KSB_SS_html_7ec532f0.gif

= 770 cm3

Thus, the volume of air inside the given toy is 770 cm3.

Example 3: A right triangle, whose perpendicular sides are 30 cm and 40 cm, is made to revolve about its hypotenuse. Find the volume of figure so obtained in terms of π.

Solution:

Let ABD be a right-angled triangle, such that AB = 30 cm and AD = 40 cm.

Using Pythagoras theorem, we have

BD2 = AB2 + AD2

BD2 =https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/470/533/10.1.13.2.1_GPL_PIY_KSB_SS_html_m20f3e42f.gif

BD = https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/470/533/10.1.13.2.1_GPL_PIY_KSB_SS_html_4cf0b492.gifcm

After revolution, we have the following figure.

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/470/533/10.1.13.2.1_GPL_PIY_KSB_SS_html_23aae2f3.png

Let OB = x and OA = y

Using Pythagoras theorem in triangle AOB, we obtain

OA2 + OB2 = AB2

x2 + y2 = 302

y2 = 900 − x2 … (1)

Using Pythagoras theorem in triangle AOD, we obtain

OA2 + OD2 = AD2

y2 + (50 − x)2 = 402

900 − x2 + 2500 + x2 − 100x = 1600 {using equation (1)}

100x = 1800

x = 18

On putting the value of x in equation (1), we obtain

y2 = 900 − x2 = 900 − 182 = 900 − 324 = 576

y = https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/470/533/10.1.13.2.1_GPL_PIY_KSB_SS_html_17c7565.gifcm

Here, y is the radius of the cones ABC and ADC.

Now, height of the cone ABC, x = 18 cm

Height of the cone ADC, 50 − x = 32 cm

Volume of the double cone = Volume of cone ABC + Volume of cone ADC

=https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/470/533/10.1.13.2.1_GPL_PIY_KSB_SS_html_md86a95.gif

=https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/470/533/10.1.13.2.1_GPL_PIY_KSB_SS_html_30a46559.gif

=https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/470/533/10.1.13.2.1_GPL_PIY_KSB_SS_html_m2a75db18.gif

= 9600π cm3

Thus, the volume of the figure so obtained is 9600π cm3.

Example 4: An iron container is cylindrical in shape as shown in the figure. The inner and outer diameters are 12 cm and 14 cm respectively. The container has a solid base of width 1.5 cm and the total height of the container is 22.5 cm. If the mass of 1 cm3 of iron is 8 gm, find the weight of the container.

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/470/533/10.1.13.2.1_GPL_PIY_KSB_SS_html_7e8e35b4.png

Solution:

The iron container consists of two cylinders.

  1. A solid cylinder and
  2. A hollow cylinder

External and internal diameters of the hollow cylinder are 14 and 12 cm respectively.

Hence, internal radius, r = 6 cm and external radius R = 7 cm

Height of the solid cylinder, h = 1.5 cm

∴ Height of the hollow cylinder, H = 22.5 − 1.5 = 21 cm

Volume of the iron used in the container = Volume of solid cylinder + Volume of hollow cylinder

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/470/533/10.1.13.2.1_GPL_PIY_KSB_SS_html_48bbd5e0.gif=

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/470/533/10.1.13.2.1_GPL_PIY_KSB_SS_html_m1055d5cc.gif=

= 231 + 858

= 1089 cm3

Mass of 1 cm3 of iron = 8 gm

Mass of 1089 cm3 of iron = 1089 × 8 = 8712 gm = 8.712 kg

Thus, the weight of the container is 8.712 kg.