10TH FOUNDATION

Chapter -1

Real Number

Theorem (Euclid’s Division Lemma): Given positive integers a and b, there exist unique integers q and r satisfying a = bq + r, 0 ≤ r < b.
To obtain the HCF of two positive integers, say c and d, with c > d, follow the steps below:
Step 1: Apply Euclid’s division lemma, to c and d. So, we find whole numbers, q and r such that c = dq + r, 0 ≤ r < d.
Step 2: If r = 0, d is the HCF of c and d. If r ≠ 0, apply the division lemma to d and r.
Step 3: Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.
This algorithm works because HCF (c, d) = HCF (d, r) where the symbol HCF (c, d) denotes the HCF of c and d, etc.

The Fundamental Theorem of Arithmetic
In your earlier classes, you have seen that any natural number can be written as a product of its prime factors. For instance, 2 = 2, 4 = 2 × 2, 253 = 11 × 23, and so on. Now, let us try and look at natural numbers from the other direction. That is, can any natural number be obtained by multiplying prime numbers? Let us see.
Take any collection of prime numbers, say 2, 3, 7, 11 and 23. If we multiply some or all of these numbers, allowing them to repeat as many times as we wish, we can produce a large collection of positive integers (In fact, infinitely many).
Let us list a few:

7 × 11 × 23 = 1771 3 × 7 × 11 × 23 = 5313
2 × 3 × 7 × 11 × 23 = 10626 23 × 3 × 73 = 8232
22 × 3 × 7 × 11 × 23 = 21252

and so on.
Theorem (Fundamental Theorem of Arithmetic): Every composite number can be expressed (factorised) as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur.
The prime factorisation of a natural number is unique, except for the order of its factors.
In general, given a composite number x, we factorise it as x = p₁p₂ ... pn, where p₁, p₂,..., p_n are primes and written in ascending order, i.e., p₁ ≤ p₂ ≤. . . ≤ p_n. If we combine the same primes, we will get powers of primes. For example,
                            32760 = 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13 = 23 × 32 × 5 × 7 × 13

Revisiting Irrational Numbers

In Class IX, you were introduced to irrational numbers and many of their properties. You studied about their existence and how the rationals and the irrationals together made up the real numbers. You even studied how to locate irrationals on the number line. However, we did not prove that they were irrationals. In this section, we will prove that 2, 3, 5  and, in general, p  is irrational, where p is a prime. One of the theorems, we use in our proof, is the Fundamental Theorem of Arithmetic.
Recall, a number‘s’ is called irrational if it cannot be written in the form  where p and q are integers and q ≠ 0. Some examples of irrational numbers, with which you are already familiar, are:

Before we prove that 2 is irrational, we need the following theorem, whose proof is based on the Fundamental Theorem of Arithmetic.

Revisiting Rational Numbers and Their Decimal Expansions

In Class IX, you studied that rational numbers have either a terminating decimal expansion or a non-terminating repeating decimal expansion. In this section, we are going to consider a rational number, say

,  and explore exactly when the decimal expansion of   is terminating and when it is non-terminating and when it is non-terminating repeating (or recurring). We do so by considering several examples.

Let us consider the following rational numbers:

(i) 0.375       (ii) 0.104        (iii) 0.0875          (iv) 23.3408.

As one would expect, they can all be expressed as rational numbers whose denominators are powers of 10. Let us try and cancel the common factors between the numerator an denominator and see what we get:

Theorem: Let x be a rational number whose decimal expansion terminates.
Then x can be expressed in the form

,  where p and q are coprime, and the prime factorization of q is of the form 2ⁿ5^m, where n, m are non-negative integers.
Theorem:   Let   be a rational number, such that the prime factorization of q is of the form 2ⁿ5^m, where n, m are non-negative integers. Then x has a decimal expansion which terminates.

Chapter -2

POLYNOMISALS

The Polynomial p(x)

   For example, 4x + 2 is a polynomial in the variable x of degree 1, 2y² – 2y + 4 is a polynomial in the variable y of degree
 is a polynomial in the variable x of degree 3 and  is a polynomial in the variable u of  degree 6. Expressions like , are not polynomials.

A polynomial of degree 1 is called a linear polynomial. For example,   are all linear polynomials. Polynomials such as 2x + 5 – x², x³ + 1, etc., are not linear polynomials.

A polynomial of degree 2 is called a quadratic polynomial. The name ‘quadratic’ has been derived from the word ‘quadratie’ which means ‘square’.

 are some examples of quadratic polynomials (whose coefficient are real numbers). More generally, any quadratic polynomial in x is of the form ax² + bx + c, where a, b, c are real numbers and a ≠0.  A polynomial of degree 3 is called a cubic polynomial. Some examples of a cubic polynomial are  In fact, the most general form of a cubic polynomial is ax³ + bx² + cx + d.

Where, a, b, c, d are real numbers and a≠0.
Now consider the polynomial p(x) = x² – 3x – 4. Then, putting x = 2 in the polynomial, we get p(2) = 2² – 3 × 2 – 4 = – 6. The value ‘– 6’, obtained by replacing x by 2 in x² – 3x – 4, is the value of x² – 3x – 4 at x = 2. Similarly, p(0) is the value of p(x) at x = 0, which is – 4.

 If p(x) is a polynomial in x, and if k is any real number, then the value obtained by replacing x by k in p(x), is called the value of p(x) at x = k, and is denoted by p(k). What is the value of p(x) = x² –3x – 4 at x = –1? We have:

p(–1) = (–1)² –{3 × (–1)} – 4 = 0
Also, note that p(4) = 4² – (3 × 4) – 4 = 0.
As p(–1) = 0 and p(4) = 0, –1 and 4 are called the zeroes of the quadratic polynomial  x² – 3x – 4. More generally, a real number k is said to be a zero of a polynomial p(x),  if p(k) = 0.

You have already studied in Class IX, how to find the zeroes of a linear polynomial. For example, if k is a zero of p(x) = 2x + 3, then p(k) = 0 gives us 2k + 3 = 0, i.e.,

 
In general, if k is a zero of p(x) = ax + b, then p(k) = ak + b = 0, i.e.,  

So, the zero of the linear polynomial ax + b is -ba= -Constant term coefficient of ax + b = 

Thus, the zero of a linear polynomial is related to its coefficients. Does this happen in the case of other polynomials too?  For example, are the zeroes of a quadratic polynomial also related to its coefficients?
In this chapter, we will try to answer these questions. We will also study the division algorithm for polynomials.

Geometrical Meaning of the Zeroes of a Polynomial
You know that a real number k is a zero of the polynomial p(x) if p(k) = 0. But why are the zeroes of a polynomial so important? To answer this, first we will see the geometrical representations of linear and quadratic polynomials and the geometrical meaning of their zeroes.

Consider first a linear polynomial ax + b, a ≠ 0. You have studied in Class IX that the graph of y = ax + b is a straight line. For example, the graph of y = 2x + 3 is a straight line passing through the points (– 2, –1) and (2, 7).

From Fig. 2.1, you can see that the graph of y = 2x + 3 intersects the x -axis mid-way between x = –1 and x = – 2, that is, at the point  
You also know that the zero of 2x + 3 is   Thus, the zero of the polynomial 2x + 3 is the x-coordinate of the point where the graph of y = 2x + 3 intersects the x-axis. In general, for a linear polynomial ax+b, a≠0,  has exactly one zero, namely, the x-coordinate of the point where the graph of y = ax + b intersects the axis.ax+b, a≠0,  The graph of y = ax + b is a straight line which Intersects the x-axis at exactly one point, namely,  Therefore, the linear polynomial

Relationship between Zeroes and Coefficients of a Polynomial

 You have already seen that zero of a linear polynomial ax + b is

 We will now try to answer the question raised in Section 2.1 regarding the relationship between zeroes and coefficients of a quadratic polynomial. For this, let us take a quadratic polynomial, say p(x) = 2x² – 8x + 6. In Class IX, you have learnt how to factorise quadratic polynomials by splitting the middle term. So, here we need to split the middle term ‘– 8x’ as a sum of two terms, whose product is 6 × 2x² = 12x². So, we write

2x² – 8x + 6 = 2x² – 6x – 2x + 6 = 2x(x – 3) – 2(x – 3)

                    = (2x – 2)(x – 3) = 2(x – 1)(x – 3)

So, the value of p(x) = 2x² – 8x + 6 is zero when x – 1 = 0 or x – 3 = 0, i.e., when x = 1 or x = 3. So, the zeroes of 2x² – 8x + 6 are 1 and 3. Observe that:
 

 
Let us take one more quadratic polynomial, say, p(x) = 3x² + 5x – 2. By the method of splitting the middle term,

    
 Hence, the value of 3x² + 5x – 2 is zero when either 3x – 1 = 0 or x + 2 = 0, i.e.,

when x=1/3 or x =-2.   So, the zeroes of  Observe that:

In general, if α* and β* are the zeroes of the quadratic polynomial p(x) = ax² + bx + c,

a≠0,  then you know that x-α and x-β     are the factors of p(x). Therefore,

 where k is a constant

Comparing the coefficients of x², x and constant terms on both the sides, we get
   

Division algorithm for Polynomials

 You know that a cubic polynomial has at most three zeroes. However, if you are given only one zero, can you find the other two? For this, let us consider the cubic polynomial

x3-3x2-x+3.  If we tell you that one of its zeroes is 1, then you know that x – 1 is a factor of x³ – 3x² – x + 3. So, you can divide x³ – 3x² – x + 3 by x – 1, as you have learnt in Class IX, to get quotient x² – 2x – 3.

Next, you could get the factors of x² – 2x – 3, by splitting the middle term, as (x + 1) (x – 3). This word give you

So, all the three zeroes of the cubic polynomial are now known to you as 1, – 1, 3.
Let us discuss the method of dividing one polynomial by another in some detail.
​​​​​​​Before noting the steps formally.

 then we can find polynomials q(x) and r(x) such that
 
Where r(x) = 0 or degree of r (x) < degree of g(x). 

Chapter -3

Pair of Linear Equations in Two Variables

Pair of Linear Equations in Two Variables
Recall, from Class IX, that the following are examples of linear equations in two variables:
2x + 3y = 5
x – 2y – 3 = 0
and    x – 0y = 2,  i.e.,  x = 2s
You also know that an equation which can be put in the form ax + by + c = 0, where a, b and c are real numbers, and a and b are not both zero, is called a linear equation in two variables x and y. (We often denote the condition a and b are not both zero by a₂ + b₂ ≠ 0). You have also studied that a solution of such an equation is a pair of values, one for x and the other for y, which makes the two sides of the equation equal.
For example, let us substitute x = 1 and y = 1 in the left hand side (LHS) of the equation 2x + 3y = 5. Then
LHS = 2(1) + 3(1) = 2 + 3 = 5,
which is equal to the right hand side (RHS) of the equation. Therefore, x = 1 and y = 1 is a solution of the equation 2x + 3y = 5. Now let us substitute x = 1 and y = 7 in the equation 2x + 3y = 5. Then,
LHS = 2(1) + 3(7) = 2 + 21 = 23 which is not equal to the RHS.
Therefore, x = 1 and y = 7 is not a solution of the equation.
Geometrically, what does this mean? It means that the point (1, 1) lies on the line representing the equation 2x + 3y = 5, and the point (1, 7) does not lie on it. So, every solution of the equation is a point on the line representing it.
In fact, this is true for any linear equation, that is, each solution (x, y) of a linear equation in two variables, ax + by + c = 0, corresponds to a point on the line representing the equation, and vice versa.
Now, consider Equations (1) and (2) given above. These equations, taken together, represent the information we have about Akhila at the fair.
These two linear equations are in the same two variables x and y. Equations like these are called a pair of linear equations in two variables.

Graphical Method of Solution of a Pair of Linear Equations
In the previous section, you have seen how we can graphically represent a pair of linear equations as two lines. You have also seen that the lines may intersect, or may be parallel, or may coincide. Can we solve them in each case? And if so, how? We shall try and answer these questions from the geometrical point of view in this section.
Let us look at the earlier examples one by one.
l In the situation of Example 1, find out how many rides on the Giant Wheel Akhila had, and how many times she played Hoopla.
In Fig. 3.2, you noted that the equations representing the situation are geometrically shown by two lines intersecting at the point (4, 2). Therefore, the point (4, 2) lies on the lines represented by both the equations x – 2y = 0 and 3x + 4y = 20. And this is the only common point.
Let us verify algebraically that x = 4, y = 2 is a solution of the given pair of equations. Substituting the values of x and y in each equation, we get
4 – 2 × 2 = 0 and 3(4) + 4(2) = 20. So, we have verified that x = 4, y = 2 is a solution of both the equations. Since (4, 2) is the only common point on both the lines, there is one and only one solution for this pair of linear equations in two variables.
We can now summarise the behavior of lines representing a pair of linear equations in two variables and the existence of solutions as follows:
(i)  The lines may intersect in a single point. In this case, the pair of equations has a unique solution (consistent pair of equations).
(ii)  The lines may be parallel. In this case, the equations have no solution (Inconsistent pair of equations).
(iii)  The lines may be coincident. In this case, the equations have infinitely many solutions [dependent (consistent) pair of equations].
Let us now go back to the pairs of linear equations formed in Examples 1, 2, and 3, and note down what kind of pair they are geometrically.

(i) x – 2y = 0 and 3x + 4y – 20 = 0                   (The lines intersect)
(ii) 2x + 3y – 9 = 0 and 4x + 6y – 18 = 0        (The lines coincide)
(iii)  x + 2y – 4 = 0 and 2x + 4y – 12 = 0       (The lines are parallel)

Algebraic Methods of Solving a Pair of Linear Equations

In the previous section, we discussed how to solve a pair of linear equations graphically. The graphical method is not convenient in cases when the point representing the solution of the linear equations has non-integral coordinates like  etc. There is every possibility of making mistakes while reading such coordinates. Is there any alternative method of finding the solution? There are several algebraic methods, which we shall now discuss.

Substitution Method: We shall explain the method of substitution by taking some examples.

Example: Solve the following pair of equations by substitution method:

7x – 15y = 2                                                           (1)
           x + 2y = 3                                                              (2)

Solution:

Step 1: We pick either of the equations and write one variable in terms of the other. Let us consider the Equation (2):

x + 2y = 3

and write it as       x = 3 – 2y                                                    (3)

Step 2: Substitute the value of x in Equation (1). We get
7(3 – 2y) – 15y = 2
i.e.,                 21 – 14y – 15y = 2
i.e.,                           – 29y = –19
Therefore,         
Step 3: Substituting this value of y in Equation (3), we get  

Therefore, the solution is 

Verification: Substituting x = 29 and y = 29, you can verify that both the Equations
(1) and (2) are satisfied.
Remark: We have substituted the value of one variable by expressing it in terms of the other variable to solve the pair of linear equations. That is why the method is known as the substitution method.

Elimination Method
Now let us consider another method of eliminating (i.e., removing) one variable. This is sometimes more convenient than the substitution method. Let us see how this method works.

Cross - Multiplication Method
So far, you have learnt how to solve a pair of linear equations in two variables by graphical, substitution and elimination methods. Here, we introduce one more algebraic method to solve a pair of linear equations which for many reasons is a very useful method of solving these equations. Before we proceed further, let us consider the following situation.
Let us now see how this method works for any pair of linear equations in two variables of the form

            a₁x + b₁y + c₁ = 0                                                             (1)

and               a₂x + b₂y + c₂ = 0                                                             (2)

To obtain the values of x and y as shown above, we follow the following steps:

Step 1: Multiply Equation (1) by b₂ and Equation (2) by b₁, to get

b₂a₁x + b₂b₁y + b₂c₁ = 0                                                (3)

b₁a₂x + b₁b₂y + b₁c₂ = 0                                                (4)

Step 2: Subtracting Equation (4) from (3), we get:

(b₂a₁ – b₁a₂) x + (b₂b₁ – b₁b₂) y + (b₂c₁ – b₁c₂) = 0

i.e.,    (b₂a₁ – b₁a₂) x = b₁c₂ – b₂c₁

     (5)

 Step 3: Substituting this value of x in (1) or (2), we get

                                                            (6)

Now, two cases arise:
Case 1:  a₁b₂ – a₂b₁ ≠ 0. In this case  Then the pair of linear equations has a unique solution.
Case 2: a₁b₂ – a₂b₁ = 0. If we write

  then a₁ = k a₂, b₁ = k b₂.
Substituting the values of a₁ and b₁ in the Equation (1), we get
                       k(a₂x + b₂y) + c₁ = 0.                                                           (7)
It can be observed that the Equations (7) and (2) can both be satisfied only if

 c₁ = k c₂, i.e.,
If c₁ = k c₂, any solution of Equation (2) will satisfy the Equation (1), and vice versa. So, if   , then there are infinitely many solutions to the pair of linear equations given by (1) and (2).
If c₁ ≠ k c₂, then any solution of Equation (1) will not satisfy Equation (2) and vice versa. Therefore the pair has no solution.
We can summarise the discussion above for the pair of linear equations given by (1) and (2) as follows:

(i) When  we get a unique solution.

(ii) When   there are infinitely many solutions.

(iii) When    there is no solution.

Note that you can write the solution given by Equations (5) and (6) in the following form:

                                                         (8)

In remembering the above result, the following diagram may be helpful to you:

The arrows between the two numbers indicate that they are to be multiplied and the second product is to be subtracted from the first.
For solving a pair of linear equations by this method, we will follow the following steps:
Step 1: Write the given equations in the form (1) and (2).
Step 2: Taking the help of the diagram above, write Equations as given in (8).
Step 3: Find x and y, provided a₁b₂ – a₁b₁ ≠ 0
Step 2 above gives you an indication of why this method is called the cross-multiplication method.

Equations Reducible to a Pair of Linear Equations in Two Variables
In this section, we shall discuss the solution of such pairs of equations which are not linear but can be reduced to linear form by making some suitable substitutions. We now explain this process through some examples.

Chapter -4

TRIANGLES

Introduction

In Class IX, you have seen that all circles with the same radii are congruent, all squares with the same side lengths are congruent and all equilateral triangles with the same side lengths are congruent.
Now consider any two (or more) circles [see Fig (i)]. Are they congruent? Since all of them do not have the same radius, they are not congruent to each other. Note that some are congruent and some are not, but all of them have the same shape. So they all are, what we call, similar. Two similar figures have the same shape but not necessarily the same size. Therefore, all circles are similar. What about two (or more) squares or two (or more) equilateral triangles [see Fig (ii) and (iii)]? As observed in the case of circles, here also all squares are similar and all equilateral triangles are similar.

From the above, we can say that all congruent figures are similar but the similar figures need not be congruent. Can a circle and a square be similar? Can a triangle and a square be similar? These questions can be answered by just looking at the figures (see Fig). Evidently these figures are not similar. (Why?)
What can you say about the two quadrilaterals ABCD and PQRS (see Fig.1)? Are they similar? These figures appear to be similar but we cannot be certain about it. Therefore, we must have some definition of similarity of figures and based on this definition some rules to decide whether the two given figures are similar or not. For this, let us look at the photographs given in Fig. 2:

You will at once say that they are the photographs of the same monument (Taj Mahal) but are in different sizes. Would you say that the three photographs are similar? Yes, they are.
What can you say about the two photographs of the same size of the same person one at the age of 10 years and the other at the age of 40 years? Are these photographs similar? These photographs are of the same size but certainly they are not of the same shape. So, they are not similar.
What does the photographer do when she prints photographs of different sizes from the same negative? You must have heard about the stamp size, passport size and postcard size photographs. She generally takes a photograph on a small size film, say of 35mm size and then enlarges it into a bigger size, say 45mm (or 55mm). Thus, if we consider any line segment in the smaller photograph (figure), its corresponding line segment in the bigger photograph (figure) will be    of that of the line segment. This really means that every line segment of the smaller photograph is enlarged (increased) in the ratio 35:45 (or 35:55). It can also be said that every line segment of the bigger photograph is reduced (decreased) in the ratio 45:35 (or 55:35). Further, if you consider inclinations (or angles) between any pair of corresponding line segments in the two photographs of different sizes, you shall see that these inclinations (or angles) are always equal. This is the essence of the similarity of two figures and in particular of two polygons. We say that:

Similarity of Triangles
You may recall that triangle is also a polygon. So, we can state the same conditions for the similarity of two triangles. That is:
Two triangles are similar, if
(i) Their corresponding angles are equal and
(ii) Their corresponding sides are in the same ratio (or proportion).
Note that if corresponding angles of two triangles are equal, then they are known as equiangular triangles. A famous Greek mathematician Thales gave an important truth relating to two equiangular triangles which is as follows: The ratio of any two corresponding sides in two equiangular triangles is always the same. It is believed that he had used a result called the Basic Proportionality Theorem (now known as the Thales Theorem) for the same.

Theorem: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Proof: We are given a triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively (see Fig).

We need to prove that   

Let us join BE and CD and then draw DM ⊥ AC and EN ⊥ AB.
  
Recall from Class IX, that area of ΔADE is denoted as are

  
  
  
  
  
Note that ∆BDE and DEC are on the same base DE and between the same parallels BC and DE.
So,                     ar(BDE) = ar(DEC)
Therefore, from (1), (2) and (3), we have:
                       

Criteria for Similarity of Triangles

In the previous section, we stated that two triangles are similar, if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion).
That is, in ΔABC and ΔDEF, if
(i) ∠A = ∠D, ∠B = ∠E, ∠C = ∠F and

 Then the two triangles are similar (see Fig).

Here, you can see that A corresponds to D, B corresponds to E and C corresponds to F. Symbolically, we write the similarity of these two triangles as ‘ΔABC ~ ΔDEF’ and read it as ‘triangle ABC is similar to triangle DEF’. The symbol ‘~’ stands for ‘is similar to’. Recall that you have used the symbol ‘≅’ for ‘is congruent to’ in Class IX.

Now a natural question arises: For checking the similarity of two triangles, say ABC and DEF, should we always look for all the equality relations of their corresponding angles (∠A = ∠D, ∠B = ∠E, ∠C = ∠F) and all the equality relations of the ratios of their corresponding sides  ? Let us examine. You may recall that in Class IX, you have obtained some criteria for congruency of two triangles involving only three pairs of corresponding parts (or elements) of the two triangles. Here also, let us make an attempt to arrive at certain criteria for similarity of two triangles involving relationship between less numbers of pairs of corresponding parts of the two triangles, instead of all the six pairs of corresponding parts. For this, let us perform the following activity:

Theorem: If in two triangles, sides of one triangle are proportional to (i.e., in the same ratio of) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar. This criterion is referred to as the SSS (Side–Side–Side) similarity criterion for two triangles.

This theorem can be proved by taking two triangles ABC and DEF such that  

Cut DP = AB and DQ = AC and join PQ.

Theorem: If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar. This criterion is referred to as the SAS (Side–Angle–Side) similarity criterion for two triangles.

As before, this theorem can be proved by taking two triangles ABC and DEF such that    (see Fig). Cut DP = AB, DQ = AC and join PQ.

Now, PQ || EF and ΔABC ≅ ΔDPQ           (How?)
So, ∠A = ∠D, ∠B = ∠P and ∠C = ∠Q
Therefore, ΔABC ~ ΔDEF                          (Why?)
We now take some examples to illustrate the use of these criteria.
Theorem: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Proof : We are given two triangles ABC and PQR such that
ΔABC ~ ΔPQR (see Fig).

We need to prove that  

For finding the areas of the two triangles, we draw altitudes AM to PN of the triangles.
 
 
 

 
 
 
  
  
 

Now using (3), we get    
     
Let us take an example to illustrate the use of this theorem.

Pythagoras Theorem
You are already familiar with the Pythagoras Theorem from your earlier classes. You had verified this theorem through some activities and made use of it in solving certain problems. You have also seen a proof of this theorem in Class IX. Now, we shall prove this theorem using the concept of similarity of triangles. In proving this, we shall make use of a result related to similarity of two triangles formed by the perpendicular to the hypotenuse from the opposite vertex of the right triangle.
Now, let us take a right triangle ABC, right angled at B. Let BD be the perpendicular to the hypotenuse AC (see Fig).
 

You may note that in ΔADB and ΔABC
                                                 ∠A = ∠A
and                                     ∠ADB = ∠ABC (Why?)
So,                                        ΔADB ~ ΔABC  (How?)                            (1)  
imilarly,                           ΔBDC ~ ΔABC (How?)                                (2)
So, from (1) and (2), triangles on both sides of the perpendicular BD are similar to the whole triangle ABC.
Also, since                   ΔADB ~ ΔABC
and                              ΔBDC ~ ΔABC
So,                     ΔADB ~ ΔBDC        (From Remark in Section 6.2)
The above discussion leads to the following theorem:
Theorem: If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.
Let us now apply this theorem in proving the Pythagoras Theorem:

Theorem: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Proof: We are given a right triangle ABC right angled at B.  

We need to prove that AC² = AB² + BC²

Let us draw    BD ⊥ AC    (see Fig.)

Now,          ΔADB ~ ΔABC    (Theorem)

Also,    ΔBDC ~ ΔABC (Theorem)

  
Adding (1) and (2),

AD . AC + CD .  AC = AB² + BC²  

or,  AC (AD + CD) = AB² + BC²  

or,  AC . AC = AB² + BC²  

or,  AC² = AB² + BC²  

Theorem: In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.
Proof: Here, we are given a triangle ABC in which AC² = AB² + BC².
We need to prove that ∠B = 90°.
To start with, we construct a ΔPQR right angled at Q such that PQ = AB and QR = BC (see Fig).

Now, from ΔPQR, we have:

 PR² = PQ² + QR²         (Pythagoras Theorem, as ∠Q = 90°)

or,                PR² = AB² + BC²   (By construction) (1)

But              AC² = AB² + BC²   (Given)                (2)
So,                        AC = PR                [From (1) and (2)] (3)
Now, in ΔABC and ΔPQR,
AB = PQ               (By construction)
BC = QR               (By construction)
AC = PR                [Proved in (3) above]
So,                      ΔABC ≅ ΔPQR                  (SSS congruence)
Therefore,        ∠B = ∠Q                    (CPCT)
But                   ∠Q = 90°          (By construction)
So,                   ∠B = 90°
Note: Also see Appendix 1 for another proof of this theorem.

Chapter:- 8

Introduction to trigonometry

Introduction

Suppose the students of a school are visiting Qutub Minar. Now, if a student is looking at the top of the Minar, a right triangle can be imagined to be made, as shown in Fig. Can the student find out the height of the Minar, without actually measuring it?

Suppose a girl is sitting on the balcony of her house located on the bank of a river. She is looking down at a flower pot placed on a stair of a temple situated nearby on the other bank of the river. A right triangle is imagined to be made in this situation as shown in Fig. If you know the height at which the person is sitting, can you find the width of the river?

Suppose a hot air balloon is flying in the air. A girl happens to spot the balloon in the sky and runs to her mother to tell her about it. Her mother rushes out of the house to look at the balloon.Now when the girl had spotted the balloon intially it was at point A. When both the mother and daughter came out to see it, it had already travelled to another point B. Can you find the altitude of B from the ground? 

In all the situations given above, the distances or heights can be found by using some mathematical techniques, which come under a branch of mathematics called ‘trigonometry’. The word ‘trigonometry’ is derived from the Greek words ‘tri’ (meaning three), ‘gon’ (meaning sides) and ‘metron’ (meaning measure). In fact, trigonometry is the study of relationships between the sides and angles of a triangle. The earliest known work on trigonometry was recorded in Egypt and Babylon. Early astronomers used it to find out the distances of the stars and planets from the Earth. Even today, most of the technologically advanced methods used in Engineering and Physical Sciences are based on trigonometrical concepts.

In this chapter, we will study some ratios of the sides of a right triangle with respect to its acute angles, called trigonometric ratios of the angle. We will restrict our discussion to acute angles only. However, these ratios can be extended to other angles also. We will also define the trigonometric ratios for angles of measure 0° and 90°. We will calculate trigonometric ratios for some specific angles and establish some identities involving these ratios, called trigonometric identities.

Trigonometric Ratios

In Section 8.1, you have seen some right triangles imagined to be formed in different situations.
Let us take a right triangle ABC as shown in Fig.
Here, ∠CAB (or, in brief, angle A) is an acute angle. Note the position of the side BC with respect to angle A. it faces ∠A. we call it the side opposite to angle A. AC is the hypotenuse of the right triangle and the side AB is a part of ∠A. so, we call it the side adjacent to angle A.
 

Note that the position of sides change when you consider angle C in place of A.
You have studied the concept of ‘ratio’ in your earlier classes. We now define certain ratios involving the sides of a right triangle, and call them trigonometric ratios.
The trigonometric ratios of the angle A in right triangle ABC are defined as follows:

Sine of      

The ratios defined above are abbreviated as sin A, cos A, tan A, cosec A, sec A and cot A respectively. Note that the ratios cosec A, sec A and cot A are respectively, the reciprocals of the ratios sin A, cos A and tan A.

Also, observe that tan A = 

From our observations, it is now clear that the values of the trigonometric ratios of an angle do not vary with the lengths of the sides of the triangle, if the angle  remains the same.

Trigonometric Ratios of Some Specific Angles

From geometry, you are already familiar with the construction of angles of 30°, 45°, 60° and 90°. In this section, we will find the values of the trigonometric ratios for these angles and, of course, for 0°.

Trigonometric Ratios of 45°
In  ∆ABC, right-angled at B, if one angle is 45°, then the other angle is also 45°, i.e., ∠A =  ∠C = 45°
So,    BC = AB      (Why?)
Now, Suppose BC = AB = a.
Then by Pythagoras Theorem, AC² = AB² + BC² = a² + a² = 2a²,And, therefore,  AC = a  .
Using the definitions of the trigonometric ratios, we have:

Trigonometric Ratios of 30° and 60°
Let us now calculate the trigonometric ratios of 30° and 60°. Consider an equilateral triangle ABC. Since each angle in an equilateral triangle is 60°, therefore,
∠A = ∠B = ∠C = 60^o
Draw the perpendicular AD from A to the side BC  
Now                                ∆ABD ≅ ∆ACD (Why?)
Therefore,                              BD = DC
And                                      ∠BAD = ∠CAD (CPCT)
Now observe that:
∆ABD is a right triangle, right-angled at D with ∠BAD = 30^o and ∠ABD = 60^o
As you know, for finding the trigonometric ratios, we need to know the lengths of the sides of the triangle. So, let us suppose that AB = 2a.

Then,                      
Now, we have:

 

       
Similarly,
 
 

Trigonometric Ratios of 0° and 90°
Let us see what happens to the trigonometric ratios of angle A, if it is made smaller and smaller in the right triangle ABC (see Fig.), till it becomes zero. As ∠A gets smaller and smaller, the length of the side BC decreases. The point C gets closer to point B, and finally when ∠A becomes very close to 0°, AC becomes almost the same as AB.

When ∠A is very close to 0^o, BC gets very close to 0 and so the value of    is very close to 0. Also, when ∠A is very close to 0⁰, AC is nearly the same as AB and so the value of cosA=ABAC is very close to 1. 
This helps us to how we can define the values of sin A cos A when A = 0⁰. We define: sin 0⁰ = 0 and cos 0⁰ = 1.
Using these, we have

 
Now, let us see what happens to the trigonometric ratios of ∠A, when it is made larger and larger in ∠ABC till it becomes 90°. As ∠A gets larger and larger, ∠C gets smaller and smaller. Therefore, as in the case above, the length of the side AB goes on decreasing. The point A gets closer to point B. Finally when ∠A is very close to 90°, ∠C becomes very close to 0° and the side AC almost coincides with side BC.

When ∠C is very close to 0^o, ∠A is very close to 90^o, side AC is nearly the same as side BC, and so sin A is very close to 1. Also when ∠A is very close to 90^o, ∠C is very close to 0^o, and the side AB is nearly zero, so cos A is very close to 0.
So, we define:
sin 90^o = 1 and cos 90^o = 0.       
We shall now give the values of all the trigonometric ratios of 0^o, 30^o 45^o, 60^o and 90^o in Table 8.1, for ready reference

Remark: From the table above you can observe that as ∠A increases from 0^o to 90^o, sin A increases from 0 to 1 and cos A decreases from 1 to 0.

Trigonometric Ratios of Complementary Angles

Recall that two angles are said to be complementary if their sum equals 90°. In ∠ABC, right-angled at B, do you see any pair of complementary angles?
Since ∠A + ∠C = 90^o, they form such a pair. We have:

 Now let us write the trigonometric ratios for ∠C = 90^o - ∠A.
For convenience, we shall write 90^o – A instead of 90^o - ∠A.
What would be the side opposite and the side adjacent to the angle 90^o – A?
You will find that AB is the side opposite and BC is the side adjacent to the angle
90^o – A. Therefore,

Now, compare the ratios in (1) and (2). Observe that:
 

So,
 
 
 
For all values of angle A lying between 0^o and 90^o. Check whether this holds for     A = 0^o or A = 90^o.

Trigonometric Identities
You may recall that an equation is called an identity when it is true for all values of the variables involved. Similarly, an equation involving trigonometric ratios of an angle is called a trigonometric identity, if it is true for all values of the angle (s) involved.
In this section, we will prove one trigonometric identity, and use it further to prove other useful trigonometric identities.
In ∆ABC, right-angled at B, we have:
AB² + BC² = AC²                           (1)
Dividing each term of (1) by AC², we get
             
i.e.,  
i.e.,                  cos² A + sin² A = 1                         (2)

This is true for all A such that 0^o ≤ A ≤ 90^o. So this is a trigonometric identity.
Let us now divide (1) by AB². We get    

Or,                
i.e.,                           1 + tan² A = sec² A                (3)
Is this equation true for A = 0^o? Yes, it is. What about A = 90^o? Well, tan A and sec A are not defined for A = 90^o. So, (3) is true for all A such that 0^o ≤ A < 90^o.
Let us see what we get on dividing (1) by BC². We get
  
i.e.,   
i.e.,   cot² A + 1 = cosec² A             (4)

Chapter -6

STATISTICS

Introduction
In fact, you went a step further by studying certain numerical representatives of the ungrouped data, also called measures of central tendency, namely, mean, median and mode. In this chapter, we shall extend the study of these three measures, i.e., mean, median and mode from ungrouped data to that of grouped data. We shall also discuss the concept of cumulative frequency, the cumulative frequency distribution and how to draw cumulative frequency curves, called ogives.

Mean of Grouped Data
The mean (or average) of observations, as we know, is the sum of the values of all the observations divided by the total number of observations. From Class IX, recall that if x₁, x₂,…., x_n are observations with respective frequencies f₁,  f₂, …., f_nx_n, then this means observation x₁ occurs f₁ times, x₂ occurs f₂ times, and so on.
Now, the sum of the values of all the observations = f₁x₁ + f₂x₂ + … + f_nx_n, and the number of observations = f₁ + f₂ + … + f_n.
So, the mean    of the data is given by

Recall that we can write this in short form by using the Greek letter ∑ (capital sigma) which means summation. That is,
                      
Which, more briefly, is written as  if it is understood that I varies from 1 to n.
So, from Table 14.4, the mean of the deviations,
Now, let us find the relation between d and x.
Since in obtaining d_i, we subtracted ‘a’ from each x_i, so, in order to get the mean  we need to add ‘a’ to . This can be explained mathematically as:

Mean of deviations,     

So,                               

Median of Grouped Data
As you have studied in Class IX, the median is a measure of central tendency which gives the value of the middle-most observation in the data. Recall that for finding the median of ungrouped data, we first arrange the data values of the observations in ascending order. Then, if n is odd, the median is the observation. And, if is even, then the

After finding the median class, we use the following formula for calculating the median.
         
Where                   l = lower limit of median class,
                                 n = Number of observations,
                                cf = Cumulative frequency of class preceding the median class,
                                f = Frequency of median class,
                              h = Class size (assuming class size to be equal).

Remarks:
1. There is an empirical relationship between the three measures of central tendency:
2 Median = Mode + 2 Mean
3. The median of grouped data with unequal class sizes can also be calculated. However, we shall not discuss it here. 

Mode of Grouped Data

Recall from Class IX, a mode is that value among the observations which occurs most often, that is, the value of the observation having the maximum frequency. Further, we discussed finding the mode of ungrouped data. Here, we shall discuss ways of obtaining a mode of grouped data. It is possible that more than one value may have the same maximum frequency. In such situations, the data is said to be multimodal.  Though grouped data can also be multimodal, we shall restrict ourselves to problems having a single mode only.
 
Where l = lower limit of the modal class,
h = size of the class interval (assuming all class sizes to be equal),
f₁ = frequency of the modal class,
f₀ = frequency of the class preceding the modal class,
f₂ = frequency of the class succeeding the modal class.
Let us consider the following examples to illustrate the use of this formula.

Graphical Representation of Cumulative Frequency Distribution
As we all know, pictures speak better than words. A graphical representation helps us in understanding given data at a glance. In Class IX, we have represented the data through bar graphs, histograms and frequency polygons. Let us now represent a cumulative frequency distribution graphically.
For example, let us consider the cumulative frequency distribution given in Table 14.13.
Recall that the values 10, 20, 30,. . ., 100 are the upper limits of the respective class intervals. To represent the data in the table graphically, we mark the upper limits of the class intervals on the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis (y-axis), choosing a convenient scale. The scale may not be the same on both the axis. Let us now plot the points corresponding to the ordered pairs given by (upper limit, corresponding cumulative frequency), i.e., (10, 5), (20, 8), (30, 12), (40, 15), (50, 18), (60, 22), (70, 29), (80, 38), (90, 45), (100, 53) on a graph paper and join them by a free hand smooth curve. The curve we get is called a cumulative frequency curve, or an ogive (of the less than type).

Chapter-7

Quadratic equations

Introduction
In Chapter 2, you have studied different types of polynomials. One type was the quadratic polynomial of the form ax² + bx + c, a ≠ 0. When we equate this polynomial to zero, we get a quadratic equation. Quadratic equations come up when we deal with many real-life situations. For instance, suppose a charity trust decides to build a prayer hall having a carpet area of 300 square metres with its length one metre more than twice its breadth. What should be the length and breadth of the hall? Suppose the breadth of the hall is x metres. Then, its length should be (2x + 1) metres. We can depict this information pictorially as shown in Fig.
Now,            area of the hall = (2x + 1). xm² = (2x² + x) m²
So,                 2x² + x = 300 (Given)
Therefore,     2x² + x – 300 = 0
So, the breadth of the hall should satisfy the equation 2x² + x – 300 = 0 which is a quadratic equation.

Quadratic Equations
A quadratic equation in the variable x is an equation of the form ax² + bx + c = 0, where a, b, c are real numbers, a ≠ 0. For example, 2x² + x – 300 = 0 is a quadratic equation.
Similarly, 2x² – 3x + 1 = 0, 4x – 3x² + 2 = 0 and 1 – x² + 300 = 0 are also quadratic equations.
In fact, any equation of the form p(x) = 0, where p(x) is a polynomial of degree 2, is a quadratic equation. But when we write the terms of p(x) in descending order of their degrees, then we get the standard form of the equation. That is, ax² + bx + c = 0, a ≠ 0 is called the standard form of a quadratic equation.
Quadratic equations arise in several situations in the world around us and in different fields of mathematics.

Relationship between Zeroes and Coefficients of a Polynomial
In general, a real number α is called a root of the quadratic equation ax² + bx + c = 0, a ≠ 0 if a α² + bα + c = 0. We also say that x = α is a solution of the quadratic equation, or that α satisfies the quadratic equation. Note that the zeroes of the quadratic polynomial ax² + bx + c and the roots of the quadratic equation ax² + bx + c = 0 are the same.
Note:- That we have found the roots of 2x² – 5x + 3 = 0 by factorizing 2x² – 5x + 3 into two linear factors and equating each factor to zero.

Solution of a Quadratic Equation by Completing the Square

In the previous section, you have learnt one method of obtaining the roots of a quadratic equation. In this section, we shall study another method.
Consider the following situation:
The product of Sunita’s age (in years) two years ago and her age four years from now is one more than twice her present age. What is her present age?
To answer this, let her present age (in years) be x. Then the product of her ages two years ago and four years from now is (x – 2) (x + 4).
Now consider the quadratic equation (x + 2)² – 9 = 0. To solve it, we can write it as        (x + 2)² = 9. Taking square roots, we get x + 2 = 3 or x + 2 = – 3.
Therefore,   x = 1 or x = –5
So, the roots of the equation (x + 2)² – 9 = 0 are 1 and – 5.
In both the examples above, the term containing x is completely inside a square, and we found the roots easily by taking the square roots. But, what happens if we are asked to solve the equation x² + 4x – 5 = 0? We would probably apply factorisation to do so, unless we realise (somehow!) that x² + 4x – 5 = (x + 2)² – 9.
So, solving x² + 4x – 5 = 0 is equivalent to solving (x + 2)² – 9 = 0, which we have seen is very quick to do. In fact, we can convert any quadratic equation to the form  (x + a)² – b² = 0 and then we can easily find its roots. Let us see if this is possible.
Look at Fig.
In this figure, we can see how x² + 4x is being converted to (x + 2)² – 4.

So, x² + 4x – 5 = 0 can be written as (x + 2)² – 9 = 0 by this process of completing the square. This is known as the method of completing the square.

In brief, this can be shown as follows:
 
 
 
 
Consider now the equation 3x² – 5x + 2 = 0. Note that the coefficient of x² is not a perfect square. So, we multiply the equation throughout by 3 to get

 
 
 
   
So, 9x² – 15x + 6 = 0 can be written as
 

So, the solution of 9x² – 15x + 6 = 0 are the same as those of  
 
(We can also write this as  where ‘± ’ denotes minus’.)
 
 
 
Consider the quadratic equation  Dividing throughout by a,
We get     
 
 
So, the roots of the given equation are the same as those of
  

If b²-4ac ≥ 0, then by taking the square in 1, we get  

So, the roots of ax² + bx + c = 0 are   

Equation will have no real roots. (Why?)
Thus, if  then the roots of the quadratic equation ax² + bx + = 0 are given by  
This formula for finding the roots of a quadratic equation is known as the quadratic formula.

Nature of Roots
In the previous section, you have seen that the roots of the equation ax² + bx + c = 0 are given by
    
If b² – 4ac > 0, we get two distinct real roots  

 
So, the roots of the equation ax² + bx + c = 0 are both  
Therefore, we say that the quadratic equation ax² + bx + c = 0 has two equal real roots in this case.
If b² – 4ac < 0, then there is no real number whose square is b² – 4ac. Therefore, there are no real roots for the given quadratic equation in this case.
Since b² – 4ac determines whether the quadratic equation ax² + bx + c = 0 has real roots or not, b² – 4ac is called the discriminant of this quadratic equation.
So, a quadratic equation ax² + bx + c = 0 has
(i) Two distinct real roots, if b² – 4ac > 0,
(ii) Two equal real roots, if b² – 4ac = 0,
(iii) No real roots, if b² – 4ac < 0.