MATHEMATICS BOOK CLASS (10th)

1. Real Numbers

Exercise 1.1 Page: 7

1. Use Euclid’s division algorithm to find the HCF of:

i. 135 and 225

ii. 196 and 38220

iii. 867 and 255

Solutions:

i. 135 and 225

As you can see, from the question 225 is greater than 135. Therefore, by Euclid’s division algorithm, we have,

225 = 135 × 1 + 90

Now, remainder 90 ≠ 0, thus again using division lemma for 90, we get,

135 = 90 × 1 + 45

Again, 45 ≠ 0, repeating the above step for 45, we get,

90 = 45 × 2 + 0

The remainder is now zero, so our method stops here. Since, in the last step, the divisor is 45, therefore, HCF (225,135) = HCF (135, 90) = HCF (90, 45) = 45.

Hence, the HCF of 225 and 135 is 45.

ii. 196 and 38220

In this given question, 38220>196, therefore the by applying Euclid’s division algorithm and taking 38220 as divisor, we get,

38220 = 196 × 195 + 0

We have already got the remainder as 0 here. Therefore, HCF(196, 38220) = 196.

Hence, the HCF of 196 and 38220 is 196.

iii. 867 and 255

As we know, 867 is greater than 255. Let us apply now Euclid’s division algorithm on 867, to get,

867 = 255 × 3 + 102

Remainder 102 ≠ 0, therefore taking 255 as divisor and applying the division lemma method, we get,

255 = 102 × 2 + 51

Again, 51 ≠ 0. Now 102 is the new divisor, so repeating the same step we get,

102 = 51 × 2 + 0

The remainder is now zero, so our procedure stops here. Since, in the last step, the divisor is 51, therefore, HCF (867,255) = HCF(255,102) = HCF(102,51) = 51.

Hence, the HCF of 867 and 255 is 51.

2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

Solution:

Let a be any positive integer and b = 6. Then, by Euclid’s algorithm, a = 6q + r, for some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5, because 0≤r<6.

Now substituting the value of r, we get,

If r = 0, then a = 6q

Similarly, for r= 1, 2, 3, 4 and 5, the value of a is 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5, respectively.

If a = 6q, 6q+2, 6q+4, then a is an even number and divisible by 2. A positive integer can be either even or odd Therefore, any positive odd integer is of the form of 6q+1, 6q+3 and 6q+5, where q is some integer.

3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

Solution:

Given,

Number of army contingent members=616

Number of army band members = 32

If the two groups have to march in the same column, we have to find out the highest common factor between the two groups. HCF(616, 32), gives the maximum number of columns in which they can march.

By Using Euclid’s algorithm to find their HCF, we get,

Since, 616>32, therefore,

616 = 32 × 19 + 8

Since, 8 ≠ 0, therefore, taking 32 as new divisor, we have,

32 = 8 × 4 + 0

Now we have got remainder as 0, therefore, HCF (616, 32) = 8.

Hence, the maximum number of columns in which they can march is 8.

4. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Solutions:

Let x be any positive integer and y = 3.

By Euclid’s division algorithm, then,

x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.

Therefore, x = 3q, 3q+1 and 3q+2

Now as per the question given, by squaring both the sides, we get,

x² = (3q)² = 9q² = 3 × 3q²

Let 3q² = m

Therefore, x²= 3m ……………………..(1)

x² = (3q + 1)² = (3q)²+1²+2×3q×1 = 9q² + 1 +6q = 3(3q²+2q) +1

Substitute, 3q²+2q = m, to get,

x²= 3m + 1 ……………………………. (2)

x²= (3q + 2)² = (3q)²+2²+2×3q×2 = 9q² + 4 + 12q = 3 (3q² + 4q + 1)+1

Again, substitute, 3q²+4q+1 = m, to get,

x²= 3m + 1…………………………… (3)

Hence, from equation 1, 2 and 3, we can say that, the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Solution:

Let x be any positive integer and y = 3.

By Euclid’s division algorithm, then,

x = 3q+r, where q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.

Therefore, putting the value of r, we get,

x = 3q

or

x = 3q + 1

or

x = 3q + 2

Now, by taking the cube of all the three above expressions, we get,

Case (i): When r = 0, then,

x²= (3q)³ = 27q³= 9(3q³)= 9m; where m = 3q³

Case (ii): When r = 1, then,

x³ = (3q+1)³ = (3q)³ +1³+3×3q×1(3q+1) = 27q³+1+27q²+9q

Taking 9 as common factor, we get,

x³ = 9(3q³+3q²+q)+1

Putting = m, we get,

Putting (3q³+3q²⁺q) = m, we get ,

x³ = 9m+1

Case (iii): When r = 2, then,

x³ = (3q+2)³= (3q)³+2³+3×3q×2(3q+2) = 27q³+54q²+36q+8

Taking 9 as common factor, we get,

x³=9(3q³+6q²+4q)+8

Putting (3q³+6q²+4q) = m, we get ,

x³ = 9m+8

Therefore, from all the three cases explained above, it is proved that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Exercise 1.2 Page: 11

1. Express each number as a product of its prime factors:

(i) 140

(ii) 156

(iii) 3825

(iv) 5005

(v) 7429

Solutions:

(i) 140

By Taking the LCM of 140, we will get the product of its prime factor.

Therefore, 140 = 2 × 2 × 5 × 7 × 1 = 2²×5×7

(ii) 156

By Taking the LCM of 156, we will get the product of its prime factor.

Hence, 156 = 2 × 2 × 13 × 3 × 1 = 2²× 13 × 3

(iii) 3825

By Taking the LCM of 3825, we will get the product of its prime factor.

Hence, 3825 = 3 × 3 × 5 × 5 × 17 × 1 = 3²×5²×17

(iv) 5005

By Taking the LCM of 5005, we will get the product of its prime factor.

Hence, 5005 = 5 × 7 × 11 × 13 × 1 = 5 × 7 × 11 × 13

(v) 7429

By Taking the LCM of 7429, we will get the product of its prime factor.

Hence, 7429 = 17 × 19 × 23 × 1 = 17 × 19 × 23

2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.

(i) 26 and 91

(ii) 510 and 92

(iii) 336 and 54

Solutions:

(i) 26 and 91

Expressing 26 and 91 as product of its prime factors, we get,

26 = 2 × 13 × 1

91 = 7 × 13 × 1

Therefore, LCM (26, 91) = 2 × 7 × 13 × 1 = 182

And HCF (26, 91) = 13

Verification

Now, product of 26 and 91 = 26 × 91 = 2366

And Product of LCM and HCF = 182 × 13 = 2366

Hence, LCM × HCF = product of the 26 and 91.

(ii) 510 and 92

Expressing 510 and 92 as product of its prime factors, we get,

510 = 2 × 3 × 17 × 5 × 1

92 = 2 × 2 × 23 × 1

Therefore, LCM(510, 92) = 2 × 2 × 3 × 5 × 17 × 23 = 23460

And HCF (510, 92) = 2

Verification

Now, product of 510 and 92 = 510 × 92 = 46920

And Product of LCM and HCF = 23460 × 2 = 46920

Hence, LCM × HCF = product of the 510 and 92.

(iii) 336 and 54

Expressing 336 and 54 as product of its prime factors, we get,

336 = 2 × 2 × 2 × 2 × 7 × 3 × 1

54 = 2 × 3 × 3 × 3 × 1

Therefore, LCM(336, 54) = = 3024

And HCF(336, 54) = 2×3 = 6

Verification

Now, product of 336 and 54 = 336 × 54 = 18,144

And Product of LCM and HCF = 3024 × 6 = 18,144

Hence, LCM × HCF = product of the 336 and 54.

3. Find the LCM and HCF of the following integers by applying the prime factorisation method.

(i) 12, 15 and 21

(ii) 17, 23 and 29

(iii) 8, 9 and 25

Solutions:

(i) 12, 15 and 21

Writing the product of prime factors for all the three numbers, we get,

12=2×2×3

15=5×3

21=7×3

Therefore,

HCF(12,15,21) = 3

LCM(12,15,21) = 2 × 2 × 3 × 5 × 7 = 420

(ii) 17, 23 and 29

Writing the product of prime factors for all the three numbers, we get,

17=17×1

23=23×1

29=29×1

Therefore,

HCF(17,23,29) = 1

LCM(17,23,29) = 17 × 23 × 29 = 11339

(iii) 8, 9 and 25

Writing the product of prime factors for all the three numbers, we get,

8=2×2×2×1

9=3×3×1

25=5×5×1

Therefore,

HCF(8,9,25)=1

LCM(8,9,25) = 2×2×2×3×3×5×5 = 1800

4. Given that HCF (306, 657) = 9, find LCM (306, 657).

Solution: As we know that,

HCF×LCM=Product of the two given numbers

Therefore,

9 × LCM = 306 × 657

LCM = (306×657)/9 = 22338

Hence, LCM(306,657) = 22338

5. Check whether 6ⁿ can end with the digit 0 for any natural number n.

Solution: If the number 6ⁿ ends with the digit zero (0), then it should be divisible by 5, as we know any number with unit place as 0 or 5 is divisible by 5.

Prime factorization of 6ⁿ = (2×3)ⁿ

Therefore, the prime factorization of 6ⁿ doesn’t contain prime number 5.

Hence, it is clear that for any natural number n, ⁶ⁿ is not divisible by 5 and thus it proves that 6ⁿ cannot end with the digit 0 for any natural number n.

6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

Solution: By the definition of composite number, we know, if a number is composite, then it means it has factors other than 1 and itself. Therefore, for the given expression;

7 × 11 × 13 + 13

Taking 13 as common factor, we get,

=13(7×11×1+1) = 13(77+1) = 13×78 = 13×3×2×13

Hence, 7 × 11 × 13 + 13 is a composite number.

Now let’s take the other number,

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

Taking 5 as a common factor, we get,

=5(7×6×4×3×2×1+1) = 5(1008+1) = 5×1009

Hence, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is a composite number.

7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?

Solution: Since, Both Sonia and Ravi move in the same direction and at the same time, the method to find the time when they will be meeting again at the starting point is LCM of 18 and 12.

Therefore, LCM(18,12) = 2×3×3×2×1=36

Hence, Sonia and Ravi will meet again at the starting point after 36 minutes.

Exercise 1.3 Page: 14

1. Prove that √5 is irrational.

Solutions: Let us assume, that √5 is rational number.

i.e. √5 = x/y (where, x and y are co-primes)

y√5= x

Squaring both the sides, we get,

(y√5)² = x²

⇒5y² = x²……………………………….. (1)

Thus, x² is divisible by 5, so x is also divisible by 5.

Let us say, x = 5k, for some value of k and substituting the value of x in equation (1), we get,

5y² = (5k)²

⇒y² = 5k²

is divisible by 5 it means y is divisible by 5.

Clearly, x and y are not co-primes. Thus, our assumption about √5 is rational is incorrect.

Hence, √5 is an irrational number.

2. Prove that 3 + 2√5 + is irrational.

Solutions: Let us assume 3 + 2√5 is rational.

Then we can find co-prime x and y (y ≠ 0) such that 3 + 2√5 = x/y

Rearranging, we get,

Since, x and y are integers, thus,

is a rational number.

Therefore, √5 is also a rational number. But this contradicts the fact that √5 is irrational.

So, we conclude that 3 + 2√5 is irrational.

3. Prove that the following are irrationals:

(i) 1/√2

(ii) 7√5

(iii) 6 + √2

Solutions:

(i) 1/√2

Let us assume 1/√2 is rational.

Then we can find co-prime x and y (y ≠ 0) such that 1/√2 = x/y

Rearranging, we get,

√2 = y/x

Since, x and y are integers, thus, √2 is a rational number, which contradicts the fact that √2 is irrational.

Hence, we can conclude that 1/√2 is irrational.

(ii) 7√5

Let us assume 7√5 is a rational number.

Then we can find co-prime a and b (b ≠ 0) such that 7√5 = x/y

Rearranging, we get,

√5 = x/7y

Since, x and y are integers, thus, √5 is a rational number, which contradicts the fact that √5 is irrational.

Hence, we can conclude that 7√5 is irrational.

(iii) 6 +√2

Let us assume 6 +√2 is a rational number.

Then we can find co-primes x and y (y ≠ 0) such that 6 +√2 = x/y⋅

Rearranging, we get,

√2 = (x/y) – 6

Since, x and y are integers, thus (x/y) – 6 is a rational number and therefore, √2 is rational. This contradicts the fact that √2 is an irrational number.

Hence, we can conclude that 6 +√2 is irrational.

Exercise 1.4 Page: 17

1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:

(i) 13/3125 (ii) 17/8 (iii) 64/455 (iv) 15/1600 (v) 29/343 (vi) 23/(2³5²) (vii) 129/(2²5⁷7⁵) (viii) 6/15 (ix) 35/50 (x) 77/210

Solutions:

Note: If the denominator has only factors of 2 and 5 or in the form of 2^m ×5ⁿ then it has terminating decimal expansion.

If the denominator has factors other than 2 and 5 then it has a non-terminating decimal expansion.

(i) 13/3125

Factorizing the denominator, we get,

3125 = 5 × 5 × 5 × 5 × 5 = 5⁵

Since, the denominator has only 5 as its factor, 13/3125 has a terminating decimal expansion.

(ii) 17/8

Factorizing the denominator, we get,

8 = 2×2×2 = 2³

Since, the denominator has only 2 as its factor, 17/8 has a terminating decimal expansion.

(iii) 64/455

Factorizing the denominator, we get,

455 = 5×7×13

Since, the denominator is not in the form of 2^m × 5ⁿ, thus 64/455 has a non-terminating decimal expansion.

(iv) 15/ 1600

Factorizing the denominator, we get,

1600 = 2⁶5²

Since, the denominator is in the form of 2^m × 5ⁿ, thus 15/1600 has a terminating decimal expansion.

(v) 29/343

Factorizing the denominator, we get,

343 = 7×7×7 = 7³ Since, the denominator is not in the form of 2^m × 5ⁿ thus 29/343 has a non-terminating decimal expansion.

(vi)23/(2³5²)

Clearly, the denominator is in the form of 2^m × 5ⁿ.

Hence, 23/ (2³5²) has a terminating decimal expansion.

(vii) 129/(2²5⁷7⁵)

As you can see, the denominator is not in the form of 2^m × 5ⁿ.

Hence, 129/ (2²5⁷7⁵) has a non-terminating decimal expansion.

(viii) 6/15

6/15 = 2/5

Since, the denominator has only 5 as its factor, thus, 6/15 has a terminating decimal expansion.

(ix) 35/50

35/50 = 7/10

Factorising the denominator, we get,

10 = 2 5

Since, the denominator is in the form of 2^m × 5ⁿ thus, 35/50 has a terminating decimal expansion.

(x) 77/210

77/210 = (7× 11)/ (30 × 7) = 11/30

Factorising the denominator, we get,

30 = 2 × 3 × 5

As you can see, the denominator is not in the form of 2^m × 5ⁿ .Hence, 77/210 has a non-terminating decimal expansion.

2. Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.

Solutions:

(i) 13/3125

13/3125 = 0.00416

(ii) 17/8

17/8 = 2.125

(iii) 64/455 has a Non terminating decimal expansion

(iv)15/ 1600

15/1600 = 0.009375

(v) 29/ 343 has a Non terminating decimal expansion

(vi)23/ (2³5²) = 23/(8×25)= 23/200

23/ (2³5²) = 0.115

(vii) 129/ (2²5⁷7⁵) has a Non terminating decimal expansion

(viii) 6/15 = 2/5

(ix) 35/50 = 7/10

35/50 = 0.7

(x) 77/210 has a non-terminating decimal expansion.

3. The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form, p q what can you say about the prime factors of q?

(i) 43.123456789

(ii) 0.120120012000120000. . .

Solutions:

(i) 43.123456789

Since it has a terminating decimal expansion, it is a rational number in the form of p/q and q has factors of 2 and 5 only.

(ii) 0.120120012000120000. . .

Since, it has non-terminating and non- repeating decimal expansion, it is an irrational number.

Since it has non-terminating but repeating decimal expansion, it is a rational number in the form of p/q and q has factors other than 2 and 5.

2. Introduction to Trigonometry

Exercise 8.1 Page: 181

1. In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:

(i) sin A, cos A

(ii) sin C, cos C

Solution:

In a given triangle ABC, right angled at B = ∠B = 90°

Given: AB = 24 cm and BC = 7 cm

According to the Pythagoras Theorem,

In a right- angled triangle, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides.

By applying Pythagoras theorem, we get

AC²=AB²+BC²

AC² = (24)²+7²

AC² = (576+49)

AC² = 625cm²

AC = √625 = 25

Therefore, AC = 25 cm

(i) To find Sin (A), Cos (A)

We know that sine (or) Sin function is the equal to the ratio of length of the opposite side to the hypotenuse side. So it becomes

Sin (A) = Opposite side /Hypotenuse = BC/AC = 7/25

Cosine or Cos function is equal to the ratio of the length of the adjacent side to the hypotenuse side and it becomes,

Cos (A) = Adjacent side/Hypotenuse = AB/AC = 24/25

(ii) To find Sin (C), Cos (C)

Sin (C) = AB/AC = 24/25

Cos (C) = BC/AC = 7/25

2. In Fig. 8.13, find tan P – cot R

Solution:

In the given triangle PQR, the given triangle is right angled at Q and the given measures are:

PR = 13cm,

PQ = 12cm

Since the given triangle is right angled triangle, to find the side QR, apply the Pythagorean theorem

According to Pythagorean theorem,

In a right- angled triangle, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides.

PR² = QR² + PQ²

Substitute the values of PR and PQ

13² = QR²+12²

169 = QR²+144

Therefore, QR² = 169−144

QR² = 25

QR = √25 = 5

Therefore, the side QR = 5 cm

To find tan P – cot R:

According to the trigonometric ratio, the tangent function is equal to the ratio of the length of the opposite side to the adjacent sides, the value of tan (P) becomes

tan (P) = Opposite side /Adjacent side = QR/PQ = 5/12

Since cot function is the reciprocal of the tan function, the ratio of cot function becomes,

Cot (R) = Adjacent side/Opposite side = QR/PQ = 5/12

Therefore,

tan (P) – cot (R) = 5/12 – 5/12 = 0

Therefore, tan(P) – cot(R) = 0

3. If sin A = 3/4, Calculate cos A and tan A.

Solution:

Let us assume a right angled triangle ABC, right angled at B

Given: Sin A = 3/4

We know that, Sin function is the equal to the ratio of length of the opposite side to the hypotenuse side.

Therefore, Sin A = Opposite side /Hypotenuse= 3/4

Let BC be 3k and AC will be 4k

where k is a positive real number.

According to the Pythagoras theorem, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides of a right angle triangle and we get,

AC²=AB² + BC²

Substitute the value of AC and BC

(4k)²=AB² + (3k)²

16k²−9k² =AB²

AB²=7k²

Therefore, AB = √7k

Now, we have to find the value of cos A and tan A

We know that,

Cos (A) = Adjacent side/Hypotenuse

Substitute the value of AB and AC and cancel the constant k in both numerator and denominator, we get

AB/AC = √7k/4k = √7/4

Therefore, cos (A) = √7/4

tan(A) = Opposite side/Adjacent side

Substitute the Value of BC and AB and cancel the constant k in both numerator and denominator, we get,

BC/AB = 3k/√7k = 3/√7

Therefore, tan A = 3/√7

4. Given 15 cot A = 8, find sin A and sec A.

Solution:

Let us assume a right angled triangle ABC, right angled at B

Given: 15 cot A = 8

So, Cot A = 8/15

We know that, cot function is the equal to the ratio of length of the adjacent side to the opposite side.

Therefore, cot A = Adjacent side/Opposite side = AB/BC = 8/15

Let AB be 8k and BC will be 15k

Where, k is a positive real number.

According to the Pythagoras theorem, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides of a right angle triangle and we get,

AC²=AB² + BC²

Substitute the value of AB and BC

AC²= (8k)² + (15k)²

AC²= 64k² + 225k²

AC²= 289k²

Therefore, AC = 17k

Now, we have to find the value of sin A and sec A

We know that,

Sin (A) = Opposite side /Hypotenuse

Substitute the value of BC and AC and cancel the constant k in both numerator and denominator, we get

Sin A = BC/AC = 15k/17k = 15/17

Therefore, sin A = 15/17

Since secant or sec function is the reciprocal of the cos function which is equal to the ratio of the length of the hypotenuse side to the adjacent side.

Sec (A) = Hypotenuse/Adjacent side

Substitute the Value of BC and AB and cancel the constant k in both numerator and denominator, we get,

AC/AB = 17k/8k = 17/8

Therefore sec (A) = 17/8

5. Given sec θ = 13/12 Calculate all other trigonometric ratios

Solution:

We know that sec function is the reciprocal of the cos function which is equal to the ratio of the length of the hypotenuse side to the adjacent side

Let us assume a right angled triangle ABC, right angled at B

sec θ =13/12 = Hypotenuse/Adjacent side = AC/AB

Let AC be 13k and AB will be 12k

Where, k is a positive real number.

According to the Pythagoras theorem, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides of a right angle triangle and we get,

AC²=AB² + BC²

Substitute the value of AB and AC

(13k)²= (12k)² + BC²

169k²= 144k² + BC²

169k²= 144k² + BC²

BC^{2 =} 169k² – 144k²

BC²= 25k²

Therefore, BC = 5k

Now, substitute the corresponding values in all other trigonometric ratios

So,

Sin θ = Opposite Side/Hypotenuse = BC/AC = 5/13

Cos θ = Adjacent Side/Hypotenuse = AB/AC = 12/13

tan θ = Opposite Side/Adjacent Side = BC/AB = 5/12

Cosec θ = Hypotenuse/Opposite Side = AC/BC = 13/5

cot θ = Adjacent Side/Opposite Side = AB/BC = 12/5

6. If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠ A = ∠ B.

Solution:

Let us assume the triangle ABC in which CD⊥AB

Give that the angles A and B are acute angles, such that

Cos (A) = cos (B)

As per the angles taken, the cos ratio is written as

AD/AC = BD/BC

Now, interchange the terms, we get

AD/BD = AC/BC

Let take a constant value

AD/BD = AC/BC = k

Now consider the equation as

AD = k BD …(1)

AC = k BC …(2)

By applying Pythagoras theorem in △CAD and △CBD we get,

CD² = BC² – BD² … (3)

CD² =AC² −AD² ….(4)

From the equations (3) and (4) we get,

AC²−AD² = BC²−BD²

Now substitute the equations (1) and (2) in (3) and (4)

K²(BC²−BD²)=(BC²−BD²) k²=1

Putting this value in equation, we obtain

AC = BC

∠A=∠B (Angles opposite to equal side are equal-isosceles triangle)

7. If cot θ = 7/8, evaluate :

(i) (1 + sin θ)(1 – sin θ)/(1+cos θ)(1-cos θ)

(ii) cot² θ

Solution:

Let us assume a △ABC in which ∠B = 90° and ∠C = θ

Given:

cot θ = BC/AB = 7/8

Let BC = 7k and AB = 8k, where k is a positive real number

According to Pythagoras theorem in △ABC we get.

AC² = AB²+BC²

AC² = (8k)²+(7k)²

AC² = 64k²+49k²

AC² = 113k²

AC = √113 k

According to the sine and cos function ratios, it is written as

sin θ = AB/AC = Opposite Side/Hypotenuse = 8k/√113 k = 8/√113 and

cos θ = Adjacent Side/Hypotenuse = BC/AC = 7k/√113 k = 7/√113

Now apply the values of sin function and cos function:

8. If 3 cot A = 4, check whether (1-tan² A)/(1+tan² A) = cos² A – sin ² A or not.

Solution:

Let △ABC in which ∠B=90°

We know that, cot function is the reciprocal of tan function and it is written as

cot(A) = AB/BC = 4/3

Let AB = 4k an BC =3k, where k is a positive real number.

According to the Pythagorean theorem,

AC²=AB²+BC²

AC²=(4k)²+(3k)²

AC²=16k²+9k²

AC²=25k²

AC=5k

Now, apply the values corresponding to the ratios

tan(A) = BC/AB = 3/4

sin (A) = BC/AC = 3/5

cos (A) = AB/AC = 4/5

Now compare the left hand side(LHS) with right hand side(RHS)

Since, both the LHS and RHS = 7/25

R.H.S. =L.H.S.

Hence, (1-tan² A)/(1+tan² A) = cos² A – sin ² A  is proved

9. In triangle ABC, right-angled at B, if tan A = 1/√3 find the value of:

(i) sin A cos C + cos A sin C

(ii) cos A cos C – sin A sin C

Solution:

Let ΔABC in which ∠B=90°

tan A = BC/AB = 1/√3

Let BC = 1k and AB = √3 k,

Where k is the positive real number of the problem

By Pythagoras theorem in ΔABC we get:

AC²=AB²+BC²

AC²=(√3 k)²+(k)²

AC²=3k²+k²

AC²=4k²

AC = 2k

Now find the values of cos A, Sin A

Sin A = BC/AC = 1/2

Cos A = AB/AC = √3/2

Then find the values of cos C and sin C

Sin C = AB/AC = √3/2

Cos C = BC/AC = 1/2

Now, substitute the values in the given problem

(i) sin A cos C + cos A sin C = (1/2) ×(1/2 )+ √3/2 ×√3/2 = 1/4 + 3/4 = 1

(ii) cos A cos C – sin A sin C = (√3/2 )(1/2) – (1/2) (√3/2 ) = 0

10. In ∆ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P

Solution:

In a given triangle PQR, right angled at Q, the following measures are

PQ = 5 cm

PR + QR = 25 cm

Now let us assume, QR = x

PR = 25-QR

PR = 25- x

According to the Pythagorean Theorem,

PR² = PQ² + QR²

Substitute the value of PR as x

(25- x) ² = 5² + x²

25² + x² – 50x = 25 + x²

625 + x²-50x -25 – x² = 0

-50x = -600

x= -600/-50

x = 12 = QR

Now, find the value of PR

PR = 25- QR

Substitute the value of QR

PR = 25-12

PR = 13

Now, substitute the value to the given problem

(1) sin p = Opposite Side/Hypotenuse = QR/PR = 12/13

(2) Cos p = Adjacent Side/Hypotenuse = PQ/PR = 5/13

(3) tan p =Opposite Side/Adjacent side = QR/PQ = 12/5

11. State whether the following are true or false. Justify your answer.

(i) The value of tan A is always less than 1.

(ii) sec A = 12/5 for some value of angle A.

(iii)cos A is the abbreviation used for the cosecant of angle A.

(iv) cot A is the product of cot and A.

(v) sin θ = 4/3 for some angle θ.

Solution:

(i) The value of tan A is always less than 1.

Answer: False

Proof: In ΔMNC in which ∠N = 90∘,

MN = 3, NC = 4 and MC = 5

Value of tan M = 4/3 which is greater than.

The triangle can be formed with sides equal to 3, 4 and hypotenuse = 5 as it will follow the Pythagoras theorem.

MC²=MN²+NC²

5²=3²+4²

25=9+16

25 = 25

(ii) sec A = 12/5 for some value of angle A

Answer: True

Justification: Let a ΔMNC in which ∠N = 90º,

MC=12k and MB=5k, where k is a positive real number.

By Pythagoras theorem we get,

MC²=MN²+NC²

(12k)²=(5k)²+NC²

NC²+25k²=144k²

NC²=119k²

Such a triangle is possible as it will follow the Pythagoras theorem.

(iii) cos A is the abbreviation used for the cosecant of angle A.

Answer: False

Justification: Abbreviation used for cosecant of angle M is cosec M. cos M is the abbreviation used for cosine of angle M.

(iv) cot A is the product of cot and A.

Answer: False

Justification: cot M is not the product of cot and M. It is the cotangent of ∠M.

(v) sin θ = 4/3 for some angle θ.

Answer: False

Justification: sin θ = Height/Hypotenuse

We know that in a right angled triangle, Hypotenuse is the longest side.

∴ sin θ will always less than 1 and it can never be 4/3 for any value of θ.

Exercise 8.2 Page: 187

1. Evaluate the following:

(i) sin 60° cos 30° + sin 30° cos 60°

(ii) 2 tan² 45° + cos² 30° – sin² 60

Solution:

(i) sin 60° cos 30° + sin 30° cos 60°

First, find the values of the given trigonometric ratios

sin 30° = 1/2

cos 30° = √3/2

sin 60° = 3/2

cos 60°= 1/2

Now, substitute the values in the given problem

sin 60° cos 30° + sin 30° cos 60° = √3/2 ×√3/2 + (1/2) ×(1/2 ) = 3/4+1/4 = 4/4 =

(ii) 2 tan² 45° + cos² 30° – sin² 60

We know that, the values of the trigonometric ratios are:

sin 60° = √3/2

cos 30° = √3/2

tan 45° = 1

Substitute the values in the given problem

2 tan² 45° + cos² 30° – sin² 60 = 2(1)² + (√3/2)²-(√3/2)²

2 tan² 45° + cos² 30° – sin² 60 = 2 + 0

2 tan² 45° + cos² 30° – sin² 60 = 2

(iii) cos 45°/(sec 30°+cosec 30°)

We know that,

cos 45° = 1/√2

sec 30° = 2/√3

cosec 30° = 2

Substitute the values, we get

Now, multiply both the numerator and denominator by √2 , we get

Therefore, cos 45°/(sec 30°+cosec 30°) = (3√2 – √6)/8

We know that,

sin 30° = 1/2

tan 45° = 1

cosec 60° = 2/√3

sec 30° = 2/√3

cos 60° = 1/2

cot 45° = 1

Substitute the values in the given problem, we get

We know that,

cos 60° = 1/2

sec 30° = 2/√3

tan 45° = 1

sin 30° = 1/2

cos 30° = √3/2

Now, substitute the values in the given problem, we get

(5cos²60° + 4sec²30° – tan²45°)/(sin² 30° + cos² 30°)

= 5(1/2)²+4(2/√3)²-1²/(1/2)²+(√3/2)²

 = (5/4+16/3-1)/(1/4+3/4)

= (15+64-12)/12/(4/4)

= 67/12

2. Choose the correct option and justify your choice :
(i) 2tan 30°/1+tan²30° =
(A) sin 60°            (B) cos 60°          (C) tan 60°            (D) sin 30°
(ii) 1-tan²45°/1+tan²45° =
(A) tan 90°            (B) 1                    (C) sin 45°            (D) 0
(iii)  sin 2A = 2 sin A is true when A =
(A) 0°                   (B) 30°                  (C) 45°                 (D) 60°

(iv) 2tan30°/1-tan²30° =
(A) cos 60°          (B) sin 60°             (C) tan 60°           (D) sin 30°

Solution:

(i) (A) is correct.

Substitute the of tan 30° in the given equation

tan 30° = 1/√3

2tan 30°/1+tan²30° = 2(1/√3)/1+(1/√3)²

= (2/√3)/(1+1/3) = (2/√3)/(4/3)

= 6/4√3 = √3/2 = sin 60°

The obtained solution is equivalent to the trigonometric ratio sin 60°

(ii) (D) is correct.

Substitute the of tan 45° in the given equation

tan 45° = 1

1-tan²45°/1+tan²45° = (1-1²)/(1+1²)

= 0/2 = 0

The solution of the above equation is 0.

(iii) (A) is correct.

To find the value of A, substitute the degree given in the options one by one

sin 2A = 2 sin A is true when A = 0°

As sin 2A = sin 0° = 0

2 sin A = 2 sin 0° = 2 × 0 = 0

or,

Apply the sin 2A formula, to find the degree value

sin 2A = 2sin A cos A

⇒2sin A cos A = 2 sin A

⇒ 2cos A = 2 ⇒ cos A = 1

Now, we have to check, to get the solution as 1, which degree value has to be applied.

When 0 degree is applied to cos value, i.e., cos 0 =1

Therefore, ⇒ A = 0°

(iv) (C) is correct.

Substitute the of tan 30° in the given equation

tan 30° = 1/√3

2tan30°/1-tan²30° =  2(1/√3)/1-(1/√3)²

= (2/√3)/(1-1/3) = (2/√3)/(2/3) = √3 = tan 60°

The value of the given equation is equivalent to tan 60°.

3. If tan (A + B) = √3 and tan (A – B) = 1/√3 ,0° < A + B ≤ 90°; A > B, find A and B.

Solution:

tan (A + B) = √3

Since √3 = tan 60°

Now substitute the degree value

⇒ tan (A + B) = tan 60°

(A + B) = 60° … (i)

The above equation is assumed as equation (i)

tan (A – B) = 1/√3

Since 1/√3 = tan 30°

Now substitute the degree value

⇒ tan (A – B) = tan 30°

(A – B) = 30° … equation (ii)

Now add the equation (i) and (ii), we get

A + B + A – B = 60° + 30°

Cancel the terms B

2A = 90°

A= 45°

Now, substitute the value of A in equation (i) to find the value of B

45° + B = 60°

B = 60° – 45°

B = 15°

Therefore A = 45° and B = 15°

4. State whether the following are true or false. Justify your answer.

(i) sin (A + B) = sin A + sin B.

(ii) The value of sin θ increases as θ increases.

(iii) The value of cos θ increases as θ increases.

(iv) sin θ = cos θ for all values of θ.

(v) cot A is not defined for A = 0°.

Solution:

(i) False.

Justification:

Let us take A = 30° and B = 60°, then

Substitute the values in the sin (A + B) formula, we get

sin (A + B) = sin (30° + 60°) = sin 90° = 1 and,

sin A + sin B = sin 30° + sin 60°

= 1/2 + √3/2 = 1+√3/2

Since the values obtained are not equal, the solution is false.

(ii) True.

Justification:

According to the values obtained as per the unit circle, the values of sin are:

sin 0° = 0

sin 30° = 1/2

sin 45° = 1/√2

sin 60° = √3/2

sin 90° = 1

Thus the value of sin θ increases as θ increases. Hence, the statement is true

(iii) False.

According to the values obtained as per the unit circle, the values of cos are:

cos 0° = 1

cos 30° = √3/2

cos 45° = 1/√2

cos 60° = 1/2

cos 90° = 0

Thus, the value of cos θ decreases as θ increases. So, the statement given above is false.

(iv) False

sin θ = cos θ, when a right triangle has 2 angles of (π/4). Therefore, the above statement is false.

(v) True.

Since cot function is the reciprocal of the tan function, it is also written as:

cot A = cos A/sin A

Now substitute A = 0°

cot 0° = cos 0°/sin 0° = 1/0 = undefined.

Hence, it is true

Exercise 8.3 Page: 189

1. Evaluate :

(i) sin 18°/cos 72°        

(ii) tan 26°/cot 64°      

(iii)  cos 48° – sin 42°      

(iv)  cosec 31° – sec 59°

Solution:

(i) sin 18°/cos 72°

To simplify this, convert the sin function into cos function

We know that, 18° is written as 90° – 18°, which is equal to the cos 72°.

= sin (90° – 18°) /cos 72°

Substitute the value, to simplify this equation

= cos 72° /cos 72° = 1

(ii) tan 26°/cot 64°

To simplify this, convert the tan function into cot function

We know that, 26° is written as 90° – 36°, which is equal to the cot 64°.

= tan (90° – 36°)/cot 64°

Substitute the value, to simplify this equation

= cot 64°/cot 64° = 1

(iii) cos 48° – sin 42°

To simplify this, convert the cos function into sin function

We know that, 48° is written as 90° – 42°, which is equal to the sin 42°.

= cos (90° – 42°) – sin 42°

Substitute the value, to simplify this equation

= sin 42° – sin 42° = 0

(iv) cosec 31° – sec 59°

To simplify this, convert the cosec function into sec function

We know that, 31° is written as 90° – 59°, which is equal to the sec 59°

= cosec (90° – 59°) – sec 59°

Substitute the value, to simplify this equation

= sec 59° – sec 59° = 0

2.  Show that:

(i) tan 48° tan 23° tan 42° tan 67° = 1

(ii) cos 38° cos 52° – sin 38° sin 52° = 0

Solution:

(i) tan 48° tan 23° tan 42° tan 67°

Simplify the given problem by converting some of the tan functions to the cot functions

We know that, tan 48° = tan (90° – 42°) = cot 42°

tan 23° = tan (90° – 67°) = cot 67°

= tan (90° – 42°) tan (90° – 67°) tan 42° tan 67°

Substitute the values

= cot 42° cot 67° tan 42° tan 67°

= (cot 42° tan 42°) (cot 67° tan 67°) = 1×1 = 1

(ii) cos 38° cos 52° – sin 38° sin 52°

Simplify the given problem by converting some of the cos functions to the sin functions

We know that,

cos 38° = cos (90° – 52°) = sin 52°

cos 52°= cos (90°-38°) = sin 38°

= cos (90° – 52°) cos (90°-38°) – sin 38° sin 52°

Substitute the values

= sin 52° sin 38° – sin 38° sin 52° = 0

3. If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.

Solution:

tan 2A = cot (A- 18°)

We know that tan 2A = cot (90° – 2A)

Substitute the above equation in the given problem

⇒ cot (90° – 2A) = cot (A -18°)

Now, equate the angles,

⇒ 90° – 2A = A- 18° ⇒ 108° = 3A

A = 108° / 3

Therefore, the value of A = 36°

4.  If tan A = cot B, prove that A + B = 90°.

Solution:

tan A = cot B

We know that cot B = tan (90° – B)

To prove A + B = 90°, substitute the above equation in the given problem

tan A = tan (90° – B)

A = 90° – B

A + B = 90°

Hence Proved.

5. If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.

Solution:

sec 4A = cosec (A – 20°)

We know that sec 4A = cosec (90° – 4A)

To find the value of A, substitute the above equation in the given problem

cosec (90° – 4A) = cosec (A – 20°)

Now, equate the angles

90° – 4A= A- 20°

110° = 5A

A = 110°/ 5 = 22°

Therefore, the value of A = 22°

6. If A, B and C are interior angles of a triangle ABC, then show that

    sin (B+C/2) = cos A/2

Solution:

We know that, for a given triangle, sum of all the interior angles of a triangle is equal to 180°

A + B + C = 180° ….(1)

To find the value of (B+ C)/2, simplify the equation (1)

⇒ B + C = 180° – A

⇒ (B+C)/2 = (180°-A)/2

⇒ (B+C)/2 = (90°-A/2)

Now, multiply both sides by sin functions, we get

⇒ sin (B+C)/2 = sin (90°-A/2)

Since sin (90°-A/2) = cos A/2, the above equation is equal to

sin (B+C)/2 = cos A/2

Hence proved.

7. Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

Solution:

Given:

sin 67° + cos 75°

In term of sin as cos function and cos as sin function, it can be written as follows

sin 67° = sin (90° – 23°)

cos 75° = cos (90° – 15°)

So, sin 67° + cos 75° = sin (90° – 23°) + cos (90° – 15°)

Now, simplify the above equation

= cos 23° + sin 15°

Therefore, sin 67° + cos 75° is also expressed as cos 23° + sin 15°

Exercise 8.4 Page: 193

1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

Solution:

To convert the given trigonometric ratios in terms of cot functions, use trigonometric formulas

We know that,

cosec²A – cot²A = 1

cosec²A = 1 + cot²A

Since cosec function is the inverse of sin function, it is written as

1/sin²A = 1 + cot²A

Now, rearrange the terms, it becomes

sin²A = 1/(1+cot²A)

Now, take square roots on both sides, we get

sin A = ±1/(√(1+cot²A)

The above equation defines the sin function in terms of cot function

Now, to express sec function in terms of cot function, use this formula

sin²A = 1/ (1+cot²A)

Now, represent the sin function as cos function

1 – cos²A = 1/ (1+cot²A)

Rearrange the terms,

cos²A = 1 – 1/(1+cot²A)

⇒cos²A = (1-1+cot²A)/(1+cot²A)

Since sec function is the inverse of cos function,

⇒ 1/sec²A = cot²A/(1+cot²A)

Take the reciprocal and square roots on both sides, we get

⇒ sec A = ±√ (1+cot²A)/cotA

Now, to express tan function in terms of cot function

tan A = sin A/cos A and cot A = cos A/sin A

Since cot function is the inverse of tan function, it is rewritten as

tan A = 1/cot A

2. Write all the other trigonometric ratios of ∠A in terms of sec A.

Solution:

Cos A function in terms of sec A:

sec A = 1/cos A

⇒ cos A = 1/sec A

sec A function in terms of sec A:

cos²A + sin²A = 1

Rearrange the terms

sin²A = 1 – cos²A

sin²A = 1 – (1/sec²A)

sin²A = (sec²A-1)/sec²A

sin A = ± √(sec²A-1)/sec A

cosec A function in terms of sec A:

sin A = 1/cosec A

⇒cosec A = 1/sin A

cosec A = ± sec A/√(sec²A-1)

Now, tan A function in terms of sec A:

sec²A – tan²A = 1

Rearrange the terms

⇒ tan²A = sec²A – 1

tan A = √(sec²A – 1)

cot A function in terms of sec A:

tan A = 1/cot A

⇒ cot A = 1/tan A

cot A = ±1/√(sec²A – 1)

3. Evaluate:

(i) (sin²63° + sin²27°)/(cos²17° + cos²73°)
(ii)  sin 25° cos 65° + cos 25° sin 65°

Solution:

(i) (sin²63° + sin²27°)/(cos²17° + cos²73°)

To simplify this, convert some of the sin functions into cos functions and cos function into sin function and it becomes,

= [sin²(90°-27°) + sin²27°] / [cos²(90°-73°) + cos²73°)]

= (cos²27° + sin²27°)/(sin²27° + cos²73°)

= 1/1 =1                       (since sin²A + cos²A = 1)

Therefore, (sin²63° + sin²27°)/(cos²17° + cos²73°) = 1

(ii) sin 25° cos 65° + cos 25° sin 65°

To simplify this, convert some of the sin functions into cos functions and cos function into sin function and it becomes,

= sin(90°-25°) cos 65° + cos (90°-65°) sin 65°

= cos 65° cos 65° + sin 65° sin 65°

= cos²65° + sin²65° = 1 (since sin²A + cos²A = 1)

Therefore, sin 25° cos 65° + cos 25° sin 65° = 1

4. Choose the correct option. Justify your choice.
(i) 9 sec²A – 9 tan²A =
(A) 1                 (B) 9              (C) 8                (D) 0
(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)
(A) 0                 (B) 1              (C) 2                (D) – 1
(iii) (sec A + tan A) (1 – sin A) =
(A) sec A           (B) sin A        (C) cosec A      (D) cos A

(iv) 1+tan²A/1+cot²A = 

      (A) sec² A                 (B) -1              (C) cot²A                (D) tan²A

Solution:

(i) (B) is correct.

Justification:

Take 9 outside, and it becomes

9 sec²A – 9 tan²A

= 9 (sec²A – tan²A)

= 9×1 = 9             (∵ sec2 A – tan2 A = 1)

Therefore, 9 sec²A – 9 tan²A = 9

(ii) (C) is correct

Justification:

(1 + tan θ + sec θ) (1 + cot θ – cosec θ)

We know that, tan θ = sin θ/cos θ

sec θ = 1/ cos θ

cot θ = cos θ/sin θ

cosec θ = 1/sin θ

Now, substitute the above values in the given problem, we get

= (1 + sin θ/cos θ + 1/ cos θ) (1 + cos θ/sin θ – 1/sin θ)

Simplify the above equation,

= (cos θ +sin θ+1)/cos θ × (sin θ+cos θ-1)/sin θ

= (cos θ+sin θ)²-1²/(cos θ sin θ)

= (cos²θ + sin²θ + 2cos θ sin θ -1)/(cos θ sin θ)

= (1+ 2cos θ sin θ -1)/(cos θ sin θ) (Since cos²θ + sin²θ = 1)

= (2cos θ sin θ)/(cos θ sin θ) = 2

Therefore, (1 + tan θ + sec θ) (1 + cot θ – cosec θ) =2

(iii) (D) is correct.

Justification:

We know that,

Sec A= 1/cos A

Tan A = sin A / cos A

Now, substitute the above values in the given problem, we get

(secA + tanA) (1 – sinA)

= (1/cos A + sin A/cos A) (1 – sinA)

= (1+sin A/cos A) (1 – sinA)

= (1 – sin²A)/cos A

= cos²A/cos A = cos A

Therefore, (secA + tanA) (1 – sinA) = cos A

(iv) (D) is correct.

Justification:

We know that,

tan²A =1/cot²A

Now, substitute this in the given problem, we get

1+tan²A/1+cot²A

= (1+1/cot²A)/1+cot²A

= (cot²A+1/cot²A)×(1/1+cot²A)

= 1/cot²A = tan²A

So, 1+tan²A/1+cot²A = tan²A

5. Prove the following identities, where the angles involved are acute angles for which the
expressions are defined.

(i) (cosec θ – cot θ)² = (1-cos θ)/(1+cos θ)

(ii) cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A

(iii) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ

     [Hint : Write the expression in terms of sin θ and cos θ]

(iv) (1 + sec A)/sec A = sin²A/(1-cos A)  

     [Hint : Simplify LHS and RHS separately]

(v) ( cos A–sin A+1)/( cos A +sin A–1) = cosec A + cot A, using the identity cosec²A = 1+cot²A.

(vii) (sin θ – 2sin³θ)/(2cos³θ-cos θ) = tan θ
(viii) (sin A + cosec A)² + (cos A + sec A)² = 7+tan²A+cot²A
(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)
[Hint : Simplify LHS and RHS separately]
(x) (1+tan²A/1+cot²A) = (1-tan A/1-cot A)² = tan²A

Solution:

(i) (cosec θ – cot θ)² = (1-cos θ)/(1+cos θ)

To prove this, first take the Left-Hand side (L.H.S) of the given equation, to prove the Right Hand Side (R.H.S)

L.H.S. = (cosec θ – cot θ)²

The above equation is in the form of (a-b)², and expand it

Since (a-b)² = a² + b² – 2ab

Here a = cosec θ and b = cot θ

= (cosec²θ + cot²θ – 2cosec θ cot θ)

Now, apply the corresponding inverse functions and equivalent ratios to simplify

= (1/sin²θ + cos²θ/sin²θ – 2cos θ/sin²θ)

= (1 + cos²θ – 2cos θ)/(1 – cos²θ)

= (1-cos θ)²/(1 – cosθ)(1+cos θ)

= (1-cos θ)/(1+cos θ) = R.H.S.

Therefore, (cosec θ – cot θ)² = (1-cos θ)/(1+cos θ)

Hence proved.

(ii)  (cos A/(1+sin A)) + ((1+sin A)/cos A) = 2 sec A

Now, take the L.H.S of the given equation.

L.H.S. = (cos A/(1+sin A)) + ((1+sin A)/cos A)

= [cos²A + (1+sin A)²]/(1+sin A)cos A

= (cos²A + sin²A + 1 + 2sin A)/(1+sin A) cos A

Since cos²A + sin²A = 1, we can write it as

= (1 + 1 + 2sin A)/(1+sin A) cos A

= (2+ 2sin A)/(1+sin A)cos A

= 2(1+sin A)/(1+sin A)cos A

= 2/cos A = 2 sec A = R.H.S.

L.H.S. = R.H.S.

(cos A/(1+sin A)) + ((1+sin A)/cos A) = 2 sec A

Hence proved.

(iii) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ

L.H.S. = tan θ/(1-cot θ) + cot θ/(1-tan θ)

We know that tan θ =sin θ/cos θ

cot θ = cos θ/sin θ

Now, substitute it in the given equation, to convert it in a simplified form

= [(sin θ/cos θ)/1-(cos θ/sin θ)] + [(cos θ/sin θ)/1-(sin θ/cos θ)]

= [(sin θ/cos θ)/(sin θ-cos θ)/sin θ] + [(cos θ/sin θ)/(cos θ-sin θ)/cos θ]

= sin²θ/[cos θ(sin θ-cos θ)] + cos²θ/[sin θ(cos θ-sin θ)]

= sin²θ/[cos θ(sin θ-cos θ)] – cos²θ/[sin θ(sin θ-cos θ)]

= 1/(sin θ-cos θ) [(sin²θ/cos θ) – (cos²θ/sin θ)]

= 1/(sin θ-cos θ) × [(sin³θ – cos³θ)/sin θ cos θ]

= [(sin θ-cos θ)(sin²θ+cos²θ+sin θ cos θ)]/[(sin θ-cos θ)sin θ cos θ]

= (1 + sin θ cos θ)/sin θ cos θ

= 1/sin θ cos θ + 1

= 1 + sec θ cosec θ = R.H.S.

Therefore, L.H.S. = R.H.S.

Hence proved

(iv)  (1 + sec A)/sec A = sin²A/(1-cos A)

First find the simplified form of L.H.S

L.H.S. = (1 + sec A)/sec A

Since secant function is the inverse function of cos function and it is written as

= (1 + 1/cos A)/1/cos A

= (cos A + 1)/cos A/1/cos A

Therefore, (1 + sec A)/sec A = cos A + 1

R.H.S. = sin²A/(1-cos A)

We know that sin²A = (1 – cos²A), we get

= (1 – cos²A)/(1-cos A)

= (1-cos A)(1+cos A)/(1-cos A)

Therefore, sin²A/(1-cos A)= cos A + 1

L.H.S. = R.H.S.

Hence proved

(v) (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A, using the identity cosec²A = 1+cot²A.

With the help of identity function, cosec²A = 1+cot²A, let us prove the above equation.

L.H.S. = (cos A–sin A+1)/(cos A+sin A–1)

Divide the numerator and denominator by sin A, we get

= (cos A–sin A+1)/sin A/(cos A+sin A–1)/sin A

We know that cos A/sin A = cot A and 1/sin A = cosec A

= (cot A – 1 + cosec A)/(cot A+ 1 – cosec A)

= (cot A – cosec²A + cot²A + cosec A)/(cot A+ 1 – cosec A) (using cosec²A – cot²A = 1

= [(cot A + cosec A) – (cosec²A – cot²A)]/(cot A+ 1 – cosec A)

= [(cot A + cosec A) – (cosec A + cot A)(cosec A – cot A)]/(1 – cosec A + cot A)

=  (cot A + cosec A)(1 – cosec A + cot A)/(1 – cosec A + cot A)

=  cot A + cosec A = R.H.S.

Therefore, (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A

Hence Proved

First divide the numerator and denominator of L.H.S. by cos A,

We know that 1/cos A = sec A and sin A/ cos A = tan A and it becomes,

= √(sec A+ tan A)/(sec A-tan A)

Now using rationalization, we get

= (sec A + tan A)/1

= sec A + tan A = R.H.S

Hence proved

(vii) (sin θ – 2sin³θ)/(2cos³θ-cos θ) = tan θ

L.H.S. = (sin θ – 2sin³θ)/(2cos³θ – cos θ)

Take sin θ as in numerator and cos θ in denominator as outside, it becomes

= [sin θ(1 – 2sin²θ)]/[cos θ(2cos²θ- 1)]

We know that sin²θ = 1-cos²θ

= sin θ[1 – 2(1-cos²θ)]/[cos θ(2cos²θ -1)]

= [sin θ(2cos²θ -1)]/[cos θ(2cos²θ -1)]

= tan θ = R.H.S.

Hence proved

(viii) (sin A + cosec A)² + (cos A + sec A)² = 7+tan²A+cot²A

L.H.S. = (sin A + cosec A)² + (cos A + sec A)²

It is of the form (a+b)², expand it

(a+b)² =a² + b² +2ab

               = (sin²A + cosec²A + 2 sin A cosec A) + (cos²A + sec²A + 2 cos A sec A)

= (sin²A + cos²A) + 2 sin A(1/sin A) + 2 cos A(1/cos A) + 1 + tan²A + 1 + cot²A

= 1 + 2 + 2 + 2 + tan²A + cot²A

= 7+tan²A+cot²A = R.H.S.

Therefore, (sin A + cosec A)² + (cos A + sec A)² = 7+tan²A+cot²A

Hence proved.

(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)

First, find the simplified form of L.H.S

L.H.S. = (cosec A – sin A)(sec A – cos A)

Now, substitute the inverse and equivalent trigonometric ratio forms

= (1/sin A – sin A)(1/cos A – cos A)

= [(1-sin²A)/sin A][(1-cos²A)/cos A]

= (cos²A/sin A)×(sin²A/cos A)

= cos A sin A

Now, simplify the R.H.S

R.H.S. = 1/(tan A+cotA)

= 1/(sin A/cos A +cos A/sin A)

= 1/[(sin²A+cos²A)/sin A cos A]

= cos A sin A

L.H.S. = R.H.S.

(cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)

Hence proved

(x)  (1+tan²A/1+cot²A) = (1-tan A/1-cot A)² = tan²A

L.H.S. = (1+tan²A/1+cot²A)

Since cot function is the inverse of tan function,

= (1+tan²A/1+1/tan²A)

= 1+tan²A/[(1+tan²A)/tan²A]

Now cancel the 1+tan²A terms, we get

= tan²A

(1+tan²A/1+cot²A) = tan²A

Similarly,

(1-tan A/1-cot A)² = tan²A

Hence proved

3. Circles

Exercise: 10.1 (Page No: 209)

1. How many tangents can a circle have?

Answer:

There can be infinite tangents to a circle. A circle is made up of infinite points which are at an equal distance from a point. Since there are infinite points on the circumference of a circle, infinite tangents can be drawn from them.

2. Fill in the blanks:

(i) A tangent to a circle intersects it in …………… point(s).

(ii) A line intersecting a circle in two points is called a ………….

(iii) A circle can have …………… parallel tangents at the most.

(iv) The common point of a tangent to a circle and the circle is called …………

Answer:

(i) A tangent to a circle intersects it in one point(s).

(ii) A line intersecting a circle in two points is called a secant.

(iii) A circle can have two parallel tangents at the most.

(iv) The common point of a tangent to a circle and the circle is called the point of contact.

3. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at

a point Q so that OQ = 12 cm. Length PQ is :

(A) 12 cm

(B) 13 cm

(C) 8.5 cm

(D) √119 cm

Answer:

In the above figure, the line that is drawn from the centre of the given circle to the tangent PQ is perpendicular to PQ.

And so, OP ⊥ PQ

Using Pythagoras theorem in triangle ΔOPQ we get,

OQ² = OP²+PQ²

(12)² = 5²+PQ²

PQ² = 144-25

PQ² = 119

PQ = √119 cm

So, option D i.e. √119 cm is the length of PQ.

4. Draw a circle and two lines parallel to a given line such that one is a tangent and the

other, a secant to the circle.

Answer:

In the above figure, XY and AB are two the parallel lines. The line segment AB is the tangent at point C while the line segment XY is the secant.

Exercise: 10.2 (Page NO: 213)

In Q.1 to 3, choose the correct option and give justification.

1. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is

(A) 7 cm

(B) 12 cm

(C) 15 cm

(D) 24.5 cm

Answer:

First, draw a perpendicular from the center O of the triangle to a point P on the circle which is touching the tangent. This line will be perpendicular to the tangent of the circle.

So, OP is perpendicular to PQ i.e. OP ⊥ PQ

From the above figure, it is also seen that △OPQ is a right angled triangle.

It is given that

OQ = 25 cm and PQ = 24 cm

By using Pythagoras theorem in △OPQ,

OQ² = OP² +PQ²

(25)² = OP²+(24)²

OP² = 625-576

OP² = 49

OP = 7 cm

So, option A i.e. 7 cm is the radius of the given circle.

2. In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to

(A) 60°

(B) 70°

(C) 80°

(D) 90°

Answer:

From the question, it is clear that OP is the radius of the circle to the tangent PT and OQ is the radius to the tangents TQ.

So, OP ⊥ PT and TQ ⊥ OQ

∴∠OPT = ∠OQT = 90°

Now, in the quadrilateral POQT, we know that the sum of the interior angles is 360°

So, ∠PTQ+∠POQ+∠OPT+∠OQT = 360°

Now, by putting the respective values we get,

∠PTQ +90°+110°+90° = 360°

∠PTQ = 70°

So, ∠PTQ is 70° which is option B.

3. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠ POA is equal to

(A) 50°

(B) 60°

(C) 70°

(D) 80°

Answer:

First, draw the diagram according to the given statement.

Now, in the above diagram, OA is the radius to tangent PA and OB is the radius to tangent PB.

So, OA is perpendicular to PA and OB is perpendicular to PB i.e. OA ⊥ PA and OB ⊥ PB

So, ∠OBP = ∠OAP = 90°

Now, in the quadrilateral AOBP,

The sum of all the interior angles will be 360°

So, ∠AOB+∠OAP+∠OBP+∠APB = 360°

Putting their values, we get,

∠AOB + 260° = 360°

∠AOB = 100°

Now, consider the triangles △OPB and △OPA. Here,

AP = BP (Since the tangents from a point are always equal)

OA = OB (Which are the radii of the circle)

OP = OP (It is the common side)

Now, we can say that triangles OPB and OPA are similar using SSS congruency.

∴△OPB ≅ △OPA

So, ∠POB = ∠POA

∠AOB = ∠POA+∠POB

2 (∠POA) = ∠AOB

By putting the respective values, we get,

=>∠POA = 100°/2 = 50°

As angle ∠POA is 50° option A is the correct option.

4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Answer:

First, draw a circle and connect two points A and B such that AB becomes the diameter of the circle. Now, draw two tangents PQ and RS at points A and B respectively.

Now, both radii i.e. AO and OB are perpendicular to the tangents.

So, OB is perpendicular to RS and OA perpendicular to PQ

So, ∠OAP = ∠OAQ = ∠OBR = ∠OBS = 90°

From the above figure, angles OBR and OAQ are alternate interior angles.

Also, ∠OBR = ∠OAQ and ∠OBS = ∠OAP (Since they are also alternate interior angles)

So, it can be said that line PQ and the line RS will be parallel to each other. (Hence Proved).

5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the center.

Solution:

First, draw a circle with center O and draw a tangent AB which touches the radius of the circle at point P.

To Proof: PQ passes through point O.

Now, let us consider that PQ doesn’t pass through point O. Also, draw a CD parallel to AB through O. Here, CD is a straight line and AB is the tangent. Refer the diagram now.

From the above diagram, PQ intersects CD and AB at R and P respectively.

As, CD ∥ AB,

Here, the line segment PQ is the line of intersection.

Now angles ORP and RPA are equal as they are alternate interior angles

So, ∠ORP = ∠RPA

And,

∠RPA = 90° (Since, PQ is perpendicular to AB)

∠ORP = 90°

Now, ∠ROP+∠OPA = 180° (Since they are co-interior angles)

∠ROP+90° = 180°

∠ROP = 90°

Now, it is seen that the △ORP has two right angles which are ∠ORP and ∠ROP. Since this condition is impossible, it can be said the supposition we took is wrong.

6. The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

Answer:

Draw the diagram as shown below.

Here, AB is the tangent that is drawn on the circle from a point A.

So, the radius OB will be perpendicular to AB i.e. OB ⊥ AB

We know, OA = 5cm and AB = 4 cm

Now, In △ABO,

OA² =AB²+BO² (Using Pythagoras theorem)

5² = 4²+BO²

BO² = 25-16

BO² = 9

BO = 3

So, the radius of the given circle i.e. BO is 3 cm.

7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Answer:

Draw two concentric circles with the center O. Now, draw a chord AB in the larger circle which touches the smaller circle at a point P as shown in the figure below.

From the above diagram, AB is tangent to the smaller circle to point P.

∴ OP ⊥ AB

Using Pythagoras theorem in triangle OPA,

OA²= AP²+OP²

5² = AP²+3²

AP² = 25-9

AP = 4

Now, as OP ⊥ AB,

Since the perpendicular from the center of the circle bisects the chord, AP will be equal to PB

So, AB = 2AP = 2×4 = 8 cm

So, the length of the chord of the larger circle is 8 cm.

8. A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD = AD + BC

Answer:

The figure given is:

From this figure we can conclude a few points which are:

(i) DR = DS

(ii) BP = BQ

(iii) AP = AS

(iv) CR = CQ

Since they are tangents on the circle from points D, B, A, and C respectively.

Now, adding the LHS and RHS of the above equations we get,

DR+BP+AP+CR = DS+BQ+AS+CQ

By rearranging them we get,

(DR+CR) + (BP+AP) = (CQ+BQ) + (DS+AS)

By simplifying,

AD+BC= CD+AB

9. In Fig. 10.13, XY and X′Y′ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X′Y′ at B. Prove that ∠ AOB = 90°.

Answer:

From the figure given in the textbook, join OC. Now, the diagram will be as-

Now the triangles △OPA and △OCA are similar using SSS congruency as:

(i) OP = OC They are the radii of the same circle

(ii) AO = AO It is the common side

(iii) AP = AC These are the tangents from point A

So, △OPA ≅ △OCA

Similarly,

△OQB ≅ △OCB

So,

∠POA = ∠COA … (Equation i)

And, ∠QOB = ∠COB … (Equation ii)

Since the line POQ is a straight line, it can be considered as a diameter of the circle.

So, ∠POA +∠COA +∠COB +∠QOB = 180°

Now, from equations (i) and equation (ii) we get,

2∠COA+2∠COB = 180°

∠COA+∠COB = 90°

∴∠AOB = 90°

10. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the center.

Answer:

First, draw a circle with centre O. Choose an external point P and draw two tangents PA and PB at point A and point B respectively. Now, join A and B to make AB in a way that it subtends ∠AOB at the center of the circle. The diagram is as follows:

From the above diagram, it is seen that the line segments OA and PA are perpendicular.

So, ∠OAP = 90°

In a similar way, the line segments OB ⊥ PB and so, ∠OBP = 90°

Now, in the quadrilateral OAPB,

∴∠APB+∠OAP +∠PBO +∠BOA = 360° (since the sum of all interior angles will be 360°)

By putting the values we get,

∠APB + 180° + ∠BOA = 360°

So, ∠APB + ∠BOA = 180° (Hence proved).

11. Prove that the parallelogram circumscribing a circle is a rhombus.

Answer:

Consider a parallelogram ABCD which is circumscribing a circle with a center O. Now, since ABCD is a parallelogram, AB = CD and BC = AD.

From the above figure, it is seen that,

(i) DR = DS

(ii) BP = BQ

(iii) CR = CQ

(iv) AP = AS

These are the tangents to the circle at D, B, C, and A respectively.

Adding all these we get,

DR+BP+CR+AP = DS+BQ+CQ+AS

By rearranging them we get,

(BP+AP)+(DR+CR) = (CQ+BQ)+(DS+AS)

Again by rearranging them we get,

AB+CD = BC+AD

Now, since AB = CD and BC = AD, the above equation becomes

2AB = 2BC

∴ AB = BC

Since AB = BC = CD = DA, it can be said that ABCD is a rhombus.

12. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.

Answer:

The figure given is as follows:

Consider the triangle ABC,

We know that the length of any two tangents which are drawn from the same point to the circle is equal.

So,

(i) CF = CD = 6 cm

(ii) BE = BD = 8 cm

(iii) AE = AF = x

Now, it can be observed that,

(i) AB = EB+AE = 8+x

(ii) CA = CF+FA = 6+x

(iii) BC = DC+BD = 6+8 = 14

Now the semi perimeter “s” will be calculated as follows

2s = AB+CA+BC

By putting the respective values we get,

2s = 28+2x

s = 14+x

By solving this we get,

= √(14+x)48x ……… (i)

Again, the area of △ABC = 2 × area of (△AOF + △COD + △DOB)

= 2×[(½×OF×AF)+(½×CD×OD)+(½×DB×OD)]

= 2×½(4x+24+32) = 56+4x …………..(ii)

Now from (i) and (ii) we get,

√(14+x)48x = 56+4x

Now, square both the sides,

48x(14+x) = (56+4x)²

48x = [4(14+x)]²/(14+x)

48x = 16(14+x)

48x = 224+16x

32x = 224

x = 7 cm

So, AB = 8+x

i.e. AB = 15 cm

And, CA = x+6 =13 cm.

13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Answer:

First draw a quadrilateral ABCD which will circumscribe a circle with its centre O in a way that it touches the circle at point P, Q, R, and S. Now, after joining the vertices of ABCD we get the following figure:

Now, consider the triangles OAP and OAS,

AP = AS (They are the tangents from the same point A)

OA = OA (It is the common side)

OP = OS (They are the radii of the circle)

So, by SSS congruency △OAP ≅ △OAS

So, ∠POA = ∠AOS

Which implies that∠1 = ∠8

Similarly, other angles will be,

∠4 = ∠5

∠2 = ∠3

∠6 = ∠7

Now by adding these angles we get,

∠1+∠2+∠3 +∠4 +∠5+∠6+∠7+∠8 = 360°

Now by rearranging,

(∠1+∠8)+(∠2+∠3)+(∠4+∠5)+(∠6+∠7) = 360°

2∠1+2∠2+2∠5+2∠6 = 360°

Taking 2 as common and solving we get,

(∠1+∠2)+(∠5+∠6) = 180°

Thus, ∠AOB+∠COD = 180°

Similarly, it can be proved that ∠BOC+∠DOA = 180°

4. Quadratic Equations

Exercise 4.1 Page: 73

1. Check whether the following are quadratic equations:

(i) (x + 1)² = 2(x – 3)

(ii) x² – 2x = (–2) (3 – x)

(iii) (x – 2)(x + 1) = (x – 1)(x + 3)

(iv) (x – 3)(2x +1) = x(x + 5)

(v) (2x – 1)(x – 3) = (x + 5)(x – 1)

(vi) x² + 3x + 1 = (x – 2)²

(vii) (x + 2)³ = 2x (x² – 1)

(viii) x³ – 4x² – x + 1 = (x – 2)³

Solutions:

(i) Given,

(x + 1)² = 2(x – 3)

By using the formula for (a+b)² = a²+2ab+b²

⇒ x² + 2x + 1 = 2x – 6

⇒ x² + 7 = 0

Since the above equation is in the form of ax² + bx + c = 0.

Therefore, the given equation is quadratic equation.

(ii) Given, x² – 2x = (–2) (3 – x)

By using the formula for (a+b)² = a²+2ab+b²

⇒ x² – 2x = -6 + 2x

⇒ x² – 4x + 6 = 0

Since the above equation is in the form of ax² + bx + c = 0.

Therefore, the given equation is quadratic equation.

(iii) Given, (x – 2)(x + 1) = (x – 1)(x + 3)

By using the formula for (a+b)² = a²+2ab+b²

⇒ x² – x – 2 = x² + 2x – 3

⇒ 3x – 1 = 0

Since the above equation is not in the form of ax² + bx + c = 0.

Therefore, the given equation is not a quadratic equation.

(iv) Given, (x – 3)(2x +1) = x(x + 5)

By using the formula for (a+b)²=a²+2ab+b²

⇒ 2x² – 5x – 3 = x² + 5x

⇒  x² – 10x – 3 = 0

Since the above equation is in the form of ax² + bx + c = 0.

Therefore, the given equation is quadratic equation.

(v) Given, (2x – 1)(x – 3) = (x + 5)(x – 1)

By using the formula for (a+b)²=a²+2ab+b²

⇒ 2x² – 7x + 3 = x² + 4x – 5

⇒ x² – 11x + 8 = 0

Since the above equation is in the form of ax² + bx + c = 0.

Therefore, the given equation is quadratic equation.

(vi) Given, x² + 3x + 1 = (x – 2)²

By using the formula for (a+b)²=a²+2ab+b²

⇒ x² + 3x + 1 = x² + 4 – 4x

⇒ 7x – 3 = 0

Since the above equation is not in the form of ax² + bx + c = 0.

Therefore, the given equation is not a quadratic equation.

(vii) Given, (x + 2)³ = 2x(x² – 1)

By using the formula for (a+b)² = a²+2ab+b²

⇒ x³ + 8 + x² + 12x = 2x³ – 2x

⇒ x³ + 14x – 6x² – 8 = 0

Since the above equation is not in the form of ax² + bx + c = 0.

Therefore, the given equation is not a quadratic equation.

(viii) Given, x³ – 4x² – x + 1 = (x – 2)³

By using the formula for (a+b)² = a²+2ab+b²

⇒  x³ – 4x² – x + 1 = x³ – 8 – 6x²  + 12x

⇒ 2x² – 13x + 9 = 0

Since the above equation is in the form of ax² + bx + c = 0.

Therefore, the given equation is quadratic equation.

2. Represent the following situations in the form of quadratic equations:

(i) The area of a rectangular plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.

(ii) The product of two consecutive positive integers is 306. We need to find the integers.

(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.

(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken

Solutions:

(i) Let us consider,

Breadth of the rectangular plot = x m

Thus, the length of the plot = (2x + 1) m.

As we know,

Area of rectangle = length × breadth = 528 m²

Putting the value of length and breadth of the plot in the formula, we get,

(2x + 1) × x = 528

⇒ 2x² + x =528

⇒ 2x² + x – 528 = 0

Therefore, the length and breadth of plot, satisfies the quadratic equation, 2x² + x – 528 = 0, which is the required representation of the problem mathematically.

(ii) Let us consider,

The first integer number = x

Thus, the next consecutive positive integer will be = x + 1

Product of two consecutive integers = x × (x +1) = 306

⇒ x² + x = 306

⇒ x² + x – 306 = 0

Therefore, the two integers x and x+1, satisfies the quadratic equation, x² + x – 306 = 0, which is the required representation of the problem mathematically.

(iii) Let us consider,

Age of Rohan’s = x  years

Therefore, as per the given question,

Rohan’s mother’s age = x + 26

After 3 years,

Age of Rohan’s = x + 3

Age of Rohan’s mother will be = x + 26 + 3 = x + 29

The product of their ages after 3 years will be equal to 360, such that

(x + 3)(x + 29) = 360

⇒ x² + 29x + 3x + 87 = 360

⇒ x² + 32x + 87 – 360 = 0

⇒ x² + 32x – 273 = 0

Therefore, the age of Rohan and his mother, satisfies the quadratic equation, x² + 32x – 273 = 0, which is the required representation of the problem mathematically.

(iv) Let us consider,

The speed of train = x  km/h

And

Time taken to travel 480 km = 480/x km/hr

As per second condition, the speed of train = (x – 8) km/h

Also given, the train will take 3 hours to cover the same distance.

Therefore, time taken to travel 480 km = 480/(x+3) km/h

As we know,

Speed × Time = Distance

Therefore,

(x – 8)(480/(x + 3) = 480

⇒ 480 + 3x – 3840/x – 24 = 480

⇒ 3x – 3840/x = 24

⇒ 3x² – 8x – 1280 = 0

Therefore, the speed of the train, satisfies the quadratic equation, 3x² – 8x – 1280 = 0, which is the required representation of the problem mathematically.

Exercise 4.2 Page: 76

1. Find the roots of the following quadratic equations by factorisation:

(i) x² – 3x – 10 = 0
(ii) 2x² + x – 6 = 0
(iii) √2 x² + 7x + 5√2 = 0
(iv) 2x² – x +1/8 = 0
(v) 100x² – 20x + 1 = 0

Solutions:

(i) Given, x² – 3x – 10 =0

Taking LHS,

=>x² – 5x + 2x – 10

=>x(x – 5) + 2(x – 5)

=>(x – 5)(x + 2)

The roots of this equation, x² – 3x – 10 = 0 are the values of x for which (x – 5)(x + 2) = 0

Therefore, x – 5 = 0 or x + 2 = 0

=> x = 5 or x = -2

(ii) Given, 2x² + x – 6 = 0

Taking LHS,

=> 2x² + 4x – 3x – 6

=> 2x(x + 2) – 3(x + 2)

=> (x + 2)(2x – 3)

The roots of this equation, 2x² + x – 6=0 are the values of x for which (x – 5)(x + 2) = 0

Therefore, x + 2 = 0 or 2x – 3 = 0

=> x = -2 or x = 3/2

(iii) √2 x² + 7x + 5√2=0

Taking LHS,

=> √2 x² + 5x + 2x + 5√2

=> x (√2x + 5) + √2(√2x + 5)= (√2x + 5)(x + √2)

The roots of this equation, √2 x² + 7x + 5√2=0 are the values of x for which (x – 5)(x + 2) = 0

Therefore, √2x + 5 = 0 or x + √2 = 0

=> x = -5/√2 or x = -√2

(iv) 2x² – x +1/8 = 0

Taking LHS,

=1/8 (16x²  – 8x + 1)

= 1/8 (16x²  – 4x -4x + 1)

= 1/8 (4x(4x  – 1) -1(4x – 1))

= 1/8 (4x – 1)²

The roots of this equation, 2x² – x + 1/8 = 0, are the values of x for which (4x – 1)²= 0

Therefore, (4x – 1) = 0 or (4x – 1) = 0

⇒ x = 1/4 or x = 1/4

(v) Given, 100x² – 20x + 1=0

Taking LHS,

= 100x² – 10x – 10x + 1

= 10x(10x – 1) -1(10x – 1)

= (10x – 1)²

The roots of this equation, 100x² – 20x + 1=0, are the values of x for which (10x – 1)²= 0

∴ (10x – 1) = 0 or (10x – 1) = 0

⇒x = 1/10 or x = 1/10

2. Solve the problems given in Example 1.

Represent the following situations mathematically:

(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.

(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ` 750. We would like to find out the number of toys produced on that day.

Solutions:

(i) Let us say, the number of marbles John have = x.

Therefore, number of marble Jivanti have = 45 – x

After losing 5 marbles each,

Number of marbles John have = x – 5

Number of marble Jivanti have = 45 – x – 5 = 40 – x

Given that the product of their marbles is 124.

∴ (x – 5)(40 – x) = 124

⇒ x² – 45x + 324 = 0

⇒ x² – 36x – 9x + 324 = 0

⇒ x(x – 36) -9(x – 36) = 0

⇒ (x – 36)(x – 9) = 0

Thus, we can say,

x – 36 = 0 or x – 9 = 0

⇒ x = 36 or x = 9

Therefore,

If, John’s marbles = 36,

Then, Jivanti’s marbles = 45 – 36 = 9

And if John’s marbles = 9,

Then, Jivanti’s marbles = 45 – 9 = 36

(ii) Let us say, number of toys produced in a day be x.

Therefore, cost of production of each toy = Rs(55 – x)

Given, total cost of production of the toys = Rs 750

∴ x(55 – x) = 750

⇒ x² – 55x + 750 = 0

⇒ x² – 25x – 30x + 750 = 0

⇒ x(x – 25) -30(x – 25) = 0

⇒ (x – 25)(x – 30) = 0

Thus, either x -25 = 0 or x – 30 = 0

⇒ x = 25 or x = 30

Hence, the number of toys produced in a day, will be either 25 or 30.

3. Find two numbers whose sum is 27 and product is 182.

Solution:

Let us say, first number be x and the second number is 27 – x.

Therefore, the product of two numbers

x(27 – x) = 182

⇒ x² – 27x – 182 = 0

⇒ x² – 13x – 14x + 182 = 0

⇒ x(x – 13) -14(x – 13) = 0

⇒ (x – 13)(x -14) = 0

Thus, either, x = -13 = 0 or x – 14 = 0

⇒ x = 13 or x = 14

Therefore, if first number = 13, then second number = 27 – 13 = 14

And if first number = 14, then second number = 27 – 14 = 13

Hence, the numbers are 13 and 14.

4. Find two consecutive positive integers, sum of whose squares is 365.

Solution:

Let us say, the two consecutive positive integers be x and x + 1.

Therefore, as per the given questions,

x² + (x + 1)² = 365

⇒ x² + x² + 1 + 2x = 365

⇒ 2x² + 2x – 364 = 0

⇒ x² + x – 182 = 0

⇒ x² + 14x – 13x – 182 = 0

⇒ x(x + 14) -13(x + 14) = 0

⇒ (x + 14)(x – 13) = 0

Thus, either, x + 14 = 0 or x – 13 = 0,

⇒ x = – 14 or x = 13

since, the integers are positive, so x can be 13, only.

∴ x + 1 = 13 + 1 = 14

Therefore, two consecutive positive integers will be 13 and 14.

5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Solution:

Let us say, the base of the right triangle be x cm.

Given, the altitude of right triangle = (x – 7) cm

From Pythagoras theorem, we know,

Base² + Altitude² = Hypotenuse²

∴ x² + (x – 7)² = 13²

⇒ x² + x² + 49 – 14x = 169

⇒ 2x² – 14x – 120 = 0

⇒ x² – 7x – 60 = 0

⇒ x² – 12x + 5x – 60 = 0

⇒ x(x – 12) + 5(x – 12) = 0

⇒ (x – 12)(x + 5) = 0

Thus, either x – 12 = 0 or x + 5 = 0,

⇒ x = 12 or x = – 5

Since sides cannot be negative, x can only be 12.

Therefore, the base of the given triangle is 12 cm and the altitude of this triangle will be (12 – 7) cm = 5 cm.

6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs.90, find the number of articles produced and the cost of each article.

Solution:

Let us say, the number of articles produced be x.

Therefore, cost of production of each article = Rs (2x + 3)

Given, total cost of production is Rs.90

∴ x(2x + 3) = 90

⇒ 2x² + 3x – 90 = 0

⇒ 2x² + 15x -12x – 90 = 0

⇒ x(2x + 15) -6(2x + 15) = 0

⇒ (2x + 15)(x – 6) = 0

Thus, either 2x + 15 = 0 or x – 6 = 0

⇒ x = -15/2 or x = 6

As the number of articles produced can only be a positive integer, therefore, x can only be 6.

Hence, number of articles produced = 6

Cost of each article = 2 × 6 + 3 = Rs 15.

Exercise 4.3 Page: 87

1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square:

(i) 2x² – 7x +3 = 0

(ii) 2x² + x – 4 = 0
(iii) 4x² + 4√3x + 3 = 0

(iv) 2x² + x + 4 = 0

Solutions:

(i) 2x² – 7x + 3 = 0

⇒ 2x² – 7x = – 3

Dividing by 2 on both sides, we get

⇒ x² -7x/2 = -3/2

⇒ x² -2 × x ×7/4 = -3/2

On adding (7/4)² to both sides of equation, we get

⇒ (x)²-2×x×7/4 +(7/4)² = (7/4)²-3/2

⇒ (x-7/4)² = (49/16) – (3/2)

⇒(x-7/4)² = 25/16

⇒(x-7/4)² = ±5/4

⇒ x = 7/4 ± 5/4

⇒ x = 7/4 + 5/4 or x = 7/4 – 5/4

⇒ x = 12/4 or x = 2/4

⇒ x = 3 or x = 1/2

(ii) 2x² + x – 4 = 0

⇒ 2x² + x = 4

Dividing both sides of the equation by 2, we get

⇒ x² +x/2 = 2

Now on adding (1/4)² to both sides of the equation, we get,

⇒ (x)² + 2 × x × 1/4 + (1/4)² = 2 + (1/4)²

⇒ (x + 1/4)² = 33/16

⇒ x + 1/4 = ± √33/4

⇒ x = ± √33/4 – 1/4

⇒ x = ± √33-1/4

Therefore, either x = √33-1/4 or x = -√33-1/4

(iii) 4x² + 4√3x + 3 = 0

Converting the equation into a²+2ab+b² form, we get,

⇒ (2x)² + 2 × 2x × √3 + (√3)² = 0

⇒ (2x + √3)² = 0

⇒ (2x + √3) = 0 and (2x + √3) = 0

Therefore, either x = -√3/2 or x = -√3/2.

(iv) 2x² + x + 4 = 0

⇒ 2x² + x = -4

Dividing both sides of the equation by 2, we get

⇒ x² + 1/2x = 2

⇒ x² + 2 × x × 1/4 = -2

By adding (1/4)² to both sides of the equation, we get

⇒ (x)² + 2 × x × 1/4 + (1/4)² = (1/4)² – 2

⇒ (x + 1/4)² = 1/16 – 2

⇒ (x + 1/4)² = -31/16

As we know, the square of numbers cannot be negative.

Therefore, there is no real root for the given equation, 2x² + x + 4 = 0.

2. Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.

(i) 2x² – 7x + 3 = 0

On comparing the given equation with ax² + bx + c = 0, we get,

a = 2, b = -7 and c = 3

By using quadratic formula, we get,

⇒ x = (7±√(49 – 24))/4

⇒ x = (7±√25)/4

⇒ x = (7±5)/4

⇒ x = (7+5)/4 or x = (7-5)/4

⇒ x = 12/4 or 2/4

∴  x = 3 or 1/2

(ii) 2x² + x – 4 = 0

On comparing the given equation with ax² + bx + c = 0, we get,

a = 2, b = 1 and c = -4

By using quadratic formula, we get,

⇒x = -1±√1+32/4

⇒x = -1±√33/4

∴ x = -1+√33/4 or x = -1-√33/4

(iii) 4x² + 4√3x + 3 = 0

On comparing the given equation with ax² + bx + c = 0, we get

a = 4, b = 4√3 and c = 3

By using quadratic formula, we get,

⇒ x = -4√3±√48-48/8

⇒ x = -4√3±0/8

∴ x = -√3/2 or x = -√3/2

(iv) 2x² + x + 4 = 0

On comparing the given equation with ax² + bx + c = 0, we get,

a = 2, b = 1 and c = 4

By using quadratic formula, we get

⇒ x = -1±√1-32/4

⇒ x = -1±√-31/4

As we know, the square of a number can never be negative. Therefore, there is no real solution for the given equation.

3. Find the roots of the following equations:

(i) x-1/x = 3, x ≠ 0
(ii) 1/x+4 – 1/x-7 = 11/30, x = -4, 7

Solution:

(i) x-1/x = 3

⇒ x² – 3x -1 = 0

On comparing the given equation with ax² + bx + c = 0, we get

a = 1, b = -3 and c = -1

By using quadratic formula, we get,

⇒ x = 3±√9+4/2

⇒ x = 3±√13/2

∴ x = 3+√13/2 or x = 3-√13/2

(ii) 1/x+4 – 1/x-7 = 11/30

⇒ x-7-x-4/(x+4)(x-7) = 11/30

⇒ -11/(x+4)(x-7) = 11/30

⇒ (x+4)(x-7) = -30

⇒ x² – 3x – 28 = 30

⇒ x² – 3x + 2 = 0

We can solve this equation by factorization method now,

⇒ x² – 2x – x + 2 = 0

⇒ x(x – 2) – 1(x – 2) = 0

⇒ (x – 2)(x – 1) = 0

⇒ x = 1 or 2

4. The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is 1/3. Find his present age.

Solution:

Let us say, present age of Rahman is x years.

Three years ago, Rehman’s age was (x – 3) years.

Five years after, his age will be (x + 5) years.

Given, the sum of the reciprocals of Rehman’s ages 3 years ago and after 5 years is equal to 1/3.

∴ 1/x-3 + 1/x-5 = 1/3

(x+5+x-3)/(x-3)(x+5) = 1/3

(2x+2)/(x-3)(x+5) = 1/3

⇒ 3(2x + 2) = (x-3)(x+5)

⇒ 6x + 6 = x² + 2x – 15

⇒ x² – 4x – 21 = 0

⇒ x² – 7x + 3x – 21 = 0

⇒ x(x – 7) + 3(x – 7) = 0

⇒ (x – 7)(x + 3) = 0

⇒ x = 7, -3

As we know, age cannot be negative.

Therefore, Rahman’s present age is 7 years.

5. In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.

Solution:

Let us say, the marks of Shefali in Maths be x.

Then, the marks in English will be 30 – x.

As per the given question,

(x + 2)(30 – x – 3) = 210

(x + 2)(27 – x) = 210

⇒ -x² + 25x + 54 = 210

⇒ x² – 25x + 156 = 0

⇒ x² – 12x – 13x + 156 = 0

⇒ x(x – 12) -13(x – 12) = 0

⇒ (x – 12)(x – 13) = 0

⇒ x = 12, 13

Therefore, if the marks in Maths are 12, then marks in English will be 30 – 12 = 18 and the marks in Maths are 13, then marks in English will be 30 – 13 = 17.

6. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.

Solution:

Let us say, the shorter side of the rectangle be x m.

Then, larger side of the rectangle = (x + 30) m

As given, the length of the diagonal is = x + 30 m

Therefore,

⇒ x² + (x + 30)² = (x + 60)²

⇒ x² + x² + 900 + 60x = x² + 3600 + 120x

⇒ x² – 60x – 2700 = 0

⇒ x² – 90x + 30x – 2700 = 0

⇒ x(x – 90) + 30(x -90) = 0

⇒ (x – 90)(x + 30) = 0

⇒ x = 90, -30

However, side of the field cannot be negative. Therefore, the length of the shorter side will be 90 m.

and the length of the larger side will be (90 + 30) m = 120 m.

7. The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

Solution:

Let us say, the larger and smaller number be x and y respectively.

As per the question given,

x² – y² = 180 and y² = 8x

⇒ x² – 8x = 180

⇒ x² – 8x – 180 = 0

⇒ x² – 18x + 10x – 180 = 0

⇒ x(x – 18) +10(x – 18) = 0

⇒ (x – 18)(x + 10) = 0

⇒ x = 18, -10

However, the larger number cannot considered as negative number, as 8 times of the larger number will be negative and hence, the square of the smaller number will be negative which is not possible.

Therefore, the larger number will be 18 only.

x = 18

∴ y² = 8x = 8 × 18 = 144

⇒ y = ±√144 = ±12

∴ Smaller number = ±12

Therefore, the numbers are 18 and 12 or 18 and -12.

8. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Solution:

Let us say, the speed of the train be x km/hr.

Time taken to cover 360 km = 360/x hr.

As per the question given,

⇒ (x + 5)(360-1/x) = 360

⇒ 360 – x + 1800-5/x = 360

⇒ x² + 5x + 10x – 1800 = 0

⇒ x(x + 45) -40(x + 45) = 0

⇒ (x + 45)(x – 40) = 0

⇒ x = 40, -45

As we know, the value of speed cannot be negative.

Therefore, the speed of train is 40 km/h.

9. Two water taps together can fill a tank in
 hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Solution:

Let the time taken by the smaller pipe to fill the tank = x hr.

Time taken by the larger pipe = (x – 10) hr

Part of tank filled by smaller pipe in 1 hour = 1/x

Part of tank filled by larger pipe in 1 hour = 1/(x – 10)

As given, the tank can be filled in
 = 75/8 hours by both the pipes together.

Therefore,

1/x + 1/x-10 = 8/75

x-10+x/x(x-10) = 8/75

⇒ 2x-10/x(x-10) = 8/75

⇒ 75(2x – 10) = 8x² – 80x

⇒ 150x – 750 = 8x² – 80x

⇒ 8x² – 230x +750 = 0

⇒ 8x² – 200x – 30x + 750 = 0

⇒ 8x(x – 25) -30(x – 25) = 0

⇒ (x – 25)(8x -30) = 0

⇒ x = 25, 30/8

Time taken by the smaller pipe cannot be 30/8 = 3.75 hours, as the time taken by the larger pipe will become negative, which is logically not possible.

Therefore, time taken individually by the smaller pipe and the larger pipe will be 25 and 25 – 10 =15 hours respectively.

10. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speeds of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.

Solution:

Let us say, the average speed of passenger train =  x km/h.

Average speed of express train = (x + 11) km/h

Given, time taken by the express train to cover 132 km is 1 hour less than the passenger train to cover the same distance. Therefore,

(132/x) – (132/(x+11)) = 1

132(x+11-x)/(x(x+11)) = 1

132 × 11 /(x(x+11)) = 1

⇒ 132 × 11 = x(x + 11)

⇒ x² + 11x – 1452 = 0

⇒ x² +  44x -33x -1452 = 0

⇒ x(x + 44) -33(x + 44) = 0

⇒ (x + 44)(x – 33) = 0

⇒ x = – 44, 33

As we know, Speed cannot be negative.

Therefore, the speed of the passenger train will be 33 km/h and thus, the speed of the express train will be 33 + 11 = 44 km/h.

11. Sum of the areas of two squares is 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares.

Solution:

Let the sides of the two squares be x m and y m.

Therefore, their perimeter will be 4x and 4y respectively

And area of the squares will be x² and y² respectively.

Given,

4x – 4y = 24

x – y = 6

x = y + 6

Also, x² + y² = 468

⇒ (6 + y²) + y² = 468

⇒ 36 + y² + 12y + y² = 468

⇒ 2y² + 12y + 432 = 0

⇒ y² + 6y – 216 = 0

⇒ y² + 18y – 12y – 216 = 0

⇒ y(y +18) -12(y + 18) = 0

⇒ (y + 18)(y – 12) = 0

⇒ y = -18, 12

As we know, the side of a square cannot be negative.

Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m.

Exercise 4.4 Page: 91

1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them;
(i) 2x² – 3x + 5 = 0
(ii) 3x² – 4√3x + 4 = 0
(iii) 2x² – 6x + 3 = 0

Solutions:

(i) Given,

2x² – 3x + 5 = 0

Comparing the equation with ax² + bx + c = 0, we get

a = 2, b = -3 and c = 5

We know, Discriminant = b² – 4ac

= ( – 3)² – 4 (2) (5) = 9 – 40

= – 31

As you can see, b² – 4ac < 0

Therefore, no real root is possible for the given equation, 2x² – 3x + 5 = 0.

(ii) 3x² – 4√3x + 4 = 0

Comparing the equation with ax² + bx + c = 0, we get

a = 3, b = -4√3 and c = 4

We know, Discriminant = b² – 4ac

= (-4√3)² – 4(3)(4)

= 48 – 48 = 0

As b² – 4ac = 0,

Real roots exist for the given equation and they are equal to each other.

Hence the roots will be –b/2a and –b/2a.

–b/2a = -(-4√3)/2×3 = 4√3/6 = 2√3/3 = 2/√3

Therefore, the roots are 2/√3 and 2/√3.

(iii) 2x² – 6x + 3 = 0

Comparing the equation with ax² + bx + c = 0, we get

a = 2, b = -6, c = 3

As we know, Discriminant = b² – 4ac

= (-6)² – 4 (2) (3)

= 36 – 24 = 12

As b² – 4ac > 0,

Therefore, there are distinct real roots exist for this equation, 2x² – 6x + 3 = 0.

= (-(-6) ± √(-6²-4(2)(3)) )/ 2(2)

= (6±2√3 )/4

= (3±√3)/2

Therefore the roots for the given equation are (3+√3)/2 and (3-√3)/2

2. Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) 2x² + kx + 3 = 0
(ii) kx (x – 2) + 6 = 0

Solutions:

(i) 2x² + kx + 3 = 0

Comparing the given equation with ax² + bx + c = 0, we get,

a = 2, b = k and c = 3

As we know, Discriminant = b² – 4ac

= (k)² – 4(2) (3)

= k² – 24

For equal roots, we know,

Discriminant = 0

k² – 24 = 0

k² = 24

k = ±√24 = ±2√6

(ii) kx(x – 2) + 6 = 0

or kx² – 2kx + 6 = 0

Comparing the given equation with ax² + bx + c = 0, we get

a = k, b = – 2k and c = 6

We know, Discriminant = b² – 4ac

= ( – 2k)² – 4 (k) (6)

= 4k² – 24k

For equal roots, we know,

b² – 4ac = 0

4k² – 24k = 0

4k (k – 6) = 0

Either 4k = 0 or k = 6 = 0

k = 0 or k = 6

However, if k = 0, then the equation will not have the terms ‘x²‘ and ‘x‘.

Therefore, if this equation has two equal roots, k should be 6 only.

3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m²? If so, find its length and breadth.

Solution:

Let the breadth of mango grove be l.

Length of mango grove will be 2l.

Area of mango grove = (2l) (l)= 2l²

2l² = 800

l² = 800/2 = 400

l² – 400 =0

Comparing the given equation with ax² + bx + c = 0, we get

a = 1, b = 0, c = 400

As we know, Discriminant = b² – 4ac

=> (0)² – 4 × (1) × ( – 400) = 1600

Here, b² – 4ac > 0

Thus, the equation will have real roots. And hence, the desired rectangular mango grove can be designed.

l = ±20

As we know, the value of length cannot be negative.

Therefore, breadth of mango grove = 20 m

Length of mango grove = 2 × 20 = 40 m

4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.

Solution:

Let’s say, the age of one friend be x years.

Then, the age of the other friend will be (20 – x) years.

Four years ago,

Age of First friend = (x – 4) years

Age of Second friend = (20 – x – 4) = (16 – x) years

As per the given question, we can write,

(x – 4) (16 – x) = 48

16x – x² – 64 + 4x = 48

 – x² + 20x – 112 = 0

x² – 20x + 112 = 0

Comparing the equation with ax² + bx + c = 0, we get

a = 1, b = -20 and c = 112

Discriminant = b² – 4ac

= (-20)² – 4 × 112

= 400 – 448 = -48

b² – 4ac < 0

Therefore, there will be no real solution possible for the equations. Hence, condition doesn’t exist.

5. Is it possible to design a rectangular park of perimeter 80 and area 400 m2? If so find its length and breadth.

Solution:

Let the length and breadth of the park be l and b.

Perimeter of the rectangular park = 2 (l + b) = 80

So, l + b = 40

Or, b = 40 – l

Area of the rectangular park = l×b = l(40 – l) = 40l – l² = 400

l² –  40l + 400 = 0, which is a quadratic equation.

Comparing the equation with ax² + bx + c = 0, we get

a = 1, b = -40, c = 400

Since, Discriminant = b² – 4ac

=(-40)² – 4 × 400

= 1600 – 1600 = 0

Thus, b² – 4ac = 0

Therefore, this equation has equal real roots. Hence, the situation is possible.

Root of the equation,

l = –b/2a

l = (40)/2(1) = 40/2 = 20

Therefore, length of rectangular park, l = 20 m

And breadth of the park, b = 40 – l = 40 – 20 = 20 m.