# RS Aggarwal Solutions Class 9 Ex 1B

## RS Aggarwal Class 9 Ex 1B Chapter 1

Question 1: SIMPLIFY;

(i) ( 4 + 2$\sqrt{2}$ ) ( 4 – 2$\sqrt{2}$ )

(ii) (5$\sqrt{5}$ + 3$\sqrt{3}$ ) (5$\sqrt{5}$3$\sqrt{3}$)

(iii) ( 6 – 6$\sqrt{6}$) ( 6 + 6$\sqrt{6}$)

(iv) (5$\sqrt{5}$2$\sqrt{2}$) (2$\sqrt{2}$3$\sqrt{3}$)

(v) (5$\sqrt{5}$3$\sqrt{3}$ )2

(vi) (3 – 3$\sqrt{3}$)2

Solution:

(i) ( 4 + 2$\sqrt{2}$ ) ( 4 – 2$\sqrt{2}$ )

= 42 – (2$\sqrt{2}$)2 [ ( a + b) ( a – b) = a2 – b2]

= 16 – 2

= 14

(ii) (5$\sqrt{5}$ + 3$\sqrt{3}$ ) (5$\sqrt{5}$3$\sqrt{3}$)

= (5$\sqrt{5}$)2 – (3$\sqrt{3}$)2

= 5 – 3

= 2

(iii) ( 6 – 6$\sqrt{6}$) ( 6 + 6$\sqrt{6}$)

= 62 – (6$\sqrt{6}$) 2

= 36 – 6

= 30

(iv) (5$\sqrt{5}$2$\sqrt{2}$) (2$\sqrt{2}$3$\sqrt{3}$)

= 5$\sqrt{5}$ x 2$\sqrt{2}$5$\sqrt{5}$ x 3$\sqrt{3}$2$\sqrt{2}$ x 2$\sqrt{2}$ + 2$\sqrt{2}$ x 3$\sqrt{3}$

= 10$\sqrt{10}$15$\sqrt{15}$ – 2 + 6$\sqrt{6}$

(v) (5$\sqrt{5}$3$\sqrt{3}$ )2

= (5$\sqrt{5}$ )2 + (3$\sqrt{3}$ )2 – 2  x 5$\sqrt{5}$ 3$\sqrt{3}$ [(a – b)2 = a2 + b2 – 2ab ]

= 5 + 3 – 215$\sqrt{15}$ = 8 – 215$\sqrt{15}$

(vi) (3 – 3$\sqrt{3}$)2

= 32  + (3$\sqrt{3}$)2 – 2  x 3 x 3$\sqrt{3}$ [ ( a – b)2 = a2 + b2 – 2ab ]

= 9 + 3 – 63$\sqrt{3}$

= 12- 63$\sqrt{3}$

Question 2: Represent 3.2$\sqrt{3.2}$ geometrically on the number line.
Solution:

Draw a line segment AB = 3.2 units and extend it to C such that BC = 1 unit.

Now, find the midpoint O of AC.

Taking O as the center and OA as the radius, draw a semicircle.

Now, draw BD $\perp$ AC, intersecting the semicircle at D.

Here,

BD = 3.2$\sqrt{3.2}$ units

Taking B as the centre and BD as the  radius, draw an arc meeting AC produced at E.

Thus, we have:

BE = BD = 3.2$\sqrt{3.2}$ units

Question 3: Represent 7.28$\sqrt{7.28}$ geometrically on the number line.
Solution:

Draw a line segment AB = 7.28 units and extend it to C such that BC = 1 unit.

Now, find the midpoint O of AC.

Taking O as the centre and OA as the radius, draw a semicircle.

Now, draw BD $\perp$ AC, intersecting the semicircle at D.

We have:

BD = 7.28$\sqrt{7.28}$ units

Now, taking B as the centre and BD as the radius, draw an arc meeting AC produced at E.

Thus, we have:

BE = BD = 7.28$\sqrt{7.28}$ units

Question 4: Mention the closure property, associative law , commutative law , the existence of Identity , existence of inverse of each real numbers for each of operations

2. multiplication on real numbers

Solution:

(i) Closure property: The sum of two real numbers is always a real number.

(ii) Associative law: (a + b) + c = a + (b + c) for all real numbers a, b and c.

(iii) Commutative law: a + b = b + a for all real numbers a and b.

(iv) Existence of additive identity: 0 is called the additive identity for real numbers.

As, for every real number a, 0 + a = a + 0 = a

(v) Existence of additive inverse: For each real number a, there exists a real number ( -a) such that a + (-a) = 0 = (-a) + a. Here, a and (-a) are the additive inverse of each other.

Multiplication properties of Real Numbers:

(i) Closure Property: The product of two real numbers is always a real number.

(ii) Associative law: (ab)c = a(bc) for all real numbers a, b and c.

(iii) Commutative law: a x b = b x a for all real numbers a and b.

(iv) Existence of multiplicative identity: 1 is called the multiplicative identity for real numbers.

As for, every real number a, 1 x a = a x 1 = a

(v) Existence of multiplicative inverse: For each real number a , there exists a real number 1a$\frac{1}{a}$ such that a 1a$\frac{1}{a}$ = 1 = 1a$\frac{1}{a}$a. Here, a and 1a$\frac{1}{a}$ are the multiplicative inverse of each other.

RATIONALISE THE DENOMINATOR OF EACH OF THE FOLLOWING.

Question 5: 7$\sqrt{7}$

Solution:

On multiplying the numerator and denominator of the given number by 7$\sqrt{7}$ , we get:

17$\frac{1}{\sqrt{7}}$ = 17$\frac{1}{\sqrt{7}}$ x 77$\frac{\sqrt{7}}{\sqrt{7}}$ =  77$\frac{\sqrt{7}}{{7}}$

Question 6:

Solution: 523$\frac{\sqrt{5}}{2\sqrt{3}}$

On Multiplying the numerator and denominator of the given number by 7$\sqrt{7}$ , we get:

523$\frac{\sqrt{5}}{2\sqrt{3}}$ = 523$\frac{\sqrt{5}}{2\sqrt{3}}$ x 33$\frac{\sqrt{3}}{\sqrt{3}}$ = 33$\frac{\sqrt{3}}{\sqrt{3}}$

Question 7: 12+3$\frac{1}{2+\sqrt{3}}$

Solution:

On Multiplying the numerator and denominator of the given number by 2 – 3$\sqrt{3}$ , we get:

12+3$\frac{1}{2+\sqrt{3}}$ = 12+3$\frac{1}{2+\sqrt{3}}$ x 2323$\frac{2 – \sqrt{3}}{2-\sqrt{3}}$

= 23(2)2(3)2$\frac{2-\sqrt{3}}{(2)^{2}-(\sqrt{3})^{2}}$

= 2343$\frac{2-\sqrt{3}}{4-3}$

= 231$\frac{2-\sqrt{3}}{1}$

= 2 – 3$\sqrt{3}$

Question 8: 152$\frac{1}{\sqrt{5}-2}$

Solution:

On Multiplying the numerator and denominator of the given number by 5$\sqrt{5}$ + 2 , we get:

152$\frac{1}{\sqrt{5}-2}$ = 152$\frac{1}{\sqrt{5}-2}$ x 5+25+2$\frac{\sqrt{5}+2}{\sqrt{5}+2}$

= 5+2(5)2(2)2$\frac{\sqrt{5}+2}{(\sqrt{5})^{2}-(2)^{2}}$

= 5+254$\frac{\sqrt{5}+2}{5-4}$

= 5+21$\frac{\sqrt{5}+2}{1}$

= 5+2$\sqrt{5}+2$

Question 9: 15+32$\frac{1}{5+3\sqrt{2}}$

Solution:

On Multiplying the numerator and denominator of the given number by 5 – 32$\sqrt{2}$ , we get:

15+32$\frac{1}{5+3\sqrt{2}}$ = 15+32$\frac{1}{5+3\sqrt{2}}$ x 532532$\frac{5-3\sqrt{2}}{5-3\sqrt{2}}$

= 532(5)2(32)2$\frac{5-3\sqrt{2}}{(5)^{2}-(3\sqrt{2})^{2}}$

= 5322518$\frac{5-3\sqrt{2}}{25-18}$

= 5327$\frac{5-3\sqrt{2}}{7}$

Question 10: 165$\frac{1}{\sqrt{6}-\sqrt{5}}$

Solution:

On Multiplying the numerator and denominator of the given number by 6$\sqrt{6}$ + 5$\sqrt{5}$ , we get:

165$\frac{1}{\sqrt{6}-\sqrt{5}}$ = 165$\frac{1}{\sqrt{6}-\sqrt{5}}$ x 6+56+5$\frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}}$

= 6+5(6)2(5)2$\frac{\sqrt{6}+\sqrt{5}}{(\sqrt{6})^{2}-(\sqrt{5})^{2}}$

= 6+5$\sqrt{6}+\sqrt{5}$

Question 11: 47+3$\frac{4}{\sqrt{7}+\sqrt{3}}$

Solution:

On Multiplying the numerator and denominator of the given number by 7$\sqrt{7}$3$\sqrt{3}$ , we get:

47+3$\frac{4}{\sqrt{7}+\sqrt{3}}$ = 47+3$\frac{4}{\sqrt{7}+\sqrt{3}}$ x 7373$\frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}-\sqrt{3}}$

= 4(73)(7)2(3)2$\frac{4(\sqrt{7}-\sqrt{3})}{(\sqrt{7})^{2}-(\sqrt{3})^{2}}$

= 4(73)73$\frac{4(\sqrt{7}-\sqrt{3})}{7-3}$

= 4(73)4$\frac{4(\sqrt{7}-\sqrt{3})}{4}$

= 73$\sqrt{7}-\sqrt{3}$

Question 12: 313+1$\frac{\sqrt{3}-1}{\sqrt{3}+1}$

Solution:

On multiplying the numerator and denominator of the given number by 3$\sqrt{3}$ – 1, we get:

313+1$\frac{\sqrt{3}-1}{\sqrt{3}+1}$ = 313+1$\frac{\sqrt{3}-1}{\sqrt{3}+1}$ x 3131$\frac{\sqrt{3}-1}{\sqrt{3}-1}$

= (31)2(3)2(1)2$\frac{(\sqrt{3}-1)^{2}}{(\sqrt{3})^{2}-(1)^{2}}$

= 3+12331$\frac{3+1-2\sqrt{3}}{3-1}$

= 4232$\frac{4-2\sqrt{3}}{2}$

= 2(23)2$\frac{2(2-\sqrt{3})}{2}$

= 2 – 3$\sqrt{3}$

Question 13: 3223+22$\frac{3-2\sqrt{2}}{3+2\sqrt{2}}$

Solution:

On multiplying the numerator and denominator of the given number by 3 – 2 2$\sqrt{2}$ , we get:

3223+22$\frac{3-2\sqrt{2}}{3+2\sqrt{2}}$ = 3223+22$\frac{3-2\sqrt{2}}{3+2\sqrt{2}}$ x 322322$\frac{3-2\sqrt{2}}{3-2\sqrt{2}}$

= (322)2(3)2(22)2$\frac{(3-2\sqrt{2})^{2}}{(3)^{2}-(2\sqrt{2})^{2}}$

= 9+812298$\frac{9+8-12\sqrt{2}}{9-8}$

= 17 – 12 2$\sqrt{2}$

Question 14: 3+131$\frac{\sqrt{3}+1}{\sqrt{3}-1}$ = a + b 3$\sqrt{3}$

Solution:

We have:

3+131$\frac{\sqrt{3}+1}{\sqrt{3}-1}$

Now, rationalizing the denominator of the given number by multiplying both the numerator and denominator with 3$\sqrt{3}$ + 1 we get:

3+131$\frac{\sqrt{3}+1}{\sqrt{3}-1}$ = 3+131$\frac{\sqrt{3}+1}{\sqrt{3}-1}$ x 3+13+1$\frac{\sqrt{3}+1}{\sqrt{3}+1}$

= (3+1)2(3)2(1)2$\frac{(\sqrt{3}+1)^{2}}{(\sqrt{3})^{2}-(1)^{2}}$

= 3+1+2331$\frac{3+1+2\sqrt{3}}{3-1}$

= 4+232$\frac{4+2\sqrt{3}}{2}$

= 2(2+3)2$\frac{2(2+\sqrt{3})}{2}$

= 2 + 3$\sqrt{3}$

Therefore,   2 + 3$\sqrt{3}$ = a + b 3$\sqrt{3}$

So, on comparing the LHS and RHS, we get

a =2 and b = 1

Question 15: 3+232$\frac{3+\sqrt{2}}{3-\sqrt{2}}$ = a + b 2$\sqrt{2}$

Solution:

We have:

3+232$\frac{3+\sqrt{2}}{3-\sqrt{2}}$

Now, rationalizing the denominator of the given number by multiplying both the denominator and the numerator with 3 + 2$\sqrt{2}$, we get:

3+232$\frac{3+\sqrt{2}}{3-\sqrt{2}}$ = 3+232$\frac{3+\sqrt{2}}{3-\sqrt{2}}$ x 3+23+2$\frac{3+\sqrt{2}}{3+\sqrt{2}}$

= (3+2)2(3)2(2)2$\frac{(3+\sqrt{2})^{2}}{(3)^{2}-(\sqrt{2})^{2}}$

= 9+2+6292$\frac{9+2+6\sqrt{2}}{9-2}$

= 11+627$\frac{11+6\sqrt{2}}{7}$

Therefore,  117$\frac{11}{7}$ + 672$\frac{6}{7}\sqrt{2}$ = a + b 2$\sqrt{2}$

So, on comparing the LHS and the RHS, we get:

a = 117$\frac{11}{7}$ and b = 67$\frac{6}{7}$

Question 16: 565+6$\frac{5-\sqrt{6}}{5+\sqrt{6}}$ a – b 6$\sqrt{6}$

Solution:

We have:

565+6$\frac{5-\sqrt{6}}{5+\sqrt{6}}$

Now, rationalizing the denominator of the given number by multiplying both the denominator and the numerator with 5 – 6$\sqrt{6}$, we get:

565+6$\frac{5-\sqrt{6}}{5+\sqrt{6}}$ = 565+6$\frac{5-\sqrt{6}}{5+\sqrt{6}}$ x 5656$\frac{5-\sqrt{6}}{5-\sqrt{6}}$

= (52)2(5)2(6)2$\frac{(5-\sqrt{2})^{2}}{(5)^{2}-(\sqrt{6})^{2}}$

= 25+6106256$\frac{25+6-10\sqrt{6}}{25-6}$

= 3110619$\frac{31-10\sqrt{6}}{19}$

= 3119$\frac{31}{19}$1019$\frac{10}{19}$ 6$\sqrt{6}$

Therefore, 3119$\frac{31}{19}$10196$\frac{10}{19}\sqrt{6}$ = a – b 6$\sqrt{6}$

So, on comparing the LHS and the RHS, we get:

a =  3119$\frac{31}{19}$ and b =  1019$\frac{10}{19}$

Question 17: 5+237+43$\frac{5+2\sqrt{3}}{7+4\sqrt{3}}$ = a – b 3$\sqrt{3}$

Solution:

5+237+43$\frac{5+2\sqrt{3}}{7+4\sqrt{3}}$

Now, rationalizing the denominator of the given number by multiplying both the denominator and the numerator with 7 – 43$\sqrt{3}$, we get:

5+237+43$\frac{5+2\sqrt{3}}{7+4\sqrt{3}}$ = 5+237+43$\frac{5+2\sqrt{3}}{7+4\sqrt{3}}$ x 743743$\frac{7-4\sqrt{3}}{7-4\sqrt{3}}$

= (3514320324)(7)2(43)2$\frac{(35-14\sqrt{3}-20\sqrt{3}-24)}{(7)^{2}-(4\sqrt{3})^{2}}$

= 11634948$\frac{11-6\sqrt{3}}{49-48}$

= 11 – 6 3$\sqrt{3}$

Therefore, 11 – 6 3$\sqrt{3}$ = a – b 3$\sqrt{3}$

So, on comparing the LHS and the RHS, we get:

a = 11 and b = 6

#### Practise This Question

In the given figure, A=60 and ABC=80 , find DPC and BQC.