# RD Sharma Class 9 Rationalization Exercise 3.1 Solutions On this page you will find Maths RD Sharma Class 9 Rationalization Exercise 3.1 Solutions. The solutions provided here are to help students practice math problems and get better at solving difficult chapter questions. Maths chapter 3 for class 9 deals with the topic of triangles and it is one of the most important chapters.

## Download RD Sharma Class 9 Rationalization Exercise 3.1 Solutions in PDF

Students can use these solutions to overcome the fear of maths and the solutions have been designed in such a way that it enables them to discover easy ways to solve different problems. These solutions can help students in refining their maths fluency and problem-solving skills. Students can go through the RD Sharma solutions for class 9 chapter 3 Rationalization Exercise 3.1 below and it will be beneficial for them.

## RD Sharma Class 9 Rationalization Exercise 3.1 Solutions

1. Simplify each of the following:
(i) $\sqrt{4} \times \sqrt{16}$
(ii) $\frac{\sqrt{1250}}{\sqrt{2}}$
Sol:
(i) $\frac{\sqrt{1250}}{\sqrt{2}}$ $=\sqrt{4 \times 16}$
$=\sqrt{64}$ $=\left(4^{3}\right)^{\frac{1}{3}}$
$=4\left(3 \times \frac{1}{3}\right)$
$=4^{1}$
$=4$   $=\sqrt{625}$ $=15\left(4 \times \frac{1}{4}\right)$
$=15$

2. Simplify the following expressions:
(i) $(4+\sqrt{7})(3+\sqrt{2})$
(ii) $(3+\sqrt{3})(5-\sqrt{2})$
(iii) $(\sqrt{5}-2)(\sqrt{3}-\sqrt{5})$
Solution:
(i) $(4+\sqrt{7})(3+\sqrt{2})$
$=12+4 \sqrt{2}+3 \sqrt{7}+\sqrt{7 \times 2}$
$=12+4 \sqrt{2}+3 \sqrt{7}+\sqrt{14}$
(ii) $(3+\sqrt{3})(5-\sqrt{2})$
$=15-3 \sqrt{2}+5 \sqrt{3}-\sqrt{3 \times 2}$
$=15-3 \sqrt{2}+5 \sqrt{3}-\sqrt{6}$
(iii) $(\sqrt{5}-2)(\sqrt{3}-\sqrt{5})$
$=\sqrt{15}-\sqrt{25}-2 \sqrt{3}+2 \sqrt{5}$
$=\sqrt{15}-\sqrt{5 \times 5}-2 \sqrt{3}+2 \sqrt{5}$
$=\sqrt{15}-5-2 \sqrt{3}+2 \sqrt{5}$

3. Simplify the following expressions:
(i) $(11+\sqrt{11})(11-\sqrt{11})$
(ii) $(5+\sqrt{7})(5-\sqrt{7})$
(iii) $(\sqrt{8}-\sqrt{2})(\sqrt{8}+\sqrt{2})$
(iv) $(3+\sqrt{3})(3-\sqrt{3})$
(v) $(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})$
Solution:
(i) $(11+\sqrt{11})(11-\sqrt{11})$
As we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
So, $11^{2}-11$
$121-11=110$
(ii) $(5+\sqrt{7})(5-\sqrt{7})$
As we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
So, $5^{2}-7$
$25-7=18$
(iii) $(\sqrt{8}-\sqrt{2})(\sqrt{8}+\sqrt{2})$
As we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$\sqrt{8 \times 8}-\sqrt{2 \times 2}=8-2$
$=6$
(iv) $(3+\sqrt{3})(3-\sqrt{3})$
As we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=9-\sqrt{3 \times 3}$
$=6$
(v) $(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})$
As we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\sqrt{5 \times 5}-\sqrt{2 \times 2}$
$=5-2$
$=3$

4. Simplify the following expressions:
(i) $(\sqrt{3}+\sqrt{7})^{2}$
(ii) $(\sqrt{5}-\sqrt{3})^{2}$
(iii) $(2 \sqrt{5}+3 \sqrt{2})^{2}$
Solution:
(i) $(\sqrt{3}+\sqrt{7})^{2}$
As we know, $(a+b)^{2}=\left(a^{2}+2 \times a \times b+b^{2}\right)$
$=\sqrt{3^{2}}+2 \times \sqrt{3} \times \sqrt{7}+\sqrt{7^{2}}$
$=3+2 \times \sqrt{3 \times 7}+7$
$=10+2 \times \sqrt{21}$
(ii) $(\sqrt{5}-\sqrt{3})^{2}$
As we know,$(a-b)^{2}=\left(a^{2}-2 \times a \times b+b^{2}\right)$
(iii) $(2 \sqrt{5}+3 \sqrt{2})^{2}$
As we know,$(a+b)^{2}=\left(a^{2}+2 \times a \times b+b^{2}\right)$
$=4 \sqrt{5 \times 5}+2 \times 2 \sqrt{5} \times 3 \sqrt{2}+9 \sqrt{2 \times 2}$
$=20+12 \sqrt{10}+18$
$=28+12 \sqrt{10}$

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