Simplify : 11 - 7 11 + 7

Given:- 11 - 7 11 + 7 We know that a - b a + b = a 2 - b 2 Therefore, 11 - 7 11 + 7 = 11 - 7 11 - 7 11 + 7 = 4 So, the answer is 4 .

Given:- 1 2 Rationalization can be defined as the process by which we eliminate radicals present in the denominators of fractions. We know that ratinalization of a b is a × b b × b = a b b . Multiply both the numerator and denominator by 2 . = 1 2 × 2 2 = 2 2 2 = 2 2 So, the required answer is 2 2 .

Given, 1 2 - 3 Rationalization can be defined as the process by which we eliminate radicals present in the denominators of fractions. Multiply both numerator and denominator by 2 - 3 = 1 2 + 3 × 2 - 3 2 - 3 = 2 - 3 2 2 - 3 2 = 2 - 3 4 - 3 = 2 - 3 1 = 2 - 3 So, the answer is 2 - 3 .

Given 5 3 - 5 Rationalization can be defined as the process by which we eliminate radicals present in the denominators of fractions. Multiply both numerator and denominator by 3 + 5 = 5 3 - 5 × 3 + 5 3 + 5 = 5 × 5 + 3 3 - 5 = 5 × 5 + 3 - 2 = - 5 2 5 + 3

Rationalize: 1 7 + 3 2

Given:- 1 7 + 3 2 Rationalization can be defined as the process by which we eliminate radicals present in the denominators of fractions. Divide both numerator and denominator by 7 - 3 2 . = 1 7 + 3 2 × 7 - 3 2 7 - 3 2 = 7 - 3 2 7 + 3 2 7 - 3 2 = 7 - 3 2 49 - 3 × 3 × 2 × 2 = 7 - 3 2 49 - 9 × 2 = 7 - 3 2 49 - 18 = 7 - 3 2 31 So, the answer is 7 - 3 2 31 .

Mark 3.5 on a number line.

Instructions to draw :- 1. First draw a line segment ' A B ' Which is of 3.5 c m . 2. Now extend the line segment to ' C ' so that ' B C ' = 1 c m . 3. Identify the midpoint of A C and mark as O . 4. Draw a semicircle considering O to be its center and O A as its radius. 5. Now draw a line perpendicular to A C such that B D ⊥ A C , which intersects the semicircle at the point D . 6. Then B D = 3.5 . 7. Having B as its center and B D as the radius draw an arc meeting A C extended to E . Then B E = B D = 3.5 .

If a - b 3 = 2 - 3 2 + 3 , then find the value of ' a ' and ' b ' .

Given, a - b 3 = 2 - 3 2 + 3 Now, Rationalize the denominator. Rationalization can be defined as the process by which we eliminate radicals present in the denominators of fractions. Multiply both numerator and denominator by 2 - 3 on RHS side, = 2 - 3 × 2 - 3 2 + 3 × 2 - 3 = 2 - 3 2 2 2 - 3 2 = 2 2 - 2 × 2 × 3 + 3 2 4 - 3 = 4 - 4 3 + 3 = 7 - 4 3 Therefore, a - b 3 = 7 - 4 3 By comparing corresponding parts of LHS and RHS we get, a = 7 , b = 4

If x = 2 + 3 , then find the value of x 2 + 1 x 2 .

Given that: x = 2 + 3 1 x = 1 2 + 3 Rationalize denominator. Multiply both numerator and denominator by 2 - 3 = 1 2 + 3 × 2 - 3 2 - 3 = 2 - 3 2 2 - 3 2 = 2 - 3 4 - 3 1 x = 2 - 3 Therefore, x 2 = 2 + 3 2 = 2 2 + 3 2 + 2 × 2 × 3 = 4 + 3 + 4 3 = 7 + 4 3 1 x 2 = 2 - 3 2 = 2 2 + 3 2 - 2 × 2 × 3 = 4 + 3 - 4 3 = 7 - 4 3 ⇒ x 2 + 1 x 2 = 7 + 4 3 + 7 - 4 3 = 7 + 4 3 + 7 - 4 3 = 7 + 7 + 4 3 - 4 3 Therefore, x 2 + 1 x 2 = 14

E M B I B E

Full Marks Mathematics 9>Chapter 1 - Number Systems>TEST YOUR SKILLS

R Chandra Solutions for Chapter: Number Systems, Exercise 4: TEST YOUR SKILLS

Author:R Chandra

R Chandra Mathematics Solutions for Exercise - R Chandra Solutions for Chapter: Number Systems, Exercise 4: TEST YOUR SKILLS

Attempt the practice questions on Chapter 1: Number Systems, Exercise 4: TEST YOUR SKILLS with hints and solutions to strengthen your understanding. Full Marks Mathematics 9 solutions are prepared by Experienced Embibe Experts.