Using suitable identity, evaluate the following : 101 × 102
Solution:
Given, 101 × 102
We have to evaluate the given expression using a suitable identity.
101 = 100 + 1
102 = 100 + 2
101 × 102 = (100 + 1) × (100 + 2)
Using algebraic identity,
(a + b)(a + c) = a² + ac + ab + bc = a² + a(b + c) + bc
(100 + 1) × (100 + 2) = (100)² + 100(1 + 2) + 1(2)
= 10000 + 100(3) + 2
= 10000 + 300 + 2
= 10000 + 302
= 10302
Therefore, (100 + 1) × (100 + 2) = 10302
✦ Try This: Using suitable identity, evaluate the following : 101³
Given, 101³
We have to evaluate the given expression using a suitable identity.
101³ = (100 + 1)³
Using algebraic identity,
(a + b)³ = a³ + b³ + 3ab(a + b)
(100 + 1)³ = (100)³ + (1)³ + 3(100)(1)(100 + 1)
= 1000000 + 1 + 300(101)
= 1000001 + 30300
= 1030301
Therefore, 101³ = 1030301
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 2
NCERT Exemplar Class 9 Maths Exercise 2.3 Problem 25(ii)
Using suitable identity, evaluate the following : 101 × 102
Summary:
Polynomials are types of expressions. On evaluating, the value of (100 + 1) × (100 + 2) is 10302
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