For this Class 8 NCERT book exercise 5.2 solution, we explore concepts dealing with the very important concepts of Squaring Numbers and Pythagorean triplets. We will dive deep into these 2 areas, solve questions from the NCERT book, and try a few more challenging questions.
Squaring Numbers:
First, we’ll solve the squares of numbers. Squaring a number means multiplying the number by itself. It sounds simple, but dealing with larger numbers becomes interesting. We’ll learn an easy method to find the squares of bigger numbers without actual multiplication. We will use the formula (a+b)² = a² + 2ab + b². This method simplifies our calculations and also clears our algebraic skills.
Pythagorean Triplets:
Next, we’ll solve Pythagorean triplets. These sets of three positive integers fit perfectly into the Pythagorean Theorem – a² + b² = c². This theorem is fundamental in geometry. It helps in understanding right-angled triangles. Finding Pythagorean triplets is like solving a puzzle. Here, we use different methods to discover these unique numbers.
We’re training our minds to think logically and creatively by exploring these areas.
Question and Answers for Class 8 Maths Exercise 5.2 Chapter 5 Squares and Square Roots
Question 1. Find the square of the following numbers.
(i) 32 (ii) 35 (iii) 86 (iv) 93 (v) 71 (vi) 46
Solution
(i) 32
= 32²
= (30 + 2)² (32 can be written as (30 + 2))
= (30 + 2)(30 + 2)
= 30(30 + 2) + 2(30 + 2)
= 30² + 30 × 2 + 2 × 30 + 2²
= 900 + 60 + 60 + 4
= 1024
(ii) 35
= 35²
= (30 + 5)² (35 can be written as (30 + 5))
= (30 + 5)(30 + 5)
= 30(30 + 5) + 5(30 + 5)
= 30² + 30 × 5 + 5 × 30 + 5²
= 900 + 150 + 150 + 25
= 1225
(iii) 86
= 86²
= (80 + 6)² (86 can be written as (80 + 6))
= (80 + 6)(80 + 6)
= 80(80 + 6) + 6(80 + 6)
= 80² + 80 × 6 + 6 × 80 + 6²
= 6400 + 480 + 480 + 36
= 7396
(iv) 93
= 93²
= (90 + 3)² (93 can be written as (90 + 3))
= (90 + 3)(90 + 3)
= 90(90 + 3) + 3(90 + 3)
= 90² + 90 × 3 + 3 × 90 + 3²
= 8100 + 270 + 270 + 9
= 8649
(v) 71
= 71²
= (70 + 1)² (71 can be written as (70 + 1))
= (70 + 1)(70 + 1)
= 70(70 + 1) + 1(70 + 1)
= 70² + 70 × 1 + 1 × 70 + 1²
= 4900 + 70 + 70 + 1
= 5041
(vi) 46
= 46²
= (40 + 6)² (46 can be written as (40 + 6))
= (40 + 6)(40 + 6)
= 40(40 + 6) + 6(40 + 6)
= 40² + 40 × 6 + 6 × 40 + 6²
= 1600 + 240 + 240 + 36
= 2116
Question 2. Write a Pythagorean triplet whose one member is.
(i) 6 (ii) 14 (iii) 16 (iv) 18
Hint:
For a natural number m > 1
(2m)² + (m² – 1)² = (m² + 1)²
Here 2m, (m² – 1), and (m² + 1) form Pythagorean triplet
(i) 6
If we take 2m = 6, then m = 6/2 = 3
Thus, m² – 1 = 3² – 1 = 9 – 1 = 8
m² + 1 = 3² + 1 = 9 + 1 = 10
Therefore, the required triplet is 6, 8, 10
(ii) 14
If we take 2m = 14, then m = 14/2 = 7
Thus, m² – 1 = 7² – 1 = 49 – 1 = 48
m² + 1 = 7² + 1 = 49 + 1 = 50
Therefore, the required triplet is 14, 48, 50
(iii) 16
If we take 2m = 16, then m = 16/2 = 8
Thus, m² – 1 = 8² – 1 = 64 – 1 = 63
m² + 1 = 8² + 1 = 64 + 1 = 65
Therefore, the required triplet is 16, 63, 65
(iv) 18
If we take 2m = 18, then m = 18/2 = 9
Thus, m² – 1 = 9² – 1 = 81 – 1 = 80
m² + 1 = 9² + 1 = 81 + 1 = 82
Therefore, the required triplet is 18, 80, 82
Challenging Questions for Exercise 5.2 Chapter 5 Squares and Square Roots
Question 1. Find the square of the following numbers using the identity (a+b)² = a² + 2ab + b².
(i) 47 (ii) 59 (iii) 76 (iv) 88 (v) 102 (vi) 115
Solution:
(i) 47² = (40 + 7)²
= 40² + 2 × 40 × 7 + 7²
= 1600 + 560 + 49
= 2209
(ii) 59² = (50 + 9)²
= 50² + 2 × 50 × 9 + 9²
= 2500 + 900 + 81
= 3481
(iii) 76² = (70 + 6)²
= 70² + 2 × 70 × 6 + 6²
= 4900 + 840 + 36
= 5776
(iv) 88² = (80 + 8)²
= 80² + 2 × 80 × 8 + 8²
= 6400 + 1280 + 64
= 7744
(v) 102² = (100 + 2)²
= 100² + 2 × 100 × 2 + 2²
= 10000 + 400 + 4
= 10404
(vi) 115² = (100 + 15)²
= 100² + 2 × 100 × 15 + 15²
= 10000 + 3000 + 225
= 13225
Question 2. Write a Pythagorean triplet whose one member is a multiple of 10.
(i) 10 (ii) 20 (iii) 30 (iv) 40 (v) 50 (vi) 60
Solution:
(i) For 10 (Using 2m = 10, m = 5)
m² – 1 = 5² – 1 = 25 – 1 = 24
m² + 1 = 5² + 1 = 25 + 1 = 26
Therefore, the required triplet is 10, 24, 26.
(ii) For 20 (Using 2m = 20, m = 10)
m² – 1 = 10² – 1 = 100 – 1 = 99
m² + 1 = 10² + 1 = 100 + 1 = 101
Therefore, the required triplet is 20, 99, 101.
(iii) For 30 (Using 2m = 30, m = 15)
m² – 1 = 15² – 1 = 225 – 1 = 224
m² + 1 = 15² + 1 = 225 + 1 = 226
Therefore, the required triplet is 30, 224, 226.
(iv) For 40 (Using 2m = 40, m = 20)
m² – 1 = 20² – 1 = 400 – 1 = 399
m² + 1 = 20² + 1 = 400 + 1 = 401
Therefore, the required triplet is 40, 399, 401.
(v) For 50 (Using 2m = 50, m = 25)
m² – 1 = 25² – 1 = 625 – 1 = 624
m² + 1 = 25² + 1 = 625 + 1 = 626
Therefore, the required triplet is 50, 624, 626.
(vi) For 60 (Using 2m = 60, m = 30)
m² – 1 = 30² – 1 = 900 – 1 = 899
m² + 1 = 30² + 1 = 900 + 1 = 901
Therefore, the required triplet is 60, 899, 901.
