GSEB Class 8 Maths Solutions Chapter 6 Square and Square Roots Exercise 6.2

Question 1. Find the square of the following numbers containing 5 in unit place?
Answer:
1. To find the square of 32:
\( (32)^2 = (30 + 2)^2 \)
\( = 30^2 + 2(30)(2) + (2)^2 \)
\( = 900 + 120 + 4 = 1024 \)

2. To find the square of 35:
\( (35)^2 = (30 + 5)^2 \)
\( = (30)^2 + 2(30)(5) + (5)^2 \)
\( = 900 + 300 + 25 \)
\( = 1200 + 25 = 1225 \)
Second method (useful when the unit digit is 5):
\( 35^2 = 3 \times (3 + 1) \times 100 + 25 \)
\( = 3 \times 4 \times 100 + 25 \)
\( = 1200 + 25 = 1225 \)

3. To find the square of 86:
\( (86)^2 = (80 + 6)^2 \)
\( = (80)^2 + 2(80)(6) + (6)^2 \)
\( = 6400 + 960 + 36 = 7396 \)

4. To find the square of 93:
\( (93)^2 = (90 + 3)^2 \)
\( = (90)^2 + 2(90)(3) + (3)^2 \)
\( = 8100 + 540 + 9 = 8649 \)

5. To find the square of 71:
\( (71)^2 = (70 + 1)^2 \)
\( = (70)^2 + 2(70)(1) + (1)^2 \)
\( = 4900 + 140 + 1 = 5041 \)

6. To find the square of 46:
\( (46)^2 = (40 + 6)^2 \)
\( = (40)^2 + 2(40)(6) + (6)^2 \)
\( = 1600 + 480 + 36 = 2116 \)
In simple words: To find the square of a number, we can use the identity \( (a+b)^2 = a^2 + 2ab + b^2 \). This means we break the number into two parts, like 32 into 30 + 2, and then apply the formula to easily calculate its square. For numbers ending in 5, there is a quicker method where you multiply the tens digit by one more than itself, then multiply by 100 and add 25.

Exam Tip: Remember the algebraic identity \( (a+b)^2 = a^2 + 2ab + b^2 \) for finding squares of numbers. For numbers ending in 5, you can use the trick: \( (10a+5)^2 = a(a+1) \times 100 + 25 \).

Question 2. Write a Pythagorean tripler whose one member is
Answer:
1. If one member is 6:
Let \( 2n = 6 \). This gives \( n = 3 \).
Now, calculate the other two members:
\( n^2 - 1 = 3^2 - 1 = 9 - 1 = 8 \).
\( n^2 + 1 = 3^2 + 1 = 9 + 1 = 10 \).
Therefore, the required Pythagorean triplet is 6, 8, 10.

2. If one member is 14:
Let \( 2n = 14 \). This gives \( n = 7 \).
Now, calculate the other two members:
\( n^2 - 1 = 7^2 - 1 = 49 - 1 = 48 \).
\( n^2 + 1 = 7^2 + 1 = 49 + 1 = 50 \).
Therefore, the required Pythagorean triplet is 14, 48, 50.

3. If one member is 16:
Let \( 2n = 16 \). This gives \( n = 8 \).
Now, calculate the other two members:
\( n^2 - 1 = 8^2 - 1 = 64 - 1 = 63 \).
\( n^2 + 1 = 8^2 + 1 = 64 + 1 = 65 \).
Therefore, the required Pythagorean triplet is 16, 63, 65.

4. If one member is 18:
Let \( 2n = 18 \). This gives \( n = 9 \).
Now, calculate the other two members:
\( n^2 - 1 = 9^2 - 1 = 81 - 1 = 80 \).
\( n^2 + 1 = 9^2 + 1 = 81 + 1 = 82 \).
The required Pythagorean triplet is 18, 80, 82.
In simple words: A Pythagorean triplet consists of three positive integers a, b, and c, such that \( a^2 + b^2 = c^2 \). We often use the formula \( 2n, n^2 - 1, \) and \( n^2 + 1 \) to generate these triplets. By setting the given member equal to \( 2n \), we can find \( n \), and then use it to calculate the other two members.

Exam Tip: Remember the general form for Pythagorean triplets: \( 2n, n^2-1, n^2+1 \). Always check if the smallest member is an even number; if it is, use \( 2n = \text{member} \). If it's odd, other formulas may be needed, but for most problems, the \( 2n \) formula works.