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Squares and Square Roots

    Exercise 5.2

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8th class > Squares and Square Roots > Exercise 5.2

Exercise 5.2

  1. Find the square of the following numbers.

Instructions

(i) 322
The square of 32 is .
322 = 30+22 = 302 + 2 × 30 × 2 + 22 = 900 + + = 1024
(ii) 352
The square of 35 is .
352 = 30+52 = 302 + 2 × 30 × 5 + 52 = 900 + + = 1225
(iii) 862
The square of 86 is .
862 = 90−42 = 902 - 2 × 90 × 4 + 42 = - + = 7396
(iv) 932
The square of 93 is .
932 = 90+32 = 902 + 2 × 90 × 3 + 32 = + + = 8649
(v) 712
The square of 71 is .
712 = 70+12 = 702 + 2 × 70 × 1 + 12 = + + = 5041
(vi) 462
The square of 46 is .
462 = 50−42 = 502 - 2 × 50 × 4 + 42 = - + = 2116
  1. Write a Pythagorean triplet whose one member is:

Instructions

(i) 6
We know that: triplets are in the form 2m, m2−1, m2+1
Let m2−1 = 6 ⇒ m2 = . Thus, the value of m an integer.
Now let 2m = 6 ⇒ m = which is an integer.
Thus, the other members are: m2−1 = 32−1 = and m2+1 = 32+1 =
One possible Pythagorean triplet is 6, 8, 10.
(ii) 14
Let m2−1 = 14 ⇒ m2 = . Thus, the value of m an integer.
Now let 2m = 14 ⇒ m = which is an integer.
Thus, the other members are: m2−1 = 72−1 = and m2+1 = 72+1 =
One possible Pythagorean triplet is 14, 48, 50.
(iii) 16
Let 2m = 16 ⇒ m = which is an integer.
Thus, the other members are: m2−1 = 82−1 = and m2+1 = 82+1 =
One possible Pythagorean triplet is 16, 63, 65.
(iv) 18
Let 2m = 18 ⇒ m = which is an integer.
Thus, the other members are: m2−1 = 92−1 = and m2+1 = 92+1 =
One possible Pythagorean triplet is 18, 80, 82.