Practice Set 2.1 | Q 6 | Page 39
Find the side and perimeter of a square whose diagonal is 10 cm ?
Solution
It is given that ABCD is a square.
∴ AB = BC = CD = DA = a (say)
According to Pythagoras theorem, in ∆ABD
[{AB}^2 + {AD}^2 = {BD}^2 ]
[ Rightarrow a^2 + a^2 = {10}^2 ]
[ Rightarrow 2 a^2 = 100]
[ Rightarrow a^2 = 50]
[ Rightarrow a = sqrt{50}]
[ Rightarrow a = 5sqrt{2} cm]
Hence, the side of the square is 5[sqrt{2}] cm.
Now,
Perimeter of a square = [4 times left( side right)]
= [4 times a]
= [4 times 5sqrt{2}]
= [20sqrt{2}] cm
Hence, the perimeter of the square is 20 [sqrt{2}] cm
Explanation:-
The problem states that ABCD is a square, so AB = BC = CD = DA = a, where a is the side of the square.
By applying the Pythagorean theorem to right triangle ABD, we have:
AB² + AD² = BD²
Substituting AB = AD = a and BD = 10, we get:
a² + a² = 10²
Simplifying this equation, we get:
2a² = 100
Dividing by 2, we get:
a² = 50
Taking the square root on both sides, we get:
a = √50
Simplifying this expression, we get:
a = 5√2 cm
Therefore, the side of the square is 5√2 cm.
The perimeter of a square is given by:
Perimeter = 4 × side
Substituting the value of a, we get:
Perimeter = 4 × 5√2
Simplifying this expression, we get:
Perimeter = 20√2 cm
Therefore, the perimeter of the square is 20√2 cm.
Chapter 2 – Pythagoras Theorem- Text Book Solution
Practice Set 2.1 | Q 6 | Page 39