Sum of the area of two …
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Yogita Ingle 6 years, 10 months ago
Sum of the areas of two squares = 468 m2
Let a and b be the sides of the two squares.
⇒a2 + b2 = 468…(1)
Also given that, the difference of their perimeters = 24m
⇒4a - 4b = 24
⇒a - b = 6
⇒a = b + 6…(2)
We need to find the sides of the two squares.
Substituting the value of a from equation (2) in equation (1), we have,
(b + 6)2 + b2 = 468
⇒b2 + 62 + 2 × b × 6 + b2 = 468
⇒2b2 + 36 + 12b = 468
⇒2b2 + 36 + 12b - 468 = 0
⇒2b2 + 12b - 432 = 0
⇒b2 + 6b - 216 = 0
⇒b2 + 18b - 12b - 216 = 0
⇒b(b + 18) - 12(b + 18) = 0
⇒(b + 18)(b - 12) = 0
⇒b + 18 = 0 or b - 12 = 0
⇒b = -18 = 0 or b = 12
Side cannot be negative and hence b = 12 m.
Therefore, a = b + 6 = 12 + 6 = 18 m.
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