Question

sum of the area of two squares is 468 m². if the difference of their perimeters 24m, find the sides of the two squares​

Answer 1

Step-by-step explanation:

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.

Answer 2

Step-by-step explanation:

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