Question

Sum of the areas of two square is 468m2. if the difference of their perimeters is 24 m, Find the sides of the two Squares. ​

Answer 1

Answer:

964m

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

Let the side of the first square be 'a'm and that of the second be

′

A

′

m.

Area of the first square =a

2

sq m.

Area of the second square =A

2

sq m.

Their perimeters would be 4a and 4A respectively.

Given 4A−4a=24

A−a=6 --(1)

A

2

+a

2

=468 --(2)

From (1), A=a+6

Substituting for A in (2), we get

(a+6)

2

+a

2

=468

a

2

+12a+36+a

2

=468

2a

2

+12a+36=468

a

2

+6a+18=234

a

2

+6a−216=0

a

2

+18a−12a−216=0

a(a+18)−12(a+18)=0

(a−12)(a+18)=0

a=12,−18

So, the side of the first square is 12 m. and the side of the second square is 18 m.