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The sum of the areas of two squares is 640 $m^{2}$ . If the difference in their perimeters is 64 m, the sides of the two squares (in m) are?

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24m and 40m

12m and 32m

24m and 16m

24m and 8m

answer is B.

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Detailed Solution

Given that the difference in the perimeter of the two squares

= 64 m

and sum of the areas of two squares

= 640 m^{2}

.
Suppose, side of the first square is

r

unit and the side of the second square is

s

unit.
We know that,
Perimeter of square

= 4 \times side

Then,

⟹ (4 \times side) - (4 \times side) = 64

⟹ 4 r - 4 s = 64

⟹ 4 (r - s) = 64

⟹ r - s = 16

⟹ s = 16 + r

….(1)
And area of square

= sid e^{2}

Then,

⟹ {(r)}^{2} + {(s)}^{2} = 640

⟹ r^{2} + {(r + 16)}^{2} = 640

{Using (1)}

⟹ r^{2} + (r^{2} + 32 r + 256) = 640

⟹ 2 r^{2} + 32 r + 256 - 640 = 0

⟹ {2 r}^{2} + 32 r - 384 = 0

⟹ r^{2} + 16 r - 192 = 0

⟹ r^{2} + 24 r - 8 r - 192 = 0

⟹ r (r + 24) - 8 (r + 24) = 0

⟹ (r - 8) + (r + 24)

⟹ r - 8 = 0 or r + 24 = 0

∴ r = 8 or r = - 24

We will take the value

r = 8

because the side of a square cannot be negative.
So, the side of the first square is 8m.
And for the second square,

∴ s = 16 + r

= 16 + 8

= 24

∴ The side of the second square is 24m.
Therefore, the sides of the two squares (in m) are 8m and 24m.
Hence, the correct option is 2.

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