Question

Sum of the areas of two squares is 640 sq. metre. The difference of their perimeter is 64 m . find the sides of the 2 squares .

Answer 1

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Answer 2

SOLUTION :  

Let the length of each side of a square be x . Then its perimeter = 4x

[Perimeter of a square = 4 × side]

Given : Difference of the perimeters of two squares = 64 m

Perimeter of second square - perimeter of first square = 64

Perimeter of second square - 4x = 64

Perimeter of second square = 64 + 4x

Length of square = perimeter of square/4

Length of each side of second square = (64 + 4x)/4  

= 4(16 + x)/4  

Length of each side of second square = (16 + x) m

Given : Sum of the area of two squares = 640 m²

Area of first square + Area of second square = 640 m²

x² + (16 + x)² = 640  

[Area of a square = side²]

x² + (16)² + x² + 2 ×16× x =  640

[(a+b)² = a² + b² + 2ab]

2x² + 256 + 32x = 640

2x² + 32x + 256 - 640 = 0

2x² + 32x - 384 = 0

2(x² + 16x - 192) = 0

x² + 16x - 192 = 0

x² + 24x - 8x - 192 = 0  

[By middle term splitting]

x(x + 24) - 8 (x + 24) = 0

(x  + 24)(x - 8) = 0

(x  + 24) = 0 or (x - 8) = 0

x = - 24  or x = 8

Since, side can't be negative ,so x ≠ - 24

Therefore, x = 8

Side of first square = (x) = 8 m

Side of second square = 16 + x  

Side of second square = 16 + 8 = 24 m

Side of second square = 24 m

Hence, the side of a square is 8 m and side of second square is 24 m.





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