Question

Sum of the areas of two squares is 400 sq. cm. If the difference of their perimeters is 16 cm, find the sides of the two squares.

Answer 1

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 = 16 m

Perimeter of second square - perimeter of first square = 16

Perimeter of second square - 4x = 16

Perimeter of second square = 16 + 4x

Length of square = perimeter of square/4

Length of each side of second square = (16 + 4x)/4  

= 4(4 + x)/4  

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

Given : Sum of the area of two squares = 400 m²

Area of first square + Area of second square = 400 m²

x² + (4 + x)² = 400

[Area of a square = side²]

x² + (4)² + x² + 2 × 4 × x =  400

[(a+b)² = a² + b² + 2ab]

2x² + 16 + 8x = 400

2x² + 8x + 16 - 400 = 0

2x² + 8x - 384 = 0

2(x² + 4x - 192) = 0

x² + 4x - 192 = 0

x² + 16x - 12x - 192 = 0  

[By middle term splitting]

x(x + 16) - 12 (x + 16) = 0

(x  + 16)(x - 12) = 0

(x  + 16) = 0 or (x - 12) = 0

x = - 16  or x = 12

Since, side can't be negative ,so x ≠ - 16

Therefore, x = 12

Side of first square = (x) = 12 m

Side of second square = 4 + x  

Side of second square = 4 + 12 = 16 m

Side of second square = 16 m

Hence, the side of a square is 12 m and side of second square is 16 m.

HOPE THIS ANSWER WILL HELP YOU

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