The sum of areas of two squares is 640 m sq.if difference of their perimeter is 64m.find the sides of two squares
Answers
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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