Question

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

Answer 1

Ans : sides of two squares are 28 m and 12 m
Let a and b sides of two squares

a^2 - b^2 = 640 ... (i)
4a - 4b = 64
a - b = 16 (ii)

a = b + 16

Putting value of a in eq (i)

(16+b)^3 - b^2 = 640
by solving,
we get b = 12m

a=16+12
a=28m

Hope it helps. ...!!

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.

HOPE THIS ANSWER WILL HELP YOU