Question

The sum of the areas of two squares is 640m2 if the difference in the perimeters be 64 find the sides of the 2 squares

Answer 1

Let sides be a and A

so, we are given

A^2 - a^2 = 640 ------------------------(1)

and 4A - 4a = 64

so, A -a = 16

so, A = a + 16 ----------------------------(2)

Solving (1) and (2)

A = 28 and a = 12

Hence, sides are 28 m and 12 m

Answer 2

Answer:

x = 24 and y = 8.

Step-by-step explanation:

Let side of first square be x and that of another be y

Ist case

area of square = (side)2

x2 + y2 =640 --------------------------   (i) 

2nd case

4x -4y =64

-> x-y=16

x=16+y

put value of x in 1

(16+y)2+y2=640

256+y2+32y+y2 =640

2y2 +32y =384

y2+16y=192

y2 + 24y-8y -192=0

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

y=8

and x=24

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