Question

sum of the areas of two squares is 640 m2.if the difference of their perimeters is64m,find the sides of the two squares.

Answer 1

let the sides of the squares be a and b respectively,

then

a²+b²=640,.........eq(1),

4a-4b=64,

a-b=16,.......................eq(2),

(a-b)²=16²,

a²+b²-2ab=256,

640-2ab=256,

then

640-256=2ab,

384=2ab,

then

ab=384/2,

ab=192,....................eq(3),


now

(a+b)²=(a-b)²+4ab,

(a+b)²=16²+4×192,

(a+b)²=256+768,

(a+b)²=1024,

then

(a+b)=√1024,

(a+b)=32,...........eq(4),


now from the eq(2) and eq(4), we get

a=24 m,

b=8 m

Answer 2

If a and b are sides of 2 squares then a²+b²=640 and diff of perimeters i.e; 4(a-b)=64

Solve to get a= 24 and b=8