the sum of the are OF 2 SQUARES IS 640 .IF THE DIFFERENCE IN PERIMETER IS 64 FIND SIDES
Answers
Answered by
1
Let the two squares has the side x and y resp.
A.T.Q
x^2+y^2 =640…………i
also
4x -4y =64
x -y =16………………ii
now in equation i
(x-y)^2+2xy=640
(16)^2 + 2xy = 640 (from ii)
256+ 2xy = 640
2xy=384……………iii
now equation i can be re written as
(x + y)^2 -2xy =640
(x + y)^2- 384=640 (from iii)
(x+y)^2=1024
x+y= 32……………iv
solving equation ii and iv
we get
x= 24
and y= 8
A.T.Q
x^2+y^2 =640…………i
also
4x -4y =64
x -y =16………………ii
now in equation i
(x-y)^2+2xy=640
(16)^2 + 2xy = 640 (from ii)
256+ 2xy = 640
2xy=384……………iii
now equation i can be re written as
(x + y)^2 -2xy =640
(x + y)^2- 384=640 (from iii)
(x+y)^2=1024
x+y= 32……………iv
solving equation ii and iv
we get
x= 24
and y= 8
Similar questions