Thread: Find Length and Width of Rectangle

If the perimeter of a rectangle is 20 feet and the diagonal is 2 (sqrt{13}) feet, then what are the length and width?

if length = x ft and breadth = y ft

then by given condition 2(x+y) = 20 .....(1)
and x^2 + y^2 = (2sqr(13))^2 = 52 .....(2)
from eq(1) x+y = 10........(3)
from eq (2) & eq(3)
2xy=(x+y)^2 -(x62+y^2) = 100-52 = 48


now (x-y)^2 = x^2+y^2 - 2xy =52 - 48 =4
or x-y =2 .......(4)

solving (3) & (4) we get

length (x) =6 ft
breadth (y) = 4 ft

Originally Posted by ips_math

if length = x ft and breadth = y ft

then by given condition 2(x+y) = 20 .....(1)
and x^2 + y^2 = (2sqr(13))^2 = 52 .....(2)
from eq(1) x+y = 10........(3)
from eq (2) & eq(3)
2xy=(x+y)^2 -(x62+y^2) = 100-52 = 48


now (x-y)^2 = x^2+y^2 - 2xy =52 - 48 =4
or x-y =2 .......(4)

solving (3) & (4) we get

length (x) =6 ft
breadth (y) = 4 ft

Much better reply.

Thanks