Question

The perimeter of a rectangle is 28 meters and its diagonal is 10 meters. Find the area of the rectangle

Answer 1

let L= length of rectangle ,w= width of rectangle ,p=perimeter of rectangle

p=28m ,d=10m ,L=? ,W=?

p=2(L+w),p/2=L+w  

w=p/2-L=28/2-L=14-L

w=14-L —————(1)

from figure in right angled triangle

d^2 =L^2+w^2 ,1o^2=(L)^2 + w^2 , from eq(1) L= w=14-L

1o^2=(L)^2 + 14-L)^2 ,

100=L^2 +196–28L+ L^2

2L^2 -28L+96=0

divide both of the equation by 2 ,we get:-

L^2 -14L+48=0

L^2 -8L-6L +48=0

L(L-8)-6(l-8)=0

(L-8)(L-6)=0

L-8=0 , L-6=0

l=8m

from equation one

w=14-L=14–8=6m

Therefore,the sides are length of 8m and width of 6m

area of rectangle is L*W= 8*6=48