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