If the perimeter and diagonal of the rectangle are 60m and 10m respectively. What is the area of the rectangle?
$412~ m^2$
$390~ m^2$
$409~ m^2$
$408~ m^2$
$400~ m^2$
Solution
Given,
Perimeter of the rectangle $= 2(l+b)=60$
Diagonal of the rectangle $=\sqrt{l^2+b^2}$
$l^2+b^2=10^2$
$l^2+b^2=100$ m
We know that,
$(l+b)^2 = l^2+b^2+2lb$
$(30)^2 = 100+2lb$
$lb = 400~m^2$
Area of the rectangle = 400 $m^2$