Q:

The area of a rectangle is 108m2 and its diagonal is 15m. Find the perimeter of the rectangle if the sides are of integer lengths.

Accepted Solution

A:
Considering the figure, In ΔABC, right angled at B, we can use Pythagoras theorem :[tex] x^2 + y^2 = 15^2 [/tex][tex] x^2 + y^2 = 225 [/tex]..................(1)Also we are given area of rectangle as 108 m²Area of rectangle = length * breadth = x * y[tex]108 =x*y[/tex].............(2)[tex]x=108/y[/tex]........(3)plugging the value of x from equation (3) in (1),[tex] x^2 + y^2 = 225 [/tex][tex] (108/y)^2 + y^2 = 225 [/tex]We can use quadratic formula to solve this.On solving this we get four values of y as :y=12, y=-12, y=9 and y=-9since length cannot be negative we have two y values as :y=12 and y=9plugging these in equation (3) to get x as x=108/y = 108/12 = 9x = 108/9 =12so we have two answers:length =9 m and breadth = 12mlength =12 m and breadth =9mPerimeter = 2(l+b)Perimeter = 2(9+12) = 42m