Question

A $34$ cm long wire is bent in to a rectangle. The length of its diagonal is $13$ cm. What are the lengths of the sides of the rectangle?

Answer 1

Perimeter of the rectangle = Length of the wire 2(Length + Breadth) = 34
$\Rightarrow$ Length + Breadth = 17
Let the length of the rectangle be x cm.
$\Rightarrow$ Breadth of the rectangle = (17 - x) cm.
Using Pythagoras theorem,
$x^2+(17-x{)}^2={13}^2$
$x^2+289-34x+x^2=169$
$2x^2-34x+120=0$
$x^2-17x+60=0$
Discriminant $=b^2-4ac=289-240=49$
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{17\pm \sqrt{49}}{2}=\frac{17\pm 7}{2}=12or5$
When length, x = 12 cm, breadth, 17 - x = 5 cm
When length, x = 5 cm, breadth, 17 - x = 12 cm
Thus, the lengths of the sides of the rectangle are 12 cm and 5 cm.