Question

Find the perimeter and area of the rectangle whose breadth is $8$ cm and a diagonal is $17$ cm.

Answer 1

In the figure, rectangle $ABCD$ has

Breadth $=BC=AD=8$ cm

Diagonal $BD=17$ cm

Let the length $=AB=CD=x$ cm

Now, triangle $DAB$ is a right-angle triangle.

So, from Pythagorus theorem,

$A{B}^2+A{D}^2=B{D}^2$

$x^2+8^2={17}^2$

$x^2=289-64$

$x^2=225$

$x=\sqrt{225}$

$x=15$ cm

Now, area of the rectangle $=length\times breadth$

$=15\times 8$

$=120$ $c{m}^2$

and perimeter $=15+15+8+8=46$ cm