Question

the length of a rectangle is twice it's width.if the length of it's diagonal is 16√5 cm,find it's area?? please please show step by step? it's urgent please?

Answer 1

Let the width of the rectangle is x.

Given that length of a rectangle is twice its width.

= > length = 2x.

Now,

Diagonal = $\sqrt{(x)^2 + (2x)^2}$

$= > \sqrt{5x^2}$.

So,

$\sqrt{5x^2} = 16\sqrt{5}$

x = 16.

Now,

width of the rectangle = 16cm.

Length of the rectangle = 32cm.

We know that Area of rectangle = Length * Breadth

= 32 * 16

= 512cm^2.

Hope this helps!

Answer 2

l =2*w

by phythagoras theorem

l^2+w^2=(16sqrt(5))^2

5w^2=1280

w^2=256

w=16

l=2*w=32

area= l*w=16*32=512