Question

In rectangle ABCD, diagonals AC and BD intersect at E. If AE=3x-28 and DE=.5x+12, find the length of AC.

Answer 1

Given:

In rectangle ABCD, diagonals AC and BD intersect at E, AE=3x-28 and DE=.5x+12.

To find:

The length of AC.

Solution:

We know that, diagonals of a rectangle are equal and they bisect each other.

In rectangle ABCD, diagonals AC and BD intersect at E, so

$AE=BE=CE=DE$          ...(i)

Taking $AE=DE$, we get

$3x-28=0.5x+12$

$3x-0.5x=28+12$

$2.5x=40$

Divide both sides by 2.5.

$x=\dfrac{40}{2.5}$

$x=16$

Now,

$AE=3x-28$

$AE=3(16)-28$

$AE=48-28$

$AE=20$

Using segment addition property,

$AC=AE+CE$

$AC=AE+AE$         [Using (i)]

$AC=20+20$

$AC=40$

Therefore, the length of AC is 40 units.