Circle O with radius 2 is inscribed in a trapezoid ABCD such that AD is parallel to BC and CD = 8. If angle A and angle B are right angles, then what is the area of ABCD?

Guest Dec 31, 2020

#1**0 **

Hello Guest!

A cool thing about circles inscribed in quadrilaterals are that the side thingys are equal. (Sorry for the bad description, here's a link)

https://mathalino.com/reviewer/plane-geometry/quadrilateral-circumscribing-circle

Let's start by setting point X as where line OP intersects line AD and point Y where line OS intersects line DC.

Let's also set the distance between XD as m and the distance YC as n.

PD = DS = m

RC = SC = n

We also know that DS + SC = 8.

m + n = 8

Because of the radius of the circle.

AP = AQ = QB = BR = 2

To find the area of a trapezoid we need the height and the average of the two bases.

The height of the trapezoid is AB = 2 + 2 (AQ + QB).

The average of the bases in the trapezoid is (AP + PD + BR + RC)/2

We know AP = 2 and that BR = 2 too.

PD + RC = m + n.

m + n = 8

Plugging everything in, (2+2+8)/2 = 6.

The average of the bases is 6 and the height is 4, making the area 24.

I hope this helped. :))))

I'm not confident about my answer so it would be amazing if someone checked.

=^._.^=

catmg Dec 31, 2020