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