In trapezoid ABCD with bases AB and CD, we have AB = 52, BC = 12, CD = 39, and DA = 5. What is the area of ABCD?

Mathgenius
Aug 24, 2018

#1**+2 **

Since two of the sides must be parallel, assume that AB is parallel to CD

Drop a perpendicular from C to AB and let the intersection be the point E

Let EB = x

Similarly, drop a perpendicular from D to AB and let the intersection be the point F

Let FA = 52 - (39 + x) = 13 - x

So....the height of the triangles = height of the trapezoid

The height of the triangle on the right =√[12^2 - x^2] = CE

The height of the triangle on the left is √ [5^2 - (13 - x)^2 ] = DF

And, as stated, CE = DF

Equating the quantities under the roots, we have

12^2 - x^2 = 5^2 - (13 - x)^2

144 - x^2 = 25 - (169 - 26x + x^2)

144 - x^2 = 25 - 169 + 26x - x^2

144 = 25 - 169 +26x

144 = -144 + 26x

288 = 26x

x = 288/26 = 144/13

So...the height of the trapezoid = √[12^2 - (144/13)^2 ] = 60/13

And the area = (1/2)(height)(sum of the bases) =

(1/2)(60/13) ( 39 + 52) =

(1/2) (60/13) (91) =

(1/2)(60) (91/13) =

(1/2)(60)(7) =

(1/2)(420) =

210 units^2

CPhill
Aug 24, 2018