+188 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?
+128407
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
