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# Help!

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420
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+187

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?

Aug 24, 2018

#1
+111321
+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

Aug 24, 2018
#2
+187
+1

Wow, that is smart. Maybe you are the genius and not me, HAHA

Mathgenius  Aug 24, 2018
#4
+111321
0

HAHAHA!!!...I'm only a "genius" at sporadic periods.....!!!!!

CPhill  Aug 24, 2018
#3
+111321
+1

Here's a pic :

Aug 24, 2018