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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
avatar+88775 
+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 

 

 

cool cool cool

CPhill  Aug 24, 2018
 #2
avatar+54 
+1

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

Mathgenius  Aug 24, 2018
 #4
avatar+88775 
0

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

 

 

cool cool cool

CPhill  Aug 24, 2018
 #3
avatar+88775 
+1

Here's a pic :

 

 

 

cool cool cool

CPhill  Aug 24, 2018

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