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In trapezoid ABCD the lengths of the bases AB and CD are 8 and 17 respectively. The legs of the trapezoid are extended beyond A  and B to meet at point . What is the ratio of the area of triangle EAB  to the area of trapezoid ABCD? Express your answer as a common fraction.

tertre  May 27, 2017
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2+0 Answers

 #1
avatar+89843 
+2

Hi Tetre,

The word trapezoid means different things in different countries, could you plase give a definition that we should use?

 

I'll just tell what I think you mean.

AB and DC are parallel.  

This means that AEB and DEC are similar triangles 

so

the ratio of the height of ABE and DEC will be   8:17

 

so  Area of  AEB = 0.5*8*8k       and area of DEC = 0.5*17* 17k    

and area of  ABCD = 0.5*17* 17k  -   0.5*8*8k

 

ratio of the area of triangle EAB  to the area of trapezoid ABCD

 

\(=\frac{0.5*8*8k}{0.5*17*17k\;-\;0.5*8*8k}\\ =\frac{8*8k}{17*17k\;-\;8*8k}\\ =\frac{8*8}{289\;-\;64}\\ =\frac{64}{225}\\\)

Melody  May 27, 2017
 #2
avatar+75356 
+1

 

 

Since the two triangles created are similar, each dimension of the larger triangle created will be  (17/8)  that of the smaller triangle

 

Call h the height of the smaller triangle......so its   area  =  (1/2)*8*h  =  4h  (1)

 

And the area of the larger triangle  = (1/2)* 17 * (17/8) h   =   (289 / 16)h    (2)

 

So  the area  of the trapezoid  =  (2) - (1)   =  (289/16) h - 4h   =  [ 289 - 64 ] / 16 * h =

 

[225 / 16] h

 

So.....the ratio of the smaller triangle to the trapezoid  =  [4h] /  ( [ 225/ 16] h )  =  64 / 225

 

Just as Melody found....!!!!

 

 

cool cool cool

 

 

 

cool cool cool

CPhill  May 27, 2017

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