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

#1**+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**+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....!!!!

CPhill
May 27, 2017