trapezoid area

rzburton · New Member Offline Joined: 31 Oct 2009 Posts: 8

look at this link image:

http://img203.imageshack.us/img203/6439/area.gif

ABCD is a trapezoid

ABE area = 256 m^2
EDC area = 25 m^2

How is area of ADE and BEC ?

**Posted:** Sat Oct 31, 2009 7:59 am

AndrewTom

Probably not a very good way of doing it but:

As the areas of triangles $CDE$ and $ABE$ are in the ratio $25: 256$ , their heights are in the ratio $5: 16$ . Let their heights be $5x$ and $16x$ . Then $DC = \frac{10}{x}$ and $AB= \frac{32}{x}$ . The area of the trpaezium is then $\frac{1}{2} (\frac{10}{x} + \frac{32}{x})21x = 441$ So the areas of the two triangles $AED$ and $BCE$ are $441- (256+25) = 160$ between them. But they have eaqual areas so this is $\frac{160}{2} = 80$ .

**Posted:** Sat Oct 31, 2009 9:21 am

randomguy64

I have a better way.

@AndrewTom: How do you know that the ratio of the bases is 5:16? You can say that the heights are in the ratio 5:16 because the triangles are similar, but you would be better off saying so for the side lengths.[/b]

**Posted:** Sat Oct 31, 2009 9:43 pm

AndrewTom

Yes, it's better.

To answer your question: Since the triangles $ABE$ and $CDE$ are similar, any pair of corresponding lengths are in the ratio $5: 16$ , because this is the scale factor of the enlagement, centred at $E$ , which transforms one triangle into the other.

**Posted:** Sun Nov 01, 2009 4:24 am

nsato

Do not double post:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=309311

**Posted:** Sun Nov 08, 2009 9:34 am