# how to solve this triangle / area of trapezoid problem

• Mar 22nd 2013, 10:30 AM
zaco519
how to solve this triangle / area of trapezoid problem
HERE IS THE DIAGRAM
AMCmath problem | Flickr - Photo Sharing!

In triangle ABC, medians CE and AD intersect at P, PE=1.5, PD =2, and DE=2.5. What is the area of AEDC?

THE CHOICES ARE
[A] 13 [B] 13.5 [C] 14 [D] 14.5 [D] 15
• Mar 22nd 2013, 01:04 PM
Henderson
Re: how to solve this triangle / area of trapezoid problem
Two things to notice/know:

The centroid (intersection of the medians) splits each median in a 2:1 ratio. You can use this to find the length of CP and AP.

Triangle DPE- with sides 1.5, 2, and 2.5- is a right triangle (It's a 3-4-5) with a right angle at P. That makes finding the area of the four triangles inside that trapezoid easy.
• Mar 22nd 2013, 01:09 PM
earboth
Re: how to solve this triangle / area of trapezoid problem
Quote:

Originally Posted by zaco519
HERE IS THE DIAGRAM
AMCmath problem | Flickr - Photo Sharing!

In triangle ABC, medians CE and AD intersect at P, PE=1.5, PD =2, and DE=2.5. What is the area of AEDC?

THE CHOICES ARE
[A] 13 [B] 13.5 [C] 14 [D] 14.5 [D] 15

1. $\displaystyle AC = 2 \cdot DE$ Why?

2. Use proportions to get PC:

$\displaystyle \frac{PC}{AC}=\frac{PE}{DE}$

With PA similarly. (For confirmation only: PC = 3, PA = 4)

3. To calculate the height x in the triangle ACP use Pythagoras:

$\displaystyle \begin{array}{rcl}x^2+y^2&=&3^2 \\ x^2+(5-y)^2&=&4^2 \end{array}$

Solve for x (For confirmation only x = 2.4)

4. The quadrilateral AEDC is a trapezium whose parallel sides are 5 and 2.5 and whose height can be calculated by using x and the known proportions.