# Triangle Ratio Problem

• Apr 14th 2008, 12:23 PM
sarahh
Triangle Ratio Problem
Given a triangle PQR, A, B, and C are midpoints of the sides of triangle PQR, and M, N, and O are the midpoints of the sides of triangle ABC. The interiors of triangle PAB, triangle AQC, triangle BCR, and triangle MNO are shaded. What fraction of the interior of triangle PQR is shaded?

I don't know how/why the answer would be 13/16?
• Apr 14th 2008, 08:33 PM
Soroban
Hello, Sarah!

Quote:

$A, B,\text{ and }C$ are midpoints of the sides of $\Delta PQR$.
$M, N,\text{ and }O$ are the midpoints of the sides of $\Delta ABC$.
The interiors of $\Delta PAB,\:\Delta AQC,\:\Delta BCR,\text{ and }\Delta MNO$ are shaded.
What fraction of the interior of $\Delta PQR$ is shaded?

Code:

                  P                   *                 /:\                 /:::\               /:::::\               /:::::::\             /::::O::::\           A* - - * - - *B           /:\  /:\  /:\           /:::\ /:::\ /:::\         /::::M*- - -*N::::\         /:::::::\  /:::::::\       /:::::::::\-/:::::::::\       * - - - - - * - - - - - *       Q          C          R
It can be shown that:
. . When the midpoints of the sides of a triangle are joined,
. . the triangle is divided into four congruent triangles.
Each triangle has one-fourth of the area.

Joining the midpoints of the side of $\Delta PQR$
. . we see that $\frac{3}{4}\text{ of }\Delta PQR$ is shaded.

The unshaded triangle, $\Delta ABC$, has $\frac{1}{4}$ the area of $\Delta PQR.$
. . and $\frac{1}{4}$ of it is shaded.

That is: . $\Delta MNO \:=\:\left(\frac{1}{4}\right)\left(\frac{1}{4}\rig ht)\:=\:\frac{1}{16}\, \text{ of }\Delta PQR$

Therefore, the total shaded area is: . $\frac{3}{4} + \frac{1}{16} \:=\:\boxed{\frac{13}{16}}\,\text{ of }\Delta PQR.$