# 2005 CEMC Gauss (Grade 7) Problems/Problem 15

## Problem

In the diagram, the area of rectangle $PQRS$ is $24$. If $TQ = TR$, what is the area of quadrilateral $PTRS$? $\text{(A)}\ 18 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 16 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 15$ $[asy] size(100); draw((0,0)--(6,0)--(6,4)--(0,4)--cycle); draw((0,4)--(6,2)); draw((5.8,1.1)--(6.2,1.1)); draw((5.8,.9)--(6.2,.9)); draw((5.8,3.1)--(6.2,3.1)); draw((5.8,2.9)--(6.2,2.9)); label("P",(0,4),NW); label("S",(0,0),SW); label("R",(6,0),SE); label("T",(6,2),E); label("Q",(6,4),NE); [/asy]$

## Solution 1

Since the area of rectangle $PQRS$ is $24$, let us assume that $PQ = 6$ and $QR = 4$. Since $QT = TR$, then $QR = 2QT$ so $QT = 2$. Therefore, triangle $PQT$ has base $PQ$ of length $6$ and height $QT$ of length 2, so its area is $\frac{1}{2}(6)(2) = \frac{1}{2}(12) = 6$. So the area of quadrilateral $PTRS$ is equal to the area of rectangle $PQRS$ (which is $24$) minus the area of triangle $PQT$ (which is $6$), or $18$. Therefore, the answer is $A$.

## Solution 2

Draw a line through $T$ parallel to $PQ$ across the rectangle parallel so that it cuts $PS$ at point $V$. Since $T$ is halfway between $Q$ and $R$, then $V$ is halfway between $P$ and $S$. Therefore, $SVTR$ is a rectangle which has an area equal to half the area of rectangle $PQRS$, or $12$. Similarly, $VPQT$ is a rectangle of area $12$, and $VPQT$ is cut in half by $PT$, so triangle $PVT$ has area $6$. Therefore, the area of $PTRS$ is equal to the sum of the area of the rectangle $SVTR$ and the area of triangle $PVT$, or $12 + 6 = 18$. The correct answer is $A$.