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 $A$ .

2005 CEMC Gauss (Grade 7) (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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CEMC Gauss (Grade 7)

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2005 CEMC Gauss (Grade 7) Problems/Problem 15

Contents

Problem

Solution 1

Solution 2

See Also