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Geometry Difficulty 3.4 AMC 10/12 Find the answer Canada

In the diagram, pentagon $PQRST$ has $PQ = 13$ , $QR =18$ , $ST=30$ , and a perimeter of 82. Also, $\angle QRS = \angle RST = \angle STP = 90^\circ$ .

The area of the pentagon $PQRST$ is

$306$
$297$
$288$
$279$
$270$

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

We extend $RQ$ to the left until it meets $PT$ at point $U$ , as shown.

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Because quadrilateral $URST$ has three right angles, then it must have four right angles and so is a rectangle.

Thus, $UT = RS$ and $UR = TS = 30$ .

Since $UR = 30$ , then $UQ = UR - QR = 30 - 18 = 12$ .

Now $\triangle PQU$ is right-angled at $U$ .

By the Pythagorean Theorem, since $PU > 0$ , we have $PU = \sqrt{PQ^2 - UQ^2} = \sqrt{13^2 - 12^2} = \sqrt{169-144} = \sqrt{25} = 5$ Since the perimeter of $PQRST$ is 82, then $13 + 18 + RS + 30 + (UT + 5) = 82$ .

Since $RS = UT$ , then $2\times RS = 82 - 13 - 18 - 30 - 5 = 16$ and so $RS = 8$ .

Finally, we can calculate the area of $PQRST$ by splitting it into $\triangle PQU$ and rectangle $URST$ .

The area of $\triangle PQU$ is $\frac{1}{2} \times UQ \times PU = \frac{1}{2} \times 12 \times 5 = 30$ .

The area of rectangle $URST$ is $RS \times TS = 8 \times 30 = 240$ .

Therefore, the area of pentagon $PQRST$ is $30 + 240 = 270$ .

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.