Bunuel wrote:

In the figure above, P, Q, R, and S are the midpoints of the sides of rectangle ABCD. What is the area of the polygon APQCRS?

A. \(12\sqrt{3}\)

B. \(9\sqrt{3}\)

C. \(6\sqrt{3}\)

D. \(3\sqrt{3}\)

E. \(24\)

Attachment:

The attachment **Polygon.png** is no longer available

In addition to the solutions offered above, we'll also show a more visual, graphical answer.

This is an Alternative approach and relies on the fact that there is only one way to draw the given figure (as the points P,Q,R,S are midpoints of a rectangle).

Connecting the points P,Q,R and S to each other gives the below image.

We can SEE that drawing the lines has divided the rectangle into 8 identical triangles.

Therefore, our total area is the area of 6 of these triangles.

To complete the calculation, notice that \(\angle QPR\) and \(\angle PRS\) are equal because PQ is parallel to RS.

(once again - since there is only one way to draw this figure, you can SEE that they are parallel)

This means that \(x - 90 + x - 120 = 90\) so \(2x = 300\) and \(x = 150\).

Then \(\triangle PQO\) is a 30-60-90 triangle with sides of \(\sqrt3\) and \(3\) so its area is \(1.5 \sqrt{3}\).

Multiplying by 6 gives \(9 \sqrt{3}\).

(B) is our answer.

Attachments

img.png [ 41.05 KiB | Viewed 868 times ]

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David

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