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In parallelogram PQRS shown, if PQ=4 and QR=6...

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by swerve » Sun Jan 07, 2018 6:30 am
Image

https://postimg.cc/image/846wtwfxh/ (zoom here)

In parallelogram PQRS shown, if PQ = 4 and QR = 6, what is the area of PQRS?

A. 8
B. 12
C. 24
D. 8√3
E. 12√3

The OA is B.

Please, can any expert explain this PS question for me? I tried to solve it of the following way but I'm not sure,

I know that the area of a parallelogram is,
$$A_p=b\cdot h$$
Then, I can determine h of,
$$\sin\ 30\ =\frac{h}{4}$$
Finally, QR will be equal to PS and with that I can determine the area, right?

I need your help. Thanks.
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by ceilidh.erickson » Sun Jan 07, 2018 3:34 pm
You never need trigonometry to solve GMAT problems!

You're right: the area of a parallelogram = b*h.

The base is easy: QR and PS are both bases, equal to 6.

The height will be the perpendicular line from point Q to the base PS. Since we're given that angle QPS = 30*, we know that we have a 30 : 60 : 90 right triangle. Remember that in a 30 : 60 : 90 triangle, the ratio of the side lengths is x : x(sqrt{3}) : 2x

If the hypotenuse PQ = 4, that represents our 2x side. The side across from the 30* angle is the x side, so the height = 2.

b*h = 6*2 = 12

The answer is B.
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by Scott@TargetTestPrep » Mon Aug 12, 2019 10:44 am
swerve wrote:Image

https://postimg.cc/image/846wtwfxh/ (zoom here)

In parallelogram PQRS shown, if PQ = 4 and QR = 6, what is the area of PQRS?

A. 8
B. 12
C. 24
D. 8√3
E. 12√3

Let T be the intersection of the height of the parallelogram and its base. We see that triangle PQT is a 30-60-90 right triangle. Since the height of the triangle (or the parallelogram), QT, is opposite the 30-degree angle, it's half the hypotenuse of the triangle, PQ. So QT = ½(PT) = ½(4) = 2.

Since the area of a parallelogram is the product of its height and its base, the area of parallelogram PQRS is 2 x 6 = 12.

Answer: B

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