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

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swerve: Legendary Member; Posts: 2499; Joined: Sun Oct 29, 2017 2:04 pm; Followed by:6 members

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

by swerve » Sun Jan 07, 2018 6:30 am

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. 8âˆš3
E. 12âˆš3

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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Source: Beat The GMAT — Problem Solving | Sun Jan 07, 2018 6:30 am

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ceilidh.erickson: GMAT Instructor; Posts: 2095; Joined: Tue Dec 04, 2012 3:22 pm; Thanked: 1443 times; Followed by:247 members

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.

Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8085; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Mon Aug 12, 2019 10:44 am

swerve wrote:

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. 8âˆš3
E. 12âˆš3

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 = Â½(PT) = Â½(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

Scott Woodbury-Stewart
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