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In the figure above, the area of the parallelogram is

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Bunuel
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In the figure above, the area of the parallelogram is [#permalink] New post  26 Apr 2019, 06:22
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A
B
C
D
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Difficulty:

15% (low)

Question Stats:

79% (01:29) correct 21% (01:25) wrong based on 2005 sessions
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In the figure above, the area of the parallelogram is

A. 40
B. \(24\sqrt{3}\)
C. 72
D. \(48\sqrt{3}\)
E. 96


PS50602.01
Quantitative Review 2020 NEW QUESTION


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2019-04-26_1721.png
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BrentGMATPrepNow
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In the figure above, the area of the parallelogram is [#permalink] New post  Updated on: 22 Jul 2020, 08:15
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Bunuel wrote:

In the figure above, the area of the parallelogram is

A. 40
B. \(24\sqrt{3}\)
C. 72
D. \(48\sqrt{3}\)
E. 96

Show Spoiler
Attachment:
2019-04-26_1721.png


Area of parallelogram = (base)(height)

Start by drawing an extra line, which also happens to be the height of the parallelogram

This creates a special 30-60-90 right triangle


When we compare the blue 30-60-90 right triangle with the purple BASE 30-60-90 right triangle, . . .

We see that the blue 30-60-90 right triangle is 4 times bigger than the purple BASE 30-60-90 right triangle, . .
So, the missing lengths are 4 and 4√3

At this point, we know the base and the height


Area of parallelogram = (base)(height)
= (12)(4√3)
= 48√3

Answer: D

Originally posted by BrentGMATPrepNow on 26 Apr 2019, 06:49.
Last edited by BrentGMATPrepNow on 22 Jul 2020, 08:15, edited 2 times in total.
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EgmatQuantExpert
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Re: In the figure above, the area of the parallelogram is [#permalink] New post  23 May 2019, 05:05
Expert Reply

Solution


Given
In this question, we are given
    • The diagram of a parallelogram, where the adjacent sides are of length 8 and 12
    • One angle between the sides is 60°

To Find
We need to determine
    • The area of the parallelogram

Approach & Working



Let us assume the parallelogram is PQRS, and QT is the perpendicular dropped from point Q to the side PS.
    • Hence, we can say that PQT is a 30°-60°-90° triangle, and the side ratio is 1: √3: 2.
    • QT: PQ = √3: 2
    Or, QT = (√3)/2 * PQ = (√3)/2 * 8 = 4√3

    • Therefore, the area of the quadrilateral = PS * QT = 12 * 4√3 = 48√3 (PS = QR)
Hence, the correct answer is option D.


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nick1816
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Re: In the figure above, the area of the parallelogram is [#permalink] New post  23 May 2019, 05:16
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Area of parallelogram= (product of adjacent sides) sinx
= 8*12*sin60= 96*[(3^1/2)/2]=48 (3^1/2)
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ScottTargetTestPrep
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Re: In the figure above, the area of the parallelogram is [#permalink] New post  27 May 2019, 19:05
Expert Reply
Bunuel wrote:

In the figure above, the area of the parallelogram is

A. 40
B. \(24\sqrt{3}\)
C. 72
D. \(48\sqrt{3}\)
E. 96


PS50602.01
Quantitative Review 2020 NEW QUESTION


Show Spoiler
Attachment:
2019-04-26_1721.png


By drawing the altitude, we see that we have a 30-60-90 triangle nested in the parallelogram, with 8 as the hypotenuse and the altitude the side opposite the 60-degree angle. Since the ratio of the sides of a 30-60-90 triangle is 1:√3:2, the altitude will be 4√3.

Therefore, the area of the parallelogram is base x height = 12 x 4√3 = 48√3.

Answer: D
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dave13
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Re: In the figure above, the area of the parallelogram is [#permalink] New post  30 Oct 2020, 08:23
BrentGMATPrepNow wrote:
Bunuel wrote:

In the figure above, the area of the parallelogram is

A. 40
B. \(24\sqrt{3}\)
C. 72
D. \(48\sqrt{3}\)
E. 96

Show Spoiler
Attachment:
2019-04-26_1721.png


Area of parallelogram = (base)(height)

Start by drawing an extra line, which also happens to be the height of the parallelogram

This creates a special 30-60-90 right triangle


When we compare the blue 30-60-90 right triangle with the purple BASE 30-60-90 right triangle, . . .

We see that the blue 30-60-90 right triangle is 4 times bigger than the purple BASE 30-60-90 right triangle, . .
So, the missing lengths are 4 and 4√3

At this point, we know the base and the height


Area of parallelogram = (base)(height)
= (12)(4√3)
= 48√3

Answer: D


BrentGMATPrepNow if parallelogram has sides equal as those of rectangle why simply transforming parallelogram into rectangle and multiplying length by width wont work to find out area of parallelogram? i remember that in some cases one can transform figures but cant remember in what cases such trick works :lol: Any idea ? :)
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BrentGMATPrepNow
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Re: In the figure above, the area of the parallelogram is [#permalink] New post  30 Oct 2020, 12:13
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dave13 wrote:
BrentGMATPrepNow if parallelogram has sides equal as those of rectangle why simply transforming parallelogram into rectangle and multiplying length by width wont work to find out area of parallelogram? i remember that in some cases one can transform figures but cant remember in what cases such trick works :lol: Any idea ? :)


I'm not aware of a simple trick (e.g., one that doesn't involve trig ratios such as sine and cosine) that would accomplish this.

Consider these two parallelograms (aka rhombuses) in which all sides have length 1
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Re: In the figure above, the area of the parallelogram is [#permalink] New post  07 Nov 2020, 13:31
area of parallelogram = base * height

We already know the base is 12; we only need to determine the height.

The left hand side is a 30:60:90 triangle, giving us a ratio of \(1:\sqrt{3}:2\)

Since the hypotenuse is 8, we can conclude that the height of the parallelogram will be \(4\sqrt{3}\)

\(4\sqrt{3} * 12 = 48\sqrt{3}\)

Answer is D.
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In the figure above, the area of the parallelogram is [#permalink] New post  19 Nov 2020, 01:25
Top Contributor
Bunuel wrote:

In the figure above, the area of the parallelogram is

A. 40
B. \(24\sqrt{3}\)
C. 72
D. \(48\sqrt{3}\)
E. 96


PS50602.01
Quantitative Review 2020 NEW QUESTION

Show Spoiler
Attachment:
The attachment 2019-04-26_1721.png is no longer available


Are of Parallelogram = \(base * height\)


The base 12 and according to Pitharorian formula of 60-30-90 angle the height \((QT)\) is \(4\sqrt{3}\)

So, the \(area= \) \(12\)*\(4\sqrt{3}\) \(=\) \(48\sqrt{3}\)

The answer is D
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poomolb
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Re: In the figure above, the area of the parallelogram is [#permalink] New post  24 Mar 2022, 05:20
area of parallelogram = base * height

We already know the base is 12; we only need to determine the height.

a ratio of 1:3√:21:3:2

the hypotenuse is 8 = height of the parallelogram will be 43√43

43√∗12=483√43∗12=483

Answer is D.
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ThatDudeKnows
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Joined: 11 May 2022
Posts: 1092
In the figure above, the area of the parallelogram is [#permalink] New post  13 May 2022, 11:28
Expert Reply
Bunuel wrote:

In the figure above, the area of the parallelogram is

A. 40
B. \(24\sqrt{3}\)
C. 72
D. \(48\sqrt{3}\)
E. 96


PS50602.01
Quantitative Review 2020 NEW QUESTION


Show Spoiler
Attachment:
2019-04-26_1721.png


The geometry on this one is pretty easy, but just for the sake of argument, let's say you have a total freak-out moment and forget 30-60-90s or that you're short on time.

Orrrrrrr that you believe in ballparking as a strategy that helps you do better on the GMAT!! ;) ;)

The area of a parallelogram is base times height. The base is 12. What about the height? Well, we know that the diagonal side is 8. So, is the height 8? No, it's less than 8. How much less than 8? A lot less or just a little less? Just a little less. Okay, let's use 7.

Cool, the area is ~12*7 = ~84. We need something between C and E.

Answer choice D.

LOL at math.

ThatDudeKnowsBallparking
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