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# In parallelogram PQRS shown, if PQ = 4 and QR = 6, what is

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Manager
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In parallelogram PQRS shown, if PQ = 4 and QR = 6, what is  [#permalink]

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09 Oct 2007, 15:08
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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
##### Most Helpful Expert Reply
Math Expert
Joined: 02 Sep 2009
Posts: 59622
Re: In parallelogram PQRS shown, if PQ = 4 and QR = 6, what is  [#permalink]

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11 Sep 2010, 14:49
4
2
Orange08 wrote:
beckee529 wrote:
this is how i got 12 by finding the 3 subparts (by drawing two lines to create two triangles and quadrilateral):

30:60:90
x: √3: 2x
2: 2√3: 4

Is it mandatory to mug up this rule regarding 30:60:90 triangle?
No other ways of solving this problem? Isn't it correct that opposite sides of angles 30 and 60 are in ratio 1:2?

Any help is appreciated.

• A right triangle where the angles are 30°, 60°, and 90°.

This is one of the 'standard' triangles you should be able recognize on sight. A fact you should commit to memory is: The sides are always in the ratio $$1 : \sqrt{3}: 2$$.
Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°).

BACK TO THE ORIGINAL QUESTION:

Now, as hypotenuse PQ (the largest side) equals to 4 then the side opposite 30 degrees (smallest side, which is also the height of the parallelogram) equals to 4/2=2. Thus area of parallelogram is height*base=2*6=12.

Answer: B.

For more on this issues check Triangles chapter of Math Book (link in my signature).

Hope it helps.
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Re: In parallelogram PQRS shown, if PQ = 4 and QR = 6, what is  [#permalink]

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09 Oct 2007, 15:24
singh_amit19 wrote:
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

Ok if you look at the first triangle. the one with one side =4..we know its a 30-60-90 triangle

so based on that we know that side opposite 30 is (sqrt(3)*x)

we know x=2 so we know that the height is sqrt(3)*2

we know the base is 6, height is sqrt(3)2 area= base*height

12sqrt(3)
Director
Joined: 11 Jun 2007
Posts: 636
Re: In parallelogram PQRS shown, if PQ = 4 and QR = 6, what is  [#permalink]

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Updated on: 09 Oct 2007, 21:05
1
fresinha12 wrote:
singh_amit19 wrote:
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

Ok if you look at the first triangle. the one with one side =4..we know its a 30-60-90 triangle

so based on that we know that side opposite 30 is (sqrt(3)*x)

we know x=2 so we know that the height is sqrt(3)*2

we know the base is 6, height is sqrt(3)2 area= base*height

12sqrt(3)

30:60:90
x: √3: 2x
2: 2√3: 4<-given to us

i get 2√3 for part of my base. the height i got was 2

height * base = 2*6 = 12
even when i did it the long way of adding the two triangles with the quad in the middle i was able to deduce the answer to 12

Originally posted by beckee529 on 09 Oct 2007, 15:30.
Last edited by beckee529 on 09 Oct 2007, 21:05, edited 1 time in total.
Senior Manager
Joined: 08 Jun 2007
Posts: 437
Re: In parallelogram PQRS shown, if PQ = 4 and QR = 6, what is  [#permalink]

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09 Oct 2007, 20:55
singh_amit19 wrote:
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

B .
easy if you know some formulas of triginometry

sin(30) = 1/2 = height/hypetenuse(4) => height = 2
area = base * height = 2*6=12
Director
Joined: 11 Jun 2007
Posts: 636
Re: In parallelogram PQRS shown, if PQ = 4 and QR = 6, what is  [#permalink]

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09 Oct 2007, 21:07
ashkrs wrote:
singh_amit19 wrote:
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

B .
easy if you know some formulas of triginometry

sin(30) = 1/2 = height/hypetenuse(4) => height = 2
area = base * height = 2*6=12

nice one! hhahaah.. dont remember much about this.. took trig well over 10 years ago freshmen year of HS
Senior Manager
Joined: 12 Jun 2006
Posts: 411
Re: In parallelogram PQRS shown, if PQ = 4 and QR = 6, what is  [#permalink]

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09 Oct 2007, 21:39
I get E.

triangle is 30:60:90 ==> x:x(sqrt)3:2x ==> 2:2(sqrt)3:4

area(triangle)=2(sqrt)3*2/2 = 2(sqrt)3

6 [parallelogram base] - 2(sqrt)3 [triangle base] = 4(sqrt)3 [trapezoid base]

area(trapezoid)=6+4(sqrt)3*2/2 = 10(sqrt)3

10(sqrt)3 + 2(sqrt)3 = 12(sqrt)3
Director
Joined: 11 Jun 2007
Posts: 636
Re: In parallelogram PQRS shown, if PQ = 4 and QR = 6, what is  [#permalink]

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09 Oct 2007, 21:49
this is how i got 12 by finding the 3 subparts (by drawing two lines to create two triangles and quadrilateral):

30:60:90
x: √3: 2x
2: 2√3: 4

h = 2,
base of triangle => b = 2√3,
base of quad => B = 6 - 2√3

area of two triangles = 2 * 1/2 b*h
= 2 [ 1/2 * 2 * 2√3 ] = 4√3

area of middle part (quad) = B * h
(6 - 2√3) * 2 = 12 - 4√3

adding the two together:
4√3 + 12 - 4√3 = 12 B
Senior Manager
Joined: 12 Jun 2006
Posts: 411
Re: In parallelogram PQRS shown, if PQ = 4 and QR = 6, what is  [#permalink]

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09 Oct 2007, 22:09
Quote:
area of middle part (quad) = B * h
(6 - 2√3) * 2 = 12 - 4√3

aren't we solving what's inside the parentheses first?
6 - 2√3 = 4√3
4√3 * 2 = 8√3
should we be using the distributive property here?
Director
Joined: 11 Jun 2007
Posts: 636
Re: In parallelogram PQRS shown, if PQ = 4 and QR = 6, what is  [#permalink]

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09 Oct 2007, 22:21
ggarr wrote:
Quote:
area of middle part (quad) = B * h
(6 - 2√3) * 2 = 12 - 4√3

aren't we solving what's inside the parentheses first?
6 - 2√3 = 4√3
4√3 * 2 = 8√3
should we be using the distributive property here?

6 - 2√3 does not = to 4√3!!
6√3 - 2√3 = 4√3
use a calculator and you will see the difference
and i did use distributive property
(6 - 2√3) * 2 =
6*2 - 2√3*2 =
12 - 4√3
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Posts: 553
Re: In parallelogram PQRS shown, if PQ = 4 and QR = 6, what is  [#permalink]

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09 Oct 2007, 23:42
singh_amit19 wrote:
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

B.

The right triangle w/ hypotenuese PQ is a 30, 60, 90 triangle with sides a, sqrt 2a and 2a.

Side 2a corresponds to 4.
2a = 4
a = 2

a is the shortest side, therefore facing the smallest angle which is P. Therefore 2 is the height.

ar = b*h = 12
Manager
Joined: 25 Jul 2010
Posts: 88
Re: In parallelogram PQRS shown, if PQ = 4 and QR = 6, what is  [#permalink]

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11 Sep 2010, 14:27
beckee529 wrote:
this is how i got 12 by finding the 3 subparts (by drawing two lines to create two triangles and quadrilateral):

30:60:90
x: √3: 2x
2: 2√3: 4

Is it mandatory to mug up this rule regarding 30:60:90 triangle?
No other ways of solving this problem? Isn't it correct that opposite sides of angles 30 and 60 are in ratio 1:2?

Any help is appreciated.
Manager
Joined: 25 Jul 2010
Posts: 88
Re: In parallelogram PQRS shown, if PQ = 4 and QR = 6, what is  [#permalink]

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11 Sep 2010, 14:55
Sounds assuring. Thanks a lot Bunuel.
Manager
Joined: 20 Jul 2010
Posts: 177
Re: In parallelogram PQRS shown, if PQ = 4 and QR = 6, what is  [#permalink]

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11 Sep 2010, 15:23
Thanks Bunuel for the triangle. I was deciding on what is sin 30 to get height. I was confused in 1/2 and sqrt(3)/2
Math Expert
Joined: 02 Sep 2009
Posts: 59622
Re: In parallelogram PQRS shown, if PQ = 4 and QR = 6, what is  [#permalink]

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11 Sep 2010, 22:32
saxenashobhit wrote:
Thanks Bunuel for the triangle. I was deciding on what is sin 30 to get height. I was confused in 1/2 and sqrt(3)/2

Trigonometry is not tested on GMAT, so any GMAT geometry question can be solved without it.

Anyway: The sine of an angle is the ratio of the length of the opposite side to the length of the hypotenuse.

Sin (30 degrees)= cos (60 degrees) = 1/2 --> in out case: sin(30 degrees)=height/PQ=height/4=1/2 --> height=2.
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Re: In parallelogram PQRS shown, if PQ = 4 and QR = 6, what is  [#permalink]

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13 Sep 2010, 06:33
Bunuel

Are we supposed to remeber the corelations for the standard triangles
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Joined: 02 Sep 2009
Posts: 59622
Re: In parallelogram PQRS shown, if PQ = 4 and QR = 6, what is  [#permalink]

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13 Sep 2010, 08:33
prashantbacchewar wrote:
Bunuel

Are we supposed to remeber the corelations for the standard triangles

Yes, I think it's good to know below 2 cases:

• A right triangle where the angles are 30°, 60°, and 90°.

This is one of the 'standard' triangles you should be able recognize on sight. A fact you should commit to memory is: The sides are always in the ratio $$1 : \sqrt{3}: 2$$.
Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°).

• A right triangle where the angles are 45°, 45°, and 90°.

This is one of the 'standard' triangles you should be able recognize on sight. A fact you should also commit to memory is: The sides are always in the ratio $$1 : 1 : \sqrt{2}$$. With the $$\sqrt{2}$$ being the hypotenuse (longest side). This can be derived from Pythagoras' Theorem. Because the base angles are the same (both 45°) the two legs are equal and so the triangle is also isosceles.
• Area of a 45-45-90 triangle. As you see from the figure above, two 45-45-90 triangles together make a square, so the area of one of them is half the area of the square. As a formula $$A=\frac{S^2}{2}$$. Where S is the length of either short side.

For more on this issues check Triangles chapter of Math Book (link in my signature).

Hope it helps.
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Re: In parallelogram PQRS shown, if PQ = 4 and QR = 6, what is  [#permalink]

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Re: In parallelogram PQRS shown, if PQ = 4 and QR = 6, what is   [#permalink] 28 Nov 2019, 05:44
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