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

Question

PS-SET22 Q8.GIF 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

Bunuel · Accepted Answer

• A right triangle where the angles are 30°, 60°, and 90°. 
 https://gmatclub.com/forum/download/file.php?id=10157 
 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.

FN · Answer

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)

beckee529 · Answer

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

ashkrs · Answer

B .
easy if you know some formulas of triginometry 

sin(30) = 1/2 = height/hypetenuse(4) => height = 2
area = base * height = 2*6=12

beckee529 · Answer

nice one! hhahaah.. dont remember much about this.. took trig well over 10 years ago freshmen year of HS

ggarr · Answer

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   - 2(sqrt)3   = 4(sqrt)3  

area(trapezoid)=6+4(sqrt)3*2/2 = 10(sqrt)3

10(sqrt)3 + 2(sqrt)3 = 12(sqrt)3

beckee529 · Answer

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   =  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

ggarr · Answer

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?

beckee529 · Answer

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

GK_Gmat · Answer

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

Orange08 · Answer

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.

Orange08 · Answer

Sounds assuring. Thanks a lot Bunuel.

saxenashobhit · Answer

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

Bunuel · Answer

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.

prashantbacchewar · Answer

Bunuel

Are we supposed to remeber the corelations for the standard triangles

Bunuel · Answer

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

• A right triangle where the angles are 30°, 60°, and 90°. 
 https://gmatclub.com/forum/download/file.php?id=10157 
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°.  
 https://gmatclub.com/forum/download/file.php?id=10158 
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.

Kimberly77 · Answer

Hi  brunel , What about if if PQ = 5 then? Will it still be 5/2 ? Thanks

bumpbot · Answer

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

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

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