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Posts: 181
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
64% (01:38) correct
36%
(01:53)
wrong
based on 304
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Attachment:
File comment: PS-SET22 Q8 PARALLELOGRAM
PS-SET22 Q8.GIF [ 2.06 KiB | Viewed 32129 times ]
In parallelogram PQRS shown, if PQ = 4 and QR = 6, what is the area of PQRS?PS-SET22 Q8.GIF [ 2.06 KiB | Viewed 32129 times ]
A. 8
B. 12
C. 24
D. 8√3
E. 12√3
Orange08
• 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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General Discussion
singh_amit19
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)
fresinha12
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
singh_amit19
B .
easy if you know some formulas of triginometry
sin(30) = 1/2 = height/hypetenuse(4) => height = 2
area = base * height = 2*6=12
ashkrs
nice one! hhahaah.. dont remember much about this.. took trig well over 10 years ago freshmen year of HS
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
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
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
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
Quote:
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?
ggarr
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
singh_amit19
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
beckee529
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.
Sounds assuring. Thanks a lot Bunuel.
Posts: 135
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
saxenashobhit
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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Posts: 150
Bunuel
Are we supposed to remeber the corelations for the standard triangles
Are we supposed to remeber the corelations for the standard triangles
prashantbacchewar
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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Kimberly77
Joined: 16 Nov 2021
Last visit: 07 Sep 2024
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Posts: 417
Hi brunel, What about if if PQ = 5 then? Will it still be 5/2 ? Thanks
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