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

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

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

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

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

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

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

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.

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.

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.

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

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