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Paralleogram Problem

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by gmat009 » Tue Oct 07, 2008 9:03 pm
Each side of a certain parallelogram has length 6. if the area of the parallelogram is 18.
which of the following is the measure of one of its angles?
A. 30’ B. 45’ C. 60’ D. 90’ E.120’

OA is A
Plz. explain
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by stop@800 » Wed Oct 08, 2008 1:43 am
area = base * ht
so ht will be 3

if you drawa a fig
you will get a triangle

with
hyp = 6
ht = 3

using trigonometry I can easily find the angle to be 30 degree...
or
the triangle is 1 sqrt(3) 2 triangle
so angle will be 30 degree
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by kris610 » Wed Oct 08, 2008 6:39 am
For the purposes of the GMAT, you don've to use trigonometry.

After you draw the triangle and find the other side, you'll see the sides are in the ratio 1:sqrt(3):2 which is a 30:60:90 triangle. you'll see that 30 is one of the angles of the //'gram.

Similarly for a 45:45:90 triangle, the sides will be in the ratio 1:1:sqrt(2).

These two ratios are good enough to solve problems of this type -you won't need any trigonometry.
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another option reduction method

by getk » Wed Oct 08, 2008 8:19 am
seems the angle is 30 degrees

It would be a rhombus coz all sides are equal

it cant be 90, as it would become a square (so area is 6x6 = 36)
if one angle is 60, then other pair of angles are 120 each.. so two options. So can avoid 60 and 120

so remaining options are 45 and 30

if its 45, then the height and base would be 6xSin(45)
two small triangles would be 2 x 1/2*b*h = b* h = 18 (so triangles itself are 18) and no space for remaning rectangle.

only option left is 30 degrees
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