Im working on an easy quant question but this book does not explain much, so I need your help! The question is:
Parallelogram ABCD has an area of 52 sq centimeters, a height of 4 centimeters and an angle with measure 60 degrees. What is the perimeter of parallelogram ABCD?
my answer was 32 but the book has 26+(16 root 3/3)
since the area is 52, i got that the sides were 13, and 4 (given)... so the total perimeter is (13*2) + (4*2)
How did they get 16 root 3 / 3? ?
I'm happy to help with this.
parallelogram, area = 52.JPG [ 24.19 KiB | Viewed 497 times ]
First of all, it sounds like you were treating the situation as if it were a rectangle, not a parallelogram. If 4 is the height, then 4 would NOT be the "slant length" of the sides. In the diagram, BE = 4 but AB and CD do not equal 4. This is a very common mistake folks make about parallelograms. The "height" is not the length of the slant.
You correctly figured out, from the area, that AD = BC = 13.
To find the length of AB = CD, we have to use the properties of the 30-60-90 triangle. Here's a blog to refresh your memory on these:http://magoosh.com/gmat/2012/the-gmats- ... triangles/
Triangle ABE is a 30-60-90 triangle. For the 30-60-90 triangle, we can remember
long leg: sqrt(3)
We know the long leg, BE = 4 and we want the hypotenuse.
long leg/hypotenuse = sqrt(3)/2
4/(AB) = sqrt(3)/2
8 = (AB)*sqrt(3)
AB = 8/sqrt(3)
Now, because we have a radical in the denominator, we are going to use a procedure you may recall from algebra two: rationalizing the denominator
. We will multiply both the numerator and the denominator by sqrt(3). This has the effect of removing the radical from the denominator.
AB = [8/sqrt(3)]*[sqrt(3)/sqrt(3)] = [8*sqrt(3)]/3
perimeter = 2*(AD) + 2*(AB) = 26 + [16*sqrt(3)]/3Voila
! The OA.
Does all this make sense?
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