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Geometry question

by sivanhas » Wed Apr 24, 2013 10:01 pm
A parallelogram with area of 18 has a side of 6. Which of the
following could be one of its inside angles?
A. 30 B. 50 C. 60 D. 70

Does anyone has an idea how to solve this?
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by Anju@Gurome » Wed Apr 24, 2013 10:22 pm
sivanhas wrote:A parallelogram with area of 18 has a side of 6. Which of the following could be one of its inside angles?
The area of a parallelogram with base B and height H (the perpendicular distance between base and the opposite side) is given by B*H

In this case, if we take B = 6, then H = 18/6 = 3

Now, we can draw different parallelograms with B = 6 and H = 3 as follows,
Image
As we can see without changing B and H we can draw different parallelograms with different inside angles ranging from 0 to 180 degrees, excluding both.
For example, in the diagram,
Green ---> 30 degrees
Blue ---> 50 degrees
Red ---> 60 degrees

So, all the options are possible.
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by rohankamath619 » Thu Apr 25, 2013 6:12 am
Hi Anju,

By all options, do you mean 70 is also possible?

Thanks!
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by Anju@Gurome » Thu Apr 25, 2013 6:19 am
rohankamath619 wrote:By all options, do you mean 70 is also possible?
Yes, as I've mentioned "we can draw different parallelograms with different inside angles ranging from 0 to 180 degrees, excluding both". That means the inside angle can have any value greater than 0 and less than 180 degrees.

Hope that helps.
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by fifafreak » Mon Apr 29, 2013 5:51 am
sivanhas wrote:A parallelogram with area of 18 has a side of 6. Which of the
following could be one of its inside angles?
A. 30 B. 50 C. 60 D. 70

Does anyone has an idea how to solve this?
Area of a parallelogram = a*b*sin(X) = 6*b*sin(X) = 18 => b*sin(X)= 3.

if X = 30; b = 6
if X = 60; b = 2_/3
if X = 90; b = 3 [for X=50, 6>b>2_/3 ; for X=70, 2_/3>b>3] ... so X can have any of the above values.
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