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Isosceles triangle: find area with only base and vertex?

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Isosceles triangle: find area with only base and vertex?

by ken3233 » Wed Mar 04, 2009 11:10 am
I have just started the preliminary stages of my math review for the GMAT. My workbook has given me a problem that has stumped me.

PROBLEM:
Given that an iscoceles triangle has a base of 20 and a vertex of 68 degrees, what is the triangle's area?

I have not been able to solve this problem. Given that we don't know the height of the triangle, it is not helpful to use the (1/2)(b)(h) formula. Further, because dividing the triangle into two halves won't give us a 30-60-90 right triangle or a 45-45-90 right triangle, the other classic formulas would not apply here.

Can anyone tell me how to figure this out?

In case anyone is interested, I am starting with these two Schaum's books, and when I am done, I will move on to more specific workers such as OG10, OG11, Kaplan, Manhattan GMAT, etc.

Schaum's Geometry
https://www.amazon.com/Schaums-Outline-G ... 0070527660

Schaum's Elementary Algebra
https://www.amazon.com/Schaums-Outline-E ... 007141083X

Five years ago, I used Schaum's math reviews to prepare for the GRE, and they were very helpful, as I obtained a 760 on the quantitative section. I obtained an 800 on the verbal.
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by DanaJ » Wed Mar 04, 2009 11:20 am
I don't think that the question provides enough info, since we don't even know if the 68 degrees refer to an equal angle or an un-equal angle. This is why you would have two cases:
a. 68 is the measurement of one of the equal angles, making the unequal angles 180 - 2*68 = 180 - 136 = 44.
b. 68 is the measurement of the un-equal angle, making the measurement of the other two angles (180 - 68)/2 = 112/2 = 56.
Both cases are mathematically possible...
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by truplayer256 » Wed Mar 04, 2009 11:20 am
You have to use trig in this problem. Since the vertex of the isoceles triangle measures 68 degrees, the other two angles measure 180-68/2 or 56 degrees. Now if you bisect the triangle into two pieces, you get a 34-56-90 triangle. From this, you can say that tan 56=x/10, where x represents the height of the triangle.
10*tan(56)=x 20*10*tan(56)/2=148.26 sq units. Correct me if I've done anything wrong. Dana's response to this problem is also correct.
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by ken3233 » Wed Mar 04, 2009 4:45 pm
Thanks for the replies. That trig would be needed to solve the problem makes sense...I'm not aware of any geometry formula that would answer the question given so little information about the triangle itself.
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by ithoughtshewas18 » Sat Mar 07, 2009 6:36 pm
Is this a difficult question in the book ? Do they mark it as difficult ?

You can use the angle measurements as ratio's to their portioned sizes of the 1:1:Sqrt 2 equation, knowing the base is 20.

Example the 56' is 31% to the whole, compared to 45 being 45 being 25%, a 6% increase would mean the normal values had it been normal, woudl be increase by 6% and the compensation would be followed up on the other angles.

You're giving nothing besides a base and a vertices, why would they give an unequal vertices measurement? I think the degree measurement gives soem indication to what % of the base value should be in the height.

I got A = 33 * 20 / 2

Since thr 34' is opposite to the 20 base. The 90 = sqrt 2 we don't have to worry about.

But the 56 is opposite to the height. So with a portion ratio

34/20 as 56/x x = 33 something.

Most defiantly wrong .
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