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Triangle , rectangle , circle.

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by sapuna » Sun Aug 03, 2014 1:30 pm
Where is my mistake ?

https://prikachi.com/images.php?images/592/7514592m.jpg

Here is what I did. From the given information we can deduce that QRS is a right triangle that is isoscales . Therefore if we put H in the middle of QS = > QH=HS = 1/2r= 0,5. Since the triangle is inscribed in the circle , QRS must measure 90 degrees and therefore QR/Qh = 1/square root of 2. From then on we easily find the area of the triangle QRS which should equal 1.h and find h. However, the number for h I find is not in the answers. What did I do wrong ?

p.s. Don`t mind my low score.. I was a noob back then a month ago when I started doing GMAT lol.
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Source: — Problem Solving |

by Matt@VeritasPrep » Sun Aug 03, 2014 5:08 pm
Here's a visual. The dot in the middle is the center of the circle.

Image

I'm assuming that the rectangle is conveniently inscribed so that the center of the circle is the center of the rectangle. Given that assumption, we know that the distance from the center to the midpoint of the length is HALF of the width, or (h/2).

Since the radius is 1, we know that the remaining distance is (1 - h/2).

From there we just use the fact that the areas are equal to find h.

The flaw (if I'm reading you correctly) in your solution is that you assumed the isosceles triangle is a 45-45-90, but that need not be the case. (Similarly, the length of the rectangle doesn't have to bisect the radius.)
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by BestGMATEliza » Sun Aug 03, 2014 8:23 pm
Matt makes a great point! A triangle that is inscribed in a circle is a right triangle IF one of the sides of the triangle is the diameter of the circle. However, since this is not the case here, you cannot assume that it is a right triangle.
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by sapuna » Sun Aug 03, 2014 10:59 pm
Ah, you are absolutely right. Silly mistake on my part. I solved it the way you guys knew. The radius = 1/2h + the height of qrs and its easy after that
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by GMATinsight » Mon Aug 04, 2014 8:19 am
Answer: Option B
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by j_shreyans » Sun Aug 24, 2014 1:53 am
Guys ,

Is there any other ways to solve this question?

Thanks ,

Shreyans
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