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Isosceles Triangle

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by cans » Thu Sep 08, 2011 9:21 am
which triangle??
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by force5 » Thu Sep 08, 2011 10:07 am
please paste the exact question even the figure is not very clear.
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by Whitney Garner » Thu Sep 08, 2011 10:58 am
tttrn333 wrote:HI

I don't understand the proof for this problem? Why is the triangle QR an isosceles?
force5 wrote:please paste the exact question even the figure is not very clear.
**NOTE: this is the complete question given in the 1800 Questions set by GMAT Hacks, and we can definitely answer this question with the information given**

Let's start with the knowledge that the lines ST and QR are parallel. With parallel lines, we can use the rules when a parallel line is cut by a transversal. For example, the transversal QT makes the angles at Q and at T congruent (let's call them x). The transversal SR makes the angles at S and at R congruent (let's call them y).

Image

Now, we're getting closer, but we don't know the relative values of x and y, other than that they sum to 40 degrees (because the 140-degree angle forms triangles with them). But let's think what the problem is asking us to find: the length of minor arc QS. To find that, we will need to use the inscribed angle R or T, to determine the central angle that forms that arc. Then, we take the fraction of 360 that is the central angle and find that same fraction of the circumference.

Arc Length / Circumference = central arc degree / 360

We know the circumference is 2(4.5)*Pi = 9Pi. Now we just need the central angle. But we actually stumbled across the answer while we thought this through - the arc QS is formed by BOTH inscribed angles T and R. This means that those angles must be congruent!

Image

And therefore all of the angles at Q, R, S, and T are congruent!

Image

So we know that each triangle has the angles 140, x and x, so 140+x+x=180, or x=20. So if the inscribed angle that forms arc QS is 20 degrees, the central angle would be 40 degrees (double the inscribed angle). Therefore:

arc Length / 9pi = 40/360

arc Length = 40(9pi)/360 = Pi.

The correct answer is B
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by pemdas » Thu Sep 08, 2011 11:01 am
measure of inscribed angle QRS = 1/2 measure of the intercepted minor arc QS.
triangle (QR-diagonals' cross point) is isosceles because two diagonals are drawn between the parallel lines in circle; these diagonals form the vertical angles, which are equal on the opposite sides. Hence angle QRS=(180`-140`)/2=20` and the angle of intercepted minor arc QS is 40`. Perimeter of circle is 2pi*r, and the length of QS=(2pi*r)*(40`/360`) or (2pi*r)/9

we know that r=4.5 and (2*4.5 *pi)/9=pi

[spoiler]answer is pi or =~3.14[/spoiler]


tttrn333 wrote:HI

I don't understand the proof for this problem? Why is the triangle QR an isosceles?
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