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).



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!



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



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