HI
I don't understand the proof for this problem? Why is the triangle QR an isosceles?
Isosceles Triangle
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- cans
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which triangle??
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tttrn333 wrote:HI
I don't understand the proof for this problem? Why is the triangle QR an isosceles?
**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**force5 wrote:please paste the exact question even the figure is not very clear.
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
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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]
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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