Bunuel wrote:
If arc PQR above is a semicircle, what is the length of diameter PR? You should know the following properties to solve this question:
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A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s hypotenuse, then that triangle is a right triangle.So, as given that PR is a diameter then angle PQR is a right angle.
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Perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle.
Thus, the perpendicular QT divides right triangle PQR into two similar triangles PQT and QRT (which are also similar to big triangle PQR). Now, in these three triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles). For example: QR/PR=QT/PQ=TR/QR. This property (sometimes along with Pythagoras) will give us the following: if we know ANY 2 values from PR, PQ, QR, PT, QT, TR then we'll be able to find other 4. We are given that QT=2 thus to find PR we need to know the length of any other line segment.
Also in such kind of triangles might be useful to equate the areas to find the length of some line segment, for example area of PQR=1/2*QT*PR=1/2*QP*QR (for more check:
http://gmatclub.com/forum/triangles-106177.html,
http://gmatclub.com/forum/geometry-problem-106009.html,
http://gmatclub.com/forum/mgmat-ds-help-94037.html,
http://gmatclub.com/forum/help-108776.html)
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(1) a = 4. Sufficient.
(2) b = 1. Sufficient.
Answer: D.
Attachment:
Semicircle2.PNG
hi
Bunuel,
chetan2uThe first thing that came to my mind while approaching is similar triangles, but I didn't clearly understand how to write down the proportions as you did, QT/PQ = TR/QR, I undetstand that once we have the proportions pythagoras can be used in the smaller triangle and b can be found.
How do you understand the proportionS QT/PQ = TR/QR So I thought the base of the bigger triangle is twice it's height then the base of the smaller one should be twice it's height, that would mean b =4, but that's clearly incorrect