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Geometry DS

Expert replies
Source: — Data Sufficiency |

by neerajkumar1_1 » Wed Oct 06, 2010 5:46 am
statement1)

see the attached image in which i have deduced the angles..


the angle which i have marked 90 with help of line is deduced by the fact that any angle subtended by the diameter is 90...
Rest was given...
since the angles are same with the traversals
LM is paralled to RQ

statement 2) no clue... spent a lot of time, could not deduce anything...

so truly i would have opted option A, but since u have mentioned the OA otherwise... cant say much.. :)
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by Yanat » Wed Oct 06, 2010 7:26 am
I feel its A. Please if someone has any other answer, do explain
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by Tani » Wed Oct 06, 2010 7:49 am
First, you know that angle PQR is 90 degrees because PR is a diameter. You also know that angle RPQ is the same for both triangles. The questions then is, can we say that triangles PON and PRQ are similar?

We know that the hypotenuse of PON is a radius and the hypotenuse of PRQ is a diameter. Therefore the larger hypotenuse is twice the smaller. If the triangles are similar, their angles will be similar and their sides proportionate. Assume the length of ON is x and the length of PN is y. The area of PON is then 1/2 xy. If the triangles are similar, since the hypotenuse is double, all sides are double and RQ is 2x and PQ is 2y. That would make the area of PQR 1/2 (2x)*(2y) or 2XY. The area of the larger triangle would have to be four times the area of the smaller. Since the area of the quadrilateral ONRQ is three times the area of triangle PON, the area of the larger triangle is, in fact, four times the area of PON. That means the triangles are similar and angle PNO and PQR are both right angles, which makes the lines parallel.
Tani Wolff
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by narik11 » Wed Oct 06, 2010 8:09 am
For option 2 it states using mid point theorem.
But i am unable to deduce it.
Can someone elaborate it.
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