A geometry problem by Jayesh Agrawal

Geometry Level 3

PQR is a triangle inscribed in a circle with PR as diameter. The length PR is as much more as length QR as the length QR is more than length PQ. find PR:PQ.

4:3 5:3 6:4 5:2

Jayesh Agrawal by Jayesh Agrawal , 32, India

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2 solutions

Vasudev Chandna
Apr 6, 2015

No need to solve it even!

Since the triangle lies in a semicircle with base as the diameter, we get that it is a right triangle.

Now take the minimum sides that right triangle can have i.e. 3, 4 and 5

Because 3^2+ 4^2= 5^2

The ratio of the diameter(or hypotenuse) to the smallest side= 5:3

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Akhilesh Singh
Sep 3, 2014

Given:PR=x+QR....(i),QR=x+PQ..........(ii).Also PR is a diameter.So,angle(PQR)=90 which implies that,PR^2=QR^2+PQ^2......(iii).Using (i),(ii) and(iii) QR=4x.Putting this back in (i) amd (ii) givesPR:PQ=5:3

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it should be level 3

Akhilesh Singh - 6 years, 9 months ago

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