Maths Olympiad Prep

Library / /123 of 241

, 2015

Geometry Difficulty 2.1 Junior Find the answer Canada

In the diagram, RR lies on line segment PSPS.

The value of xx is

Pick one

Solution

Solution 1

SRQ\angle SRQ is an exterior angle of PQR\triangle PQR.

Thus, SRQ=RPQ+PQR=50+90=140\angle SRQ = \angle RPQ + \angle PQR = 50^\circ + 90^\circ = 140^\circ.

Therefore, x=140x^\circ = 140^\circ and so x=140x=140.

Solution 2

The sum of the angles of PQR\triangle PQR is 180180^\circ, and so PRQ=180RPQPQR=1805090=40\angle PRQ = 180^\circ - \angle RPQ - \angle PQR = 180^\circ - 50^\circ - 90^\circ = 40^\circ Since PRQ\angle PRQ and SRQ\angle SRQ are supplementary, then x+40=180x^\circ + 40^\circ = 180^\circ, and so x=18040=140x = 180 - 40 = 140.

Want a route through all this instead of an archive? The track puts 2,444 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.