PS, QT and RU are medians of triangle PQR. PS lies along the line y = x+3, QT lies along the line y = 2x+4.
The length of PQ is 60 and ∠PRQ = 90^o.

Determine the area of triangle PQR.

Having used the coordinates of P, Q, S, T as variables, solved
simultaneous quadratics and struggled with decimals for some
time, I got the answer 399.999 and guessed that something
was afoot.
Here’s a shorter way (much fun – thanks KS).

Denote the lengths |RQ| and |RP| by a and b respectively.

By Pythagoras,               a² + b² = 60²                 (1)

Let G be the intersection of the medians, so that GQ and GS
have gradients 2 and 1 respectively. Using these as tangent
ratios:  
            tan/QGS = (2 – 1)/(1 + 2*1) = 1/3          (2)

From triangle QGS:   /GSR = /QGS + /SQG  (exterior angle…)

Taking tangents of both sides..

    tan/GSR = (tan/QGS + tan/SQG)/(1 - tan/QGS*tan/SQG)

Using (2) and length ratios in triangles PSR and RQT,

we get         2b/a  = (1/3 + b/(2a))/(1 – (b/(6a))

which simplifies (eventually) to    ab = 2(a² + b²)/9

Using (1)                                   ab = 2*60²/9 = 800

Therefore area of triangle PQR = ab/2 = 400

Posted by Harry on 2015-12-27 16:45:56