Consider a right angled triangle ABC situated in the XY plane, where AB is the hypotenuse.

Given that:

(i) AB = 60, and:
(ii) The medians through A and B lie along the lines y=x+3 and y=2x+4 respectively.

Determine the area of triangle ABC.

The acute angle, q, between the two lines with gradients 1 and 2
is given by tan(q) = (2 - 1)/(1 + 2x1) = 1/3.

The point of intersection of the lines, G, is the centroid of triangle
ABC. Let a, b and c be the position vectors of A, B and C,
relative to G, and a, b and c be their magnitudes.

Since /ACB = 90^o, C lies on a circle with diameter AB. So the

median from C is a radius of length 30, and GC = c = 20.

Since G is the centroid:               a + b + c  = 0               (1)

Since /ACB = 90^o:                      (a – c).(b – c) = 0          (2)

From (2):                      a.b – c.(a + b) + c² = 0

and using (1):                            a.b = -2c²

which gives:                   a*b*cos(180^o - q) = -800

                                    a*b*cos(q) = 800

Since the medians of a triangle divide it into 6 triangles with
equal areas, it follows that

Area of ABC = 3*Area of ABG      = (3/2)*a*b*sin(q)

                                                = (3/2)*a*b*cos(q)*tan(q)

                                                = (3/2)*800*1/3

                                                = 400

Posted by Harry on 2014-12-11 10:21:13