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|Triangle Area Trial (Posted on 2014-12-07)
Consider a right angled triangle ABC situated in the XY plane, where AB is the hypotenuse.
(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.
No Solution Yet
Submitted by K Sengupta
Rating: 5.0000 (1 votes)
Comment 3 of 3 |
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 = 90o, 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 = 90o: (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(180o
- 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)
Posted by Harry
on 2014-12-11 10:21:13
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