singhmaharaj wrote:Hi,
As far as I know, the property is
if two similar triangles have corresponding lengths in the ratio a:b, then their areas will be in the ratio (a)^2 : (b)^2
This isn't quite right.
Suppose that a = 6 and b = 2, and further suppose that the height of triangle A is 12 and the height of triangle B is 4. The area of A is thus (6*12)/2, or 36, and the area of triangle B is thus (2*4)/2, or 4.
Notice that the larger triangle has an area equal to (a/b)² times the area of the smaller triangle. This will always be true, as we can demonstrate algebraically.
Suppose that B is the smaller triangle, and that B has a base of b units and a height of h units. Since A and B are similar, A's base = b*k, where k is some number greater than 1, and A's height = h*k, where k is the same multiplier that it was before (since the triangles are similar).
Then the area of A = (bk)(hk)/2, and the area of B = (b*h)/2. Thus A has an area equal to the area of B times k², or the square of the ratio between the two corresponding bases.
