SOLUTION: Prove that the ratio of the areas of two similer triangles is equal to the squares of the ratio of their corresponding sides

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Question 1006494: Prove that the ratio of the areas of two similer triangles is equal to the squares of the ratio of their corresponding sides
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
if the triangles are similar, then all corresponding parts are proportional which means the ratios of the corresponding parts are the same.

the altitude of a triangle counts as a corresponding part.

let the area of triangle 1 be 2/3 * b * h
b is the base
h is the height.

let the area of triangle 2 be 2/3 * xb * xh

x is the common ratio.

A1 = area of triangle 1 = 1/2 * b * h

A2 = area of triangle 2 = 1/2 * xb * xh

the ratio of the area of A2 to the area of A1 is (1/2 * xb * xh) / (1/2 * b * h)

since xb * xh is the same as x^2 * b * h, the ratio of A2 to A1 becomes:

A2 / A1 = (1/2 * x^2 * b * h) / (1/2 * b * h)

the (1/2 * b * h) in the numerator cancels out with the (1/2 * b * h) in the denominator and you are left with:

A2 / A1 = x^2

x is the common ratio of the corresponding parts.

x^2 is the square of the ratio of the coresponding sides.