Question

Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

Hint:

For similar triangles, all corresponding sides and heights maintain the same ratio.

Answer 1

Given two similar triangles \( \triangle ABC \sim \triangle PQR \), we need to prove:
\[ \frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle PQR)} = \frac{AB^2}{PQ^2} = \frac{BC^2}{QR^2} = \frac{AC^2}{PR^2}. \] Proof:
For two similar triangles, their corresponding angles are equal, and their sides are proportional:
\[ \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} = k. \] Area of a triangle:
\[ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}. \] Since height is also proportional:
\[ h_1 = k h_2. \] Thus, the ratio of the areas is:
\[ \frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle PQR)} = \frac{\frac{1}{2} AB \times h_1}{\frac{1}{2} PQ \times h_2}. \] \[ = \frac{AB}{PQ} \times \frac{h_1}{h_2} = k \times k = k^2. \] Hence proved.