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:

Remember: For similar triangles, not only the corresponding sides but also the altitudes are in the same ratio. Hence, areas are proportional to the squares of corresponding sides.

Answer 1

Step 1: Write the given condition of similarity.}
Let \( \triangle ABC \sim \triangle DEF \). Since the triangles are similar, their corresponding angles are equal and their corresponding sides are proportional. Therefore, \[ \frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{FD} \]
Step 2: Write the formula for area of a triangle.}
We know that the area of a triangle can be written as \[ \text{ar}(\triangle) = \frac{1}{2} \times \text{base} \times \text{height} \] Let the altitudes corresponding to sides \( AB \) and \( DE \) be \( h_1 \) and \( h_2 \) respectively. Then, \[ \text{ar}(\triangle ABC) = \frac{1}{2} \cdot AB \cdot h_1 \] and \[ \text{ar}(\triangle DEF) = \frac{1}{2} \cdot DE \cdot h_2 \]
Step 3: Take the ratio of the areas.}
So, \[ \frac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle DEF)} = \frac{\frac{1}{2} \cdot AB \cdot h_1}{\frac{1}{2} \cdot DE \cdot h_2} = \frac{AB}{DE} \cdot \frac{h_1}{h_2} \]
Step 4: Use the property of similar triangles.}
In similar triangles, the ratio of corresponding altitudes is equal to the ratio of corresponding sides. Hence, \[ \frac{h_1}{h_2} = \frac{AB}{DE} \] Substituting this in the above expression, we get \[ \frac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle DEF)} = \frac{AB}{DE} \cdot \frac{AB}{DE} \]
Step 5: Simplify the result.}
Therefore, \[ \frac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle DEF)} = \left(\frac{AB}{DE}\right)^2 \] Similarly, we may also write \[ \frac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle DEF)} = \left(\frac{BC}{EF}\right)^2 = \left(\frac{CA}{FD}\right)^2 \] Thus, the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.