Question

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Answer 1

Hint:
First, we will calculate the area of two similar triangles and then divide them. Then the similarities of two triangles are used to find the ratios of the corresponding sides.

Complete step by step solution:
Let us assume the two triangles are $$\mathrm{\Delta }\text{PQR}$$ and $$\mathrm{\Delta }\text{ABC}$$.

We will use the formula to find the area of triangle, $$\text{Area = }{\displaystyle \frac{1}{2}}\times \text{Base}\times \text{Height}$$.
Now, we will find the area of the triangles $$\mathrm{\Delta }\text{ABC}$$ and $$\mathrm{\Delta }\text{PQR}$$ from the above diagram.
$$\text{Area of }\mathrm{\Delta }\text{PQR}={\displaystyle \frac{1}{2}}\times \text{QR}\times \text{PS ......}\left(1\right)$$
$$\text{Area of }\mathrm{\Delta }\text{ABC}={\displaystyle \frac{1}{2}}\times \text{BC}\times \text{AD ......}\left(2\right)$$
Dividing $$\left(1\right)$$ by $$\left(2\right)$$, we get
$$\begin{gathered} {\displaystyle \frac{\text{Area of }\mathrm{\Delta }\text{PQR}}{\text{Area of }\mathrm{\Delta }\text{ABC}}}={\displaystyle \frac{{\displaystyle \frac{1}{2}}\times \text{QR}\times \text{PS}}{{\displaystyle \frac{1}{2}}\times \text{BC}\times \text{AD}}} \\ ={\displaystyle \frac{\text{QR}\times \text{PS}}{\text{BC}\times \text{AD}}}\text{ ......}\left(3\right) \end{gathered}$$
Since we know that $$\mathrm{\angle }\text{ABC}$$ and $$\mathrm{\angle }\text{PQR}$$ are angles of similar triangles, so $$\mathrm{\angle }\text{ABC}=\mathrm{\angle }\text{PQR}$$ and both right angled triangles $$\mathrm{\angle }\text{ADB}$$ and $$\mathrm{\angle }\text{PSQ}$$ are equal.
Therefore, $$\mathrm{\Delta }\text{PQS}\sim \mathrm{\Delta }\text{ABD}$$.
$$\Rightarrow {\displaystyle \frac{\text{PS}}{\text{AD}}}={\displaystyle \frac{\text{PQ}}{\text{AB}}}\text{ ......}\left(4\right)$$
Substituting this value in equation $$\left(3\right)$$, we get
$${\displaystyle \frac{\text{Area of }\mathrm{\Delta }\text{PQR}}{\text{Area of }\mathrm{\Delta }\text{ABC}}}={\displaystyle \frac{\text{QR}}{\text{BC}}}\times {\displaystyle \frac{\text{PQ}}{\text{AB}}}\text{ ......}\left(5\right)$$
Since we know that the triangles $$\mathrm{\Delta }\text{PQR}$$ and $$\mathrm{\Delta }\text{ABC}$$ are similar,
$${\displaystyle \frac{\text{PQ}}{\text{AB}}}={\displaystyle \frac{\text{QR}}{\text{BC}}}={\displaystyle \frac{\text{PR}}{\text{AC}}}$$
Using this value in equation $$\left(5\right)$$, we get
$$\begin{gathered} {\displaystyle \frac{\text{Area of }\mathrm{\Delta }\text{PQR}}{\text{Area of }\mathrm{\Delta }\text{ABC}}}={\displaystyle \frac{\text{QR}}{\text{BC}}}\times {\displaystyle \frac{\text{QR}}{\text{BC}}} \\ ={\left({\displaystyle \frac{\text{QR}}{\text{BC}}}\right)}^2 \end{gathered}$$
Also from equation $$\left(5\right)$$, we get
$$\begin{gathered} {\displaystyle \frac{\text{Area of }\mathrm{\Delta }\text{PQR}}{\text{Area of }\mathrm{\Delta }\text{ABC}}}={\left({\displaystyle \frac{\text{QR}}{\text{BC}}}\right)}^2 \\ ={\left({\displaystyle \frac{\text{PQ}}{\text{AB}}}\right)}^2 \\ ={\left({\displaystyle \frac{\text{RP}}{\text{CA}}}\right)}^2 \end{gathered}$$

Thus, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Hence, proved.

Note:
In this question, students should write the sides of the triangles appropriately. Since the general area of any triangle is $$\text{Area = }{\displaystyle \frac{1}{2}}\times \text{Base}\times \text{Height}$$, so we need to construct the perpendicular triangles for height. Students should know that when two triangles are similar then the ratio of their corresponding sides are same with the ratio of their corresponding altitudes and sides. The measurement of their corresponding angles is also the same.