9.2 Area of Similar Figures

9.2 Area of Similar Figures

There are two parallelograms with corresponding bases $6$ cm and $8$ cm and corresponding altitudes $3$ cm and $4$ cm respectively. The ratio between their lengths is $3 : 4$ written as:

The area of smaller parallelogram is: $A_1 = \text{base} \times \text{altitude}$
$= 6 \times 3 = 18 \text{ cm}^2$

The area of larger parallelogram is: $A_2 = \text{base} \times \text{altitude}$
$= 8 \times 4 = 32 \text{ cm}^2$

The ratio of their areas is:

Where $A_1$ and $A_2$ are areas and $\ell_1$ and $\ell_2$ are any two corresponding lengths of similar figures.

Hence the ratio of the areas of any two similar figures is equal to the square of the ratio of any two corresponding lengths of the figures.

Since each length is $k$ times of the other, we take $\dfrac{\ell_1}{\ell_2} = k$, then $\dfrac{A_1}{A_2} = k^2$. i.e. Area $A_1$ is $k^2$ times the area $A_2$. $k$ is called scale factor.

Example 6

Find the unknown value in the following:
(i) [Diagram]
(ii) [Diagram]
(iii) The quadrilaterals $PQRS$ and $XYZW$ are similar where $m\overline{PQ} = 35$ cm and $m\overline{XY} = 25$ cm.
(iv) [Diagram]
(v) [Diagram]

Solution: (i) Since two pairs of corresponding angles are equal i.e., triangles are similar. We use the formula for ratio of areas of similar figures.

Here $\ell_1 = 2.4$ cm, $\ell_2 = 1.5$ cm, $A_2 = 25 \text{ cm}^2$, $A_1 = ?$

(ii) Apply formula:

Here $\ell_1 = 10$ cm, $\ell_2 = 25$ cm, $A_1 = 40 \text{ cm}^2$, $A_2 = ?$

(iii) It is given that the quadrilateral $PQRS$ is similar to quadrilateral $XYZW$.

Here $\ell_1 = 35$ cm, $\ell_2 = 25$ cm, $A_1 = ?$, $A_2 = 98 \text{ cm}^2$

(iv) Since two pairs of corresponding angles in both triangles are equal, so triangles are similar.

Here $\ell_1 = ?$, $\ell_2 = 3$ cm, $A_1 = 13.5 \text{ cm}^2$, $A_2 = 24 \text{ cm}^2$

(v) For similar spheres

Here $r_1 = ?$, $r_2 = 7$ cm, $A_1 = 153 \text{ cm}^2$, $A_2 = 833 \text{ cm}^2$

Example 7

Two polygons are similar with a ratio of corresponding sides being $\dfrac{3}{5}$. If the area of the smaller polygon is $54 \text{ cm}^2$, find the area of the larger polygon.

Solution: The ratio of the areas of two similar polygons is the square of the ratio of corresponding sides. So, $\dfrac{\text{Area of larger polygon}}{\text{Area of smaller polygon}} = \left(\dfrac{5}{3}\right)^2 = \dfrac{25}{9}$

Therefore, Area of larger polygon $= \dfrac{25}{9} \times 54 = 150 \text{ cm}^2$

Example 8

Given that $\overline{BC} \parallel \overline{DE}$, prove that the triangles $ABC$ and $ADE$ are similar.

(i) If $m\overline{AB} = 3$ cm and $m\overline{BD} = 1.2$ cm, find the ratio of area of $\Delta ABC$ to the area of $\Delta ADE$.
(ii) If area of $\Delta ADE$ is $125 \text{ cm}^2$, find the area of $\Delta ABC$ and area of trapezium $BCED$.

Solution: Since $m\angle A = m\angle A$ (common), $m\angle B = m\angle D$ and $m\angle C = m\angle E$ (Corresponding angles of parallel lines $\overline{BC}$ and $\overline{DE}$). Hence $\Delta ABC$ is similar to $\Delta ADE$.

(i) Ratio of sides $= \dfrac{m\overline{AB}}{m\overline{AD}} = \dfrac{3}{3 + 1.2} = \dfrac{4.2}{3} = \dfrac{7}{5}$

(ii) Area of $\Delta ADE = 125 \text{ cm}^2$

Area of trapezium $BCED$ = Area of $\Delta ABC$ - Area of $\Delta ADE$