Triangles ABC and DEF are similar.

Solution:

Given: ΔABC and ΔDEF are similar triangles

To find:

(i) If area of $\triangle \mathrm{ABC}=16 \mathrm{~cm}^{2}$, area of $\triangle \mathrm{DEF}=25 \mathrm{~cm}^{2}$ and $\mathrm{BC}=2.3 \mathrm{~cm}$, Find $\mathrm{EF}$.

(ii) If area of $\triangle \mathrm{ABC}=9 \mathrm{~cm}^{2}$, area of $\triangle \mathrm{DEF}=64 \mathrm{~cm}^{2}$ and $\mathrm{DE}=5.1 \mathrm{~cm}$, Find $\mathrm{AB}$.

(iii) If $\mathrm{AC}=19 \mathrm{~cm}$ and $\mathrm{DF}=8 \mathrm{~cm}$, find the ratio of the area of two triangles.

(iv) If area of $\triangle \mathrm{ABC}=36 \mathrm{~cm}^{2}$, area of $\triangle \mathrm{DEF}=64 \mathrm{~cm}^{2}$ and $\mathrm{DE}=6.2 \mathrm{~cm}$, Find $\mathrm{AB}$.

(v) If $\mathrm{AB}=1.2 \mathrm{~cm}$ and $\mathrm{DE}=1.4 \mathrm{~cm}$, find the ratio of the area of two triangles.

(i) We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

$\frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\triangle \mathrm{DEF})}=\left(\frac{\mathrm{BC}}{\mathrm{EF}}\right)^{2}$

$\frac{16}{25}=\left(\frac{2.3}{E F}\right)^{2}$

$\Rightarrow \frac{4}{5}=\frac{2.3}{\mathrm{EF}}$

$\mathrm{EF}=2.875 \mathrm{~cm}$

(ii) $\frac{\operatorname{ar}(\triangle \mathrm{ABC})}{\operatorname{ar}(\triangle \mathrm{DEF})}=\left(\frac{\mathrm{AB}}{\mathrm{DE}}\right)^{2}$

$\frac{9}{64}=\left(\frac{\mathrm{AB}}{\mathrm{DE}}\right)^{2}$

$\Rightarrow \frac{3}{8}=\frac{\mathrm{AB}}{5.1}$

$\mathrm{AB}=1.9125 \mathrm{~cm}$

(iii) $\frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{DEF})}=\left(\frac{\mathrm{AC}}{\mathrm{DF}}\right)^{2}$

$\frac{a r(\Delta \mathrm{ABC})}{a r(\Delta \mathrm{DEF})}=\left(\frac{19}{8}\right)^{2}$

$\frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{DEF})}=\left(\frac{361}{64}\right)$

(iv) $\frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{DEF})}=\left(\frac{\mathrm{AB}}{\mathrm{DE}}\right)^{2}$

$\frac{36}{64}=\left(\frac{\mathrm{AB}}{\mathrm{DE}}\right)^{2}$

$\Rightarrow \frac{6}{8}=\frac{\mathrm{AB}}{6.2}$

$\mathrm{AB}=4.65 \mathrm{~cm}$

(v) $\frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{DEF})}=\left(\frac{\mathrm{AB}}{\mathrm{DE}}\right)^{2}$

$\frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{DEF})}=\left(\frac{1.2}{1.4}\right)^{2}$

$\frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{DEF})}=\frac{36}{49}$