Question

In the given figure, $DE\parallel BC$

(i) If DE = 4 cm, BC= 6 cm and Area ($△$ ADE)= 16 $c{m}^2$, find the area of $△$ ABC.

(ii) If DE = 4 cm BC = 8 cm and Area ($△$ ADE)= 25 $c{m}^2$, find the area of $△$ ABC.

(iii) If DE : BC = 3 : 5. Calculate the ratio of the areas of $△$ ADE and the area of BCED.

Answer 1

In $△ABC$, $DE\parallel BC$

$\Rightarrow$ $\angle B=\angle D$ [ Corresponding angles ]

$\Rightarrow$ $\angle C=\angle E$ [ Corresponding angles ]

$\Rightarrow$ $\angle A=\angle A$ [ Common angle]

$\therefore$ $4△ABC\sim △ADE$ [ By AAA criteria ]

$\left(i\right)$ $\frac{area\left(△ABC\right)}{area\left(△ADE\right)}={\left(\frac{BC}{DE}\right)}^2$ [ Ratio of areas of two similar triangles is equal the ratio of squares of their corresponding sides. ]

$\therefore$ $\frac{area\left(△ABC\right)}{16}={\left(\frac{6}{4}\right)}^2$

$\therefore$ $area\left(△ABC\right)=\frac{36\times 16}{16}$

$\therefore$ $area\left(△ABC\right)=36c{m}^2$

$\left(ii\right)$ $\frac{area\left(△ABC\right)}{area\left(△ADE\right)}={\left(\frac{BC}{DE}\right)}^2$ [ Ratio of areas of two similar triangles is equal the ratio of squares of their corresponding sides. ]

$\therefore$ $\frac{area\left(△ABC\right)}{25}={\left(\frac{8}{4}\right)}^2$

$\therefore$ $area\left(△ABC\right)=\frac{64\times 25}{16}$

$\therefore$ $area\left(△ABC\right)=100c{m}^2$

$\left(iii\right)$ $\frac{area\left(△ADE\right)}{area\left(△ABC\right)}={\left(\frac{DE}{BC}\right)}^2$ [Ratio of areas of two similar triangles is equal the ratio of squares of their corresponding sides. ]

$\frac{area\left(△ADE\right)}{area\left(△ABC\right)}={\left(\frac{3}{5}\right)}^2$

$\frac{area\left(△ADE\right)}{area\left(△ABC\right)}=\left(\frac{9}{25}\right)$

$\Rightarrow$ $\frac{25}{9}\times area\left(△ADE\right)=area\left(△ABC\right)$

$\Rightarrow$ $AreaoftrapeziumBCED=area\left(△ABC\right)-area\left(△ADE\right)$

$\therefore$ $\frac{area\left(△ADE\right)}{AreaofBCED}=\frac{area\left(△ADE\right)}{area\left(△ABC\right)-area\left(△ADE\right)}$

$AreaofBCED$ $=\frac{area\left(△ADE\right)}{(\frac{25}{9}-1)area\left(△ADE\right)}$

$=\frac{1}{\frac{16}{9}}$

$=\frac{9}{16}$