Question

In this figure, triangle ABC is right - angled at B. Given that AB = 9 cm, AC = 15 cm and D and E are the mid-points of the sides AB and AC respectively, calculate the area of $\Delta ADE$.

• A. $13c{m}^2$
• B. $12c{m}^2$
• C. $13.5c{m}^2$
• D. $14c{m}^2$

Answer 1

The correct option is C

$13.5c{m}^2$

By pythogoras theorem, $A{B}^2+B{C}^2=A{C}^2$

$\Rightarrow B{C}^2=A{C}^2-A{B}^2$

$\Rightarrow B{C}^2={15}^2-9^2$

$\Rightarrow B{C}^2=225-81=144$

$\Rightarrow BC=12cm$

As D and E are mid-points of AB and AC, by mid-point theorem, $DE\parallel BC$ and $DE=\frac{1}{2}BC=6cm$

Since, $DE\parallel BC,wehave\Delta ADEcongruentto\Delta ABC$

$\frac{Area\left(\Delta ADE\right)}{Area\left(\Delta ABC\right)}=\frac{D{E}^2}{B{C}^2}$

$\Rightarrow \frac{Area\left(\Delta ADE\right)}{\frac{1}{2}\times AB\times BC}=\frac{(6{)}^2}{B{C}^2}$

$\Rightarrow \frac{Area\left(\Delta ADE\right)}{\frac{1}{2}\times 9\times 12}=\frac{36}{144}$

$\Rightarrow Area\left(\Delta ADE\right)=\frac{1}{4}\times 9\times 6$

$\Rightarrow Area\left(\Delta ADE\right)=\frac{27}{4}=13.5c{m}^2$