Question

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

(i) The length of BC

(ii) The area of $\Delta ADE$

Answer 1

In $\Delta ABC,\angle B={90}^{\circ }$

AC = 15 cm, AB = 9 cm

D and E are the mid points of sides AB and AC respectively and D, E are joined.

To find:

(i) Length of BC

(ii) Area of $\Delta ADE$

(i) In ABC, $\angle B={90}^{\circ }$

$\therefore A{C}^2=A{B}^2+B{C}^2$ (Pythagonas Theorem)

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

$=(15{)}^2-(9{)}^2=225-81$

$=144=(12{)}^2$

$\therefore$ BC = 12 cm

(ii) $\because$ D and E are the mid points of AB and AC

$\therefore$ DE || BC and $DE=\frac{1}{2}BC$

$\Rightarrow DE=\frac{1}{2}\times 12=6cm$

Now area of $\Delta ADE=\frac{1}{2}DE\times AD$

$=\frac{1}{2}\times 6\times \frac{9}{2}$ $(\because$ D is mid point of AB)

$=\frac{27}{2}=13.5c{m}^2$