Question:
In figure, Triangle ABC is a right angled triangle 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 ΔADE.
Solution:
In ΔABC, ∠B = 900,
By using Pythagoras theorem
$A C^{2}=A B^{2}+B C^{2}$
$\Rightarrow 15^{2}=9^{2}+B C^{2}$
$\Rightarrow B C=\sqrt{15^{2}-9^{2}}$
$\Rightarrow \mathrm{BC}=\sqrt{225-81}$
$\Rightarrow \mathrm{BC}=\sqrt{144}=12 \mathrm{~cm}$
In ΔABC,
D and E are mid-points of AB and AC
∴ DE ∥ BC, DE = (1/2) BC [By mid−point theorem]
AD = DB = AB/2 = 9/2 = 4.5 cm [∵ D is the mid−point of AB]]
Area of ΔADE = 1/2 ∗AD∗DE
= 1/2 ∗4.5∗6
$=13.5 \mathrm{~cm}^{2}$
