In figure given below, triangle ABC is right angled at B. Given that AB = 9 cm, AC = 15 cm and D, E are mid-points of the sides AB and AC respectively, calculate (i) the length of BC (ii) the area of △ADE.

(i) In right angle triangle ABC,

By pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ 15² = 9² + BC²

⇒ BC² = 225 - 81

⇒ BC² = 144

⇒ BC = $\sqrt{144}$ = 12 cm.

Hence, BC = 12 cm.

(ii) Given, D and E are midpoints of AB and AC.

∴ By mid-point theorem,

DE = $\dfrac{1}{2}BC$ = 6 cm.

Area of △ADE = $\dfrac{1}{2} \times AD \times DE = \dfrac{1}{2} \times 4.5 \times 6$ = 13.5 cm²

Hence, area of △ADE = 13.5 cm².