In the figure given below, ∠BAC = 90°, ADC = 90°, AD = 6 cm, CD = 8 cm and BC = 26 cm. Find

(i) AC

(ii) AB

(iii) area of the shaded region.

(i) By pythagoras theorem,

In right angle triangle ADC,

⇒ AC² = AD² + DC²

⇒ AC² = 6² + 8²

⇒ AC² = 36 + 64

⇒ AC² = 100

⇒ AC = $\sqrt{100}$ = 10 cm.

Hence, AC = 10 cm.

(ii) By pythagoras theorem,

In right angle triangle ABC,

⇒ BC² = AB² + AC²

⇒ 26² = AB² + 10²

⇒ AB² = 676 - 100

⇒ AB² = 576

⇒ AB = $\sqrt{576}$ = 24 cm.

Hence, AB = 24 cm.

(iii) Area of shaded region = Area of △ABC - Area of △ADC

$= \dfrac{1}{2} \times AB \times AC - \dfrac{1}{2} \times AD \times DC \\[1em] = \dfrac{1}{2} \times 24 \times 10 - \dfrac{1}{2} \times 6 \times 8 \\[1em] = 120 - 24 \\[1em] = 96 \text{ cm}^2.$

Hence, area of shaded region = 96 cm².