Question

✬ HeY There ✬

• In the given figure, ABC is a right angled triangle, right angled at A, in which AB = 6 cm, BC = 10 cm and I is incentre of ∆ABC. Find the area of the shaded region.
(Take π = 3.14)​

Answer 1

Required Answer:

ᴀʙᴄ ɪs ᴀ ʀɪɢʜᴛ ᴀɴɢʟᴇᴅ ᴛʀɪᴀɴɢʟᴇ ᴡʜᴇʀᴇ Angle A = 90°

BC = 10 cm and AB = 6 cm

Let O be the centre and (r) be the radius of the in-circle.

AB, BC and CA are the tangents to the circle at P, M and N.

Therefore IP = IM = IN = r (radius of the circle)

In ∆BAC,

BC² = AB² + AC²(by Pythagoras theorem)

10² = 6² + AC²

AC² = 100-36

= 64

Therefore, AC = 8 cm.

Area of ∆ABC = $\dfrac{1}{2}$ BH

= $\dfrac{1}{2}$ × AC × AB

= $\dfrac{1}{2}$× 8 × 6

= 24 sq.cm

Area of ∆ABC = Area of ∆IAB + Area of ∆IBC + Area of ∆ICA.

➪24 = $\dfrac{1}{2}$r(AB) + $\dfrac{1}{2}$r(BC) + $\dfrac{1}{2}$r(CA)

➪24 = $\dfrac{1}{2}$r(AB+BC+CA)

➪24 = $\dfrac{1}{2}$r(6+8+10)

➪24 = 12r

Therefore, r = 24/12

= 2cm.

Area of the circle = πr²

= 22/7 × 2²

= 12.56sq.cm

Area of shaded region = Area of ∆ABC - Area of the circle

= 24 - 12.56

= 11.44sq.cm

Answer 2

Answer:

Given :-

• ABC is a right angled triangle at A, in which AB = 6 cm, BC = 10 cm and I is incenter of ∆ABC. ( π = 3.14 )

To Find :-

• What is the area of the shaded region.

Solution :-

Given :

In ∆ABC, ∠A = 90°, AB = 6 cm, BC = 10 cm.

By using Phythagorus Theorem,

✦ BC² = AC² + AB²

⇒ AC² = BC² - AB²

⇒ AC² = (10)² - (6)²

⇒ AC² = 100 - 36

⇒ AC² = 64

⇒ AC = $\sqrt{64}$

⇒ AC = 8 cm

Hence, Area of ∆ABC,

⇒ $\dfrac{1}{2}$ × AC × AB

⇒ $\dfrac{1}{2}$ × 8 × 6

⇒ 24 cm²

Now, let r be the radius of the circle of center I

➢ Area of ∆ICB,

⇒ $\dfrac{1}{2}$ × 10 × r cm²

⇒ 5r cm²

➢ Area of ∆IAB,

⇒ $\dfrac{1}{2}$ × 6 × r cm²

⇒ 3r cm²

➢ Area of ∆ICA,

⇒ $\dfrac{1}{2}$ × 8 × cm²

⇒ 4r cm²

Again, we know that,

❖ Area of (∆ICB + ∆IAB + ∆ICA) = Area of ∆ABC

⇒ 5r + 3r + 4r = 24 cm

⇒ 12r = 24 cm

⇒ r = $\sf\dfrac{\cancel{24}}{\cancel{12}}$

⇒ 2 cm

Area of incircle = πr²

⇒ 3.14 × 2 × 2

⇒ 12.56 cm²

❖ Shaded area = Area of ∆ABC - Area of incircle

⇒ (24 - 12.56) cm²

$\implies$ 11.44 cm²

$\therefore$ The area of shaded region is 11.44 cm² .

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