Question

HEllo guys ❤️❤️

In a fig . , a circle is inscribed in an equilateral tri. ABC of side 12 cm . Find the radius of inscribed circle & the area of shaded region.

BEST ANSWER WILL BE MARKED AS BRAINLIST.....✌️✌️☺️​

Answer 1

$\huge\underline\mathcal\purple{Solution}$

Given that ABC is an equilateral triangle of side 12 cm .

Let r be the radius of the circle .Join O(Centre ) to A , B and C .Also join OD , OE amd OF .

Also , AB , BC and AC are tangents to the circle .

=> OD _|_BC, OE_|_ AB and OF_|_AC .

Hence , Area of triangle ABC = ar(AOB )+ ar ( BOC ) + ar ( AOC )

$= > \frac{ \sqrt{3} }{4} {a}^{2} = \frac{1}{2} \times AB \times r + \frac{1}{2} \times BC \times r + \frac{1}{2} \times AC\times r$

$= > \frac{ \sqrt{3} }{4} ( {12})^{2} = \frac{1}{2} \times 1 \times r + \frac{1}{2} \times 12 \times r + \frac{1}{2} \times 12 \times r$

$= > \sqrt{3} \times 3 \times 12 = 6r + 6r + 6r$

$= 36 \sqrt{3} = 18r$

$= > r = \frac{36 \sqrt{3} }{18}$

$= > r = 2 \sqrt{3}$

$= > r = 2 \times 1.73$

$= > r = 3.46cm$

Hence, Radius of the circle = 3.46 cm .

Now ,

Area of the shaded region = area of equilateral triangle - area of circle

$= \frac{ \sqrt{3} }{4} {a}^{2} - \pi {r}^{2}$

$= \frac{ \sqrt{3} }{4} \times 12 \times 12 - \pi( {2 \sqrt{3} })^{2}$

$= 36 \sqrt{3} - 3.14 \times 12$

$= 36 \times 1.73 - 3.14 \times 12$

$= 12(3 \times 1.73 - 3.14)$

$= 12(5.19 - 3.14)$

$= 12 \times 2.05$

$= 24.60 {cm}^{2}$

Hence , the area of shaded region = 24.60cm^2 .

Answer 2

We know center of incircle lies intersection point of all three altitude . So we assume that point " O " .

And Given : AB = BC = CA = 12 cm

And

π = 3.14 and 3√ = 1.73

Let radius of incircle = r cm

So,

Area of equilateral triangle = 3√4(Side)2

So,

Area of ABC = 3√4×12×12 = 3√×3×12 = 363√ = 36×1.73 = 62.28 cm2

And

Area of ∆ ABC = Area of ∆ AOB + Area of ∆ BOC + Area of ∆ COA

And

We know Area of triangle = 12×Base× Height , So

Area of ∆ ABC = 12×AB×r + 12×BC×r + 12×CA×r

Substitute values , we get

⇒12×12×r + 12×12×r + 12×12×r = 62.28⇒6r + 6r + 6r = 62.28⇒18r = 62.28⇒r = 62.2818⇒r =3.46 ( Ans )

And

Shaded area = Area of ∆ ABC - Area of incircle

We know area of circle = π r2

So,

Area of incircle = 227×3.46×3.46 = 263.37527 = 37.625 cm2

So,

Shaded area = 62.28 - 37.625 = 24.655 cm2