Question

plzzz solve this question............​

Answer 1

CONSTRUCTION :-

Draw, OD perpendicular AB, OE perpendicular BC, and OF perpendicular AC. join OA , OB and OC.

SOLUTION :-

$\leadsto \tt Area \: of \: equilateral \: Δ \: = \: \frac{ \sqrt{3} }{4}(side) { }^{2} \\ \\ \leadsto \tt \: Area \: of \: Δabc \: = \: ar(Δaob) + ar(Δboc) + ar(Δaoc) \\ \\ \leadsto \tt \: \frac{ \sqrt{3} }{4} (12) {}^{2} = \frac{ab \times r}{2} + \frac{bc \times r}{2} + \frac{ac \times r}{2} \\ \tiny \: \tt \red [ \: ar. \: of \: equilateral \: Δ \: = \frac{ \sqrt{3} }{4} a {}^{2} \red ] \\ \\ \leadsto \tt \: \frac{ \sqrt{3} }{4} (144) \: = \frac{r}{2} (ab + bc + ac) \\ \\ \leadsto \tt \: \frac{ \sqrt{3} }{ \cancel4} ( \cancel{144)} \: = \frac{r}{2} (ab + bc + ac) \: \\ \\ \leadsto \tt \: \: 2 \times 36 \sqrt{3} = r(12 + 12 + 12) \\ \tiny \: \red[ \tt\: as \: we \: know \: that , \: equilateral \: Δ \: all \: sides \: are \: equal \red ] \\ \\ \leadsto \tt \: 72 \sqrt{3} = 36r \\ \\ \leadsto \tt \: r \: = (\frac{72}{36} ) \sqrt{3} \\ \\ \leadsto \tt \: \: r \: = 2 \times 1.73 \: \: \: \: \: \: \: \: (.°. \sqrt{3} = 1.73) \\ \\ \leadsto \tt \: { \boxed{ \tt \: r \: = \: 3.46cm}} \\ \\ \leadsto \tt \: Area \: of \: \: the \: shaded \: region \: = \: ar. \: of \: Δabc \: - \: ar.\: of \: circle \: \\ \\ \leadsto \tt \: \frac{ \sqrt{3} }{4} (12) {}^{2} - 3.14 \times (2 \sqrt{3} ) {}^{2} \: \: \: \: \: \: [ \: ar. \: of \: circle \: = \: \pi \: r {}^{2} ] \: \\ \\ \leadsto \tt \: \frac{(1.73)}{4} \times \: 144 - 3.14 \times (12) \\ \\ \leadsto \tt \: 62.28 - 37.68 \\ \\ \leadsto \tt \: { \boxed{ \tt \: 24.6cm {}^{2} }} \\ \\ \leadsto \tt \: { \boxed{ \boxed{ \red{ \underline{ \tt{the \: \: area \: \: of \: \: the \: shaded \: \: region \: \: is \: \: 24.6cm {}^{2}. }}}}}}$

Answer 2

Given :-

• Side of Equaliteral ∆ = 12 cm.

Formula used :-

• inradius of Equilateral ∆ = (side/2√3).
• Area of circle = πr².
• Area of Equaliteral ∆ = (√3/4)(side)²

Solution :-

→ inradius of Circle = (12/2√3) = (6/√3) = (6/1.73) = 3.46,cm.

So,

→ Area of circle = πr² = (3.14) * (3.46)²

→ Area = 37.59cm²

Hence, Area of incircle will be 37.59cm². (Approx).

____________________

now,

→ Area of Equaliteral ∆ = (√3/4) * 12 * 12

→ Area = 1.73 * 3 * 12

→ Area = 62.28cm².

so,

Shaded Area = Area of Eq.∆ - Area of in-circle.

→ Shaded Area = 62.28 - 37.59 = 24.59cm² (Approx).

Hence , Area of shaded Region is 24.59cm² .(Approx).

_____________________