Question

In the following figure, ABC is an equilateral triangle of side 8 cm. A, B and C are the centres of circular arcs of radius 4 cm. Find the area of the shaded region correct upto 2 decimal places. (Take π =3.142 and √3 = 1.732).​

Answer 1

Answer:

The area of shaded region is 2.576 cm² .

Step-by-step explanation:

Given :

Side of equilateral ∆ PQR , a = 8 cm

Radius of each circular arcs, r = 4 cm

Sector angle , θ = 60°

[Each Angle in equilateral triangle is 60°]

Area of shaded region, A = Area of equilateral ∆PQR -  3 × Area of sector

A = √3/4 × side² - 3 [θ/360° × πr²]  

A = √3/4 × 8² - 3 [60°/360° × π× 4²]

A = √3/4 × 64 - 3 [1/6 × π× 16]

A = √3/4 × 64 - 3 [1/6 × π× 16]

A = √3 × 16 - 3[8π/3]

A = √3 × 16 - 8π

A = 1.732 × 16 - 8 × 3.142

[Given √3 = 1.732 & π =  3.142]

A = 27.712 - 25.136

A = 2.576 cm²

Area of shaded region = 2.576 cm²

Hence, the area of shaded region is 2.576 cm² .

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

$\huge\mathfrak\purple {Solution}$

Given :

Side of an equilateral triangle = 8 cm .

radius of the sector = 4 cm .

angle of the sector = 60°

Area of shaded region = ??

Solve :

Area of shaded region = area of equilateral triangle - area of 3 sectors

$= \frac{ \sqrt{3} }{4} {side}^{2} - 3( \frac{theta}{360} \times \pi {r}^{2} )$

$= \frac{ \sqrt{3} }{4} \times {8}^{2} - 3( \frac{60}{360} \times \pi {4}^{2} )$

$= \frac{ \sqrt{3} }{4} \times 64 - 3( \frac{1}{6} \times 16\pi)$

$= \sqrt{3} \times 16 - (3 \times \frac{1}{3} \times 8\pi)$

$= \sqrt{ 3} \times 16 - 8\pi$

Now by using ,

$\pi = 3.142$

and

$\sqrt{3 } = 1.732$

$= 1.732 \times 16 - 8 \times 3.142$

$= 27.712 - 25.136$

$= 2.576$

Hence , Area of shaded region

$= 2.576 {cm}^{2}$