Question

In the given figure, ABC is an equilateral triangle whose side is 23cm. A circle is drawn which passes
through the midpoints D, E and F of its sides. The area of the shaded region is


pls answer this question​

Answer 1

Correct Question :

In the given figure, ABC is an equilateral triangle whose side is 2√3cm.

A circle is drawn which passes through the midpoints D, E and F of its sides. Then find the area of the shaded region.

Answer :

The area of shaded region is

$\frac{1}{4} (4\pi - 3 \sqrt{3} ) {cm}^{2}$

Explanation :

Given,

• ABC is an equilateral triangle whose side is of length 2√3 cm.

• D, E, F are the mid points of the sides AB , BC , CA respectively.

• A circle is drawn passing through the mid points D, E, F.

Solution :

Radius of incircle is

Radius of incircle =

$\frac{s}{2 \sqrt{3} } cm$

• here, s is the length of the side of traingle.

here , S = 2√3 cm

$\implies \: r \: = \frac{2 \sqrt{3} }{2 \sqrt{3} }$

$\implies \: r = \cancel \frac{2 \sqrt{3} }{2 \sqrt{3} }$

$\implies \: r = 1cm$

=> Here, area of incircle

$\pi {r}^{2}$

$\implies\pi {r}^{2} = \pi {(1)}^{2}$

$\implies \: \pi {r}^{2} = \pi \: sq.cm$

Therefore, the area of the incircle is π sq. cm

Consider , triangles ABC and DEF

by mid-point theorem,

DE = 1/2 (AC)

DF = 1/2 (BC)

EF = 1/2 (AB)

But, AB = BC = CA

therefore,

DE = DF = EF

Now, here three sides in triangle DEF are equal,

therefore, triangle DEF is also an equilateral triangle.

We know that ,

Area of equilateral triangle =

$\frac{ \sqrt{3} }{4} {s}^{2}$

here,

$s = \frac{2 \sqrt{3} }{2} cm$

=> Area of triangle =

$\implies \: \frac{ \sqrt{3} }{4} \times {( \frac{2 \sqrt{3} }{2} )}^{2}$

$\implies \: \frac{ \sqrt{3} \times 4 \times 3}{4 \times 4}$

$\implies \: \frac{ \sqrt{3} \times 3}{4} \times \cancel \frac{4}{4}$

$\implies \: \frac{3 \sqrt{3} }{4} \: sq.cm$

Therefore, the area of triangle is (3√3)/4.

Area of shaded region = Area of circle - Area of triangle.

$a(s) = \pi - \frac{3 \sqrt{3} }{4}$

$a(s) = \frac{4\pi - 3 \sqrt{3} }{4}$

$a(s) = \frac{1}{4} (4\pi - 3 \sqrt{3} ) \: sq.cm$

Therefore, The area of shaded region is

$\frac{1}{4} (4\pi - 3 \sqrt{3} ) \: sq.cm$

Answer 2

Correct Question :-

• In the given figure, ABC is an equilateral triangle whose side is 2√3cm. A circle is drawn which passesthrough the midpoints D, E and F of its sides. The area of the shaded region is

Formula used :-

• inradius of Equaliteral ∆ = (a/2√3) .
• Circum-Radius of Equaliteral ∆ = (a/√3) .
• Area of Equaliteral ∆ = (√3/4) * (side)²
• Area of circle = π * (radius)²

Solution :-

→ Side of Equaliteral ∆ABC = 2√3 cm.

→ Radius of Shaded Circle = Radius of inscribe circle inside a Equaliteral ∆ = (a/2√3) = (2√3)/(2√3) = 1cm. ------- Equation(1)

_____________

Now, As we can see , Their is a Equaliteral ∆ inside circle .

→ Radius of Circum-circle = 1 cm.

So,

→ (side)/√3 = Radius of Circum-circle

→ (side)/√3 = 1

→ (side) = √3 cm. = Side of Equaliteral ∆DEF. ---------- Equation (2).

______________

So, From Both Equations :-

→ Shaded Area = Area of circle - Area of ∆DEF

→ Shaded Area = [πr²] - [(√3/4) (a)² ]

→ Shaded Area = [π(1)²] - [(√3/4) (√3)² ]

→ Shaded Area = π - (3√3/4)

→ Shaded Area = (1/4)[ 4π - 3√3 ] cm². (Ans).