Question

The area of the incircle of an equilateral triangle of side 42 cm is
(a)223 cm²
(b)231 cm²
(c)462 cm²
(d)924 cm²

Answer 1

Answer:

The area of circle is 462 cm².

Among the given options option (c) 462 cm² is the correct answer.

Step-by-step explanation:

Given :  

Let ABC is an equilateral triangle of side 42 cm.  

Join OA, OB, and OC. O is the incentre of a circle.

OP, OR & OQ are Radius of a circle and they are equal .

Let OP = OR =  OQ = r  

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

(½ × AB × OR) + (½ × BC × OP) + (½ × AC × OQ) = √3/4 × side²

[Area of ∆ = ½ × base × height , Area of equilateral ∆ = √3/4 side²]

√3/4× (42)² = (½ × 42 × r) + (½ × 42 × r) + (½ × 42 × r)  

√3/4 × 1764  = (½ × 42 × r) (1 + 1 +1 )  

441√3  = (½ × 42 × r) × 3

441√3  = 63r

r = (√3 × 441) /63

r = 7√3

Radius of the inscribed circle =   7√3  cm.

Area of circle,A = πr²

A = 22/7 × (7√3)²

A = 22/7 × 49 × 3

A = 22 × 7 × 3

A = 22 × 21

A = 462 cm²

Area of circle = 462 cm²

Hence, the area of circle is 462 cm².

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

Answer:

Option (C) is correct.

Explanation:

Given ABC is an equilateral triangle.

side (a) = 42 cm

height (h) = AQ = (√3a)/2

Radius of incircle (r) = OQ

= h/3

= (√3a/3×2)

$\boxed {r =\frac{a}{2\sqrt{3}}}$

Now ,

Area of incircle (A) = πr²

=(22/7)×(a/2√3)²

= (22×a²)/(7×4×3)

= (22×42×42)/(7×4×3)

After cancellation,we get

= 22×21

= 462 cm²

Therefore,

Area of the incircle (A) = 462cm²

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