Question

the area of a circle inscribed in an equilateral triangle is 154 cm^2 then find the perimeter of the equilateral triangle​

Answer 1

Given Area of circle=154cm

2

⇒πr

2

=154cm

2

⇒r

2

=

22

154×7

=49cm

⇒r=

49

=7cm

ABC is an equilateral △, h is the altitude of △ABC

0 is the incenter of △ABC, and this is the point of intersection of the angular bisectors. Hence, these bisectors are also the altitude and medians whose point of intersection divides the medians in the ratio 2:1

∴∠ADB=90° & OD=

3

1

AD & OD is radius of circle. Then,

r=

3

h

⇒h=3r=3×7=21cm

Let each side of an equilateral triangle be 'a', then altitude of an equilateral triangle is (

2

3

) times its side. So that

h=

2

3

a⇒a=

3

2h

=

3

2×21

=

3

2×21

3

=14

3

cm

Perimeter of triangle ABC=3a=3×14

3

cm=42

3

cm=42×1.73

=72.66cm

2

hey mate here is your answer

hope it helps you

Answer 2

ANSWER ❤️

Area of the circle = 154 sq cm

$⇒\pi r^2 = 154πr2=154$

$⇒\frac{22}{7} r^2 = 154722r2=154$

 

$⇒ r^2 = 49r2=49$

⇒r = 7 cm

Let the side of the triangle = a cm

$So, s = \frac{3a}{2}23a$

But the radius of the incircle, r = \frac{\Delta}{s}r=sΔ

  Where Δ = Area of the triangle and s = semi-perimeter

$7 = \frac{ \frac{ \sqrt{3} }{4} a^2 }{ \frac{3a}{2} }7=23a43a2$

$7 = \frac{ \sqrt{3}a^2 }{2*3a}$

$7 = \frac{ \sqrt{3}a}{6}$

$a = \frac{42}{ \sqrt{3} }$

$a = \frac{42 \sqrt{3}}{3 } = 14 \sqrt{3}$

$Perimeter of triangle, 3a = 3*14 \sqrt{3} = 42 \sqrt{3}$

Perimeter of triangle = 42(1.73) = 72.66 cm

✔️ Hope my answer helps you✔️

❤️ Follow me ❤️

-itz brainlyqueen23 here