GATE Mechanical 2021 Set 2 | GA | Question: 8

Question

The ratio of the area of the inscribed circle to the area of the circumscribed circle of an equilateral triangle is ___________

• A. $\frac{1}{8}$

• A. $\frac{1}{6}$

100. $\frac{1}{4}$

500. $\frac{1}{2}$

Comments

Can we use the concept of centroid

centroid divides each median into segments with a 2:1 ratio (or that the centroid is 2/3 along the median)

$\frac {r}{R} = \frac {1}{2}$

Area_r (inner) / Area_R(outer) = $(\frac {r}{R}) ^2$ = $(\frac {1}{2})^2$ = $\frac {1}{4}$

---

@5a1n1amarjeet I think yes

 

https://byjus.com/question-answer/in-what-ratio-does-the-centroid-of-the-triangle-divide-each-median-of-the-triangle/

---

Answer here!   

Answer 1

 

Given that, the  $\triangle ABC$ is an equilateral triangle.

Let the radius of the inscribed circle be $r\;\text{cm},$ and radius of the circumscribed circle be $R\;\text{cm}.$

In $\triangle OBD,\angle OBD = 30^{\circ},\angle ODB = 90^{\circ}$

$\implies \sin 30^{\circ} = \dfrac{r}{R} $

$\implies\dfrac{1}{2}= \dfrac{r}{R}$

$\implies R = 2r$

Let the area of the inscribed circle be $A_{I}\;\text{cm}^{2}$ and the area of the circumscribed circle  be $A_{C}\;\text{cm}^{2}.$

$\implies \dfrac{A_{I}}{A_{C}} = \dfrac{\pi r^{2}}{\pi R^{2}} = \dfrac{r^{2}}{(2r)^{2}} = \dfrac{1}{4}.$

So, the correct answer is $(C).$

Answer 2

Given that, the  $\triangle ABC$ is an equilateral triangle.

Let the radius of the inscribed circle be $r\;\text{cm},$ and radius of the circumscribed circle be $R\;\text{cm}.$

In $\triangle OBD,\angle OBD = 30^{\circ},\angle ODB = 90^{\circ}$

$\implies \sin 30^{\circ} = \dfrac{r}{R} $

$\implies\dfrac{1}{2}= \dfrac{r}{R}$

$\implies R = 2r$

Let the area of the inscribed circle be $A_{I}\;\text{cm}^{2}$ and the area of the circumscribed circle  be $A_{C}\;\text{cm}^{2}.$

$\implies \dfrac{A_{I}}{A_{C}} = \dfrac{\pi r^{2}}{\pi R^{2}} = \dfrac{r^{2}}{(2r)^{2}} = \dfrac{1}{4}.$

So, the correct answer is $(C).$

Comments

@Lakshman Patel RJIT what is AD is it angle bisector ??

Answer 3

Let the radius of the inscribed circle be $r\;\text{cm}$, the radius of the circumscribed circle be $R\;\text{cm},$ and the side of an equilateral triangle be $a  \;\text{cm}.$

Area of the circumscribed circle $ = \dfrac{abc}{4R} = \dfrac{a^{3}}{4R}\;\text{cm}^{2}.\;[\because \text{here}, a = b = c]$

Area of the inscribed circle $A= rS\;\text{cm}^{2}$, where $S = \frac{a+b+c}{2} = \frac{3a}{2}.$

Now, the required ratio $ = \dfrac{A_{IC}}{A_{CC}} = \dfrac{\pi r^{2}}{\pi R^{2}} = \left(\dfrac{r}{R}\right)^{2} \rightarrow (1)$

Now, we have area of the equilateral triangle is equal to the area of inscribed circle $:\dfrac{a^{2}\sqrt{3}}{4} = \dfrac{3ar}{2}$

$\implies r = \dfrac{\sqrt{3}\;a}{6}$

And, area of the equilateral triangle is equal to the area of circumscribed circle $:\dfrac{a^{2}\sqrt{3}}{4} = \dfrac{a^{3}}{4R}$

$\implies R = \dfrac{a}{\sqrt{3}}$

From the equation $(1),$ we get.

$\dfrac{A_{IC}}{A_{CC}} = \left(\dfrac{r}{R}\right)^{2}$

$\implies \dfrac{A_{IC}}{A_{CC}}  = \left( \dfrac{\frac{\sqrt{3}\;a}{6}}{\frac{a}{\sqrt{3}}}\right)^{2} = \left[\left(\frac{\sqrt{3}\;a}{6}\right)\left(\frac{\sqrt{3}}{a}\right)\right]^{2} = \left(\dfrac{1}{2}\right)^{2} = \dfrac{1}{4}.$

References:

• https://opencurriculum.org/5492/circumscribed-and-inscribed-circles/
• https://en.wikipedia.org/wiki/Equilateral_triangle

Comments

Area of equilateral triangle isn’t equal to that of inscribed or circumscribed circles.

If so you can directly cancel out both area to get ratio as simply 1 (which is wrong) instead of so many substitutions as you did :P

Answer 4

Striaght and Simple:

It’s a equilateral Triangle

so,The incentre,circumcentre at the same point , as well as Centroid also.

Let consider the Median length = x

And we know it will be divided into 2:1 ratio

so, Circumradius is 2x/3

and inradius = x/3

Now we got the radius

Bigger circle Area = pi * (2x/3)^2

and Incircle Area = pi* (x/3)^2

Taking the Ration it will be 1:4