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Area of circle inscribed in a triangle

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KarunMendiratta
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Area of circle inscribed in a triangle [#permalink] New post   05 May 2021, 00:26
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Quantity A
Quantity B
Area of circle inscribed in a right triangle ABC with sides 6, 8 and 10
Area of circle inscribed in an equilateral triangle DEF with each side 4


A) Quantity A is greater
B) Quantity B is greater
C) The two quantities are equal
D) The relationship cannot be determined from the information given
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KarunMendiratta
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Re: Area of circle inscribed in a triangle [#permalink] New post   05 May 2021, 23:11
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KarunMendiratta wrote:
Quantity A
Quantity B
Area of circle inscribed in a right triangle ABC with sides 6, 8 and 10
Area of circle inscribed in an equilateral triangle DEF with each side 4


A) Quantity A is greater
B) Quantity B is greater
C) The two quantities are equal
D) The relationship cannot be determined from the information given


Explanation:

Join all three vertices with the centre (O) of the circle in triangle ABC

Ar. of triangle ABC = Ar. of AOB + Ar. of BOC + Ar. of AOC
\(\frac{1}{2}(6)(8) = \frac{1}{2}(6)(r) + \frac{1}{2}(8)(r) + \frac{1}{2}(10)(r) = 12r\)

\(24 = 12r\)
\(r = 2\)

Whenever a circle is inscribed inside an equilateral triangle with each side \(a\), its radius is given by \(r = \frac{a}{2\sqrt{3}}\)

i.e. radius of circle inscribed in triangle DEF = \(\frac{4}{2\sqrt{3}}\)

Col. A: \(π(2)^2\)
Col. B: \(π(\frac{4}{2\sqrt{3}})^2\)

Col. A: \(4π\)
Col. B: \(\frac{4π}{3} = 1.33π\)

Hence, option A
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Re: Area of circle inscribed in a triangle [#permalink] New post   06 May 2021, 00:01
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The inner radius of equilateral triangle is a/2√3 because of which A is bigger
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Re: Area of circle inscribed in a triangle [#permalink] New post   06 May 2021, 00:03
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aliceeee wrote:
The inner radius of equilateral triangle is a/2√3 because of which A is bigger


Absolutely correct. My bad!
I have made the necessary changes
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Re: Area of circle inscribed in a triangle [#permalink] New post   01 Jun 2021, 22:35
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KarunMendiratta wrote:
KarunMendiratta wrote:
Quantity A
Quantity B
Area of circle inscribed in a right triangle ABC with sides 6, 8 and 10
Area of circle inscribed in an equilateral triangle DEF with each side 4


A) Quantity A is greater
B) Quantity B is greater
C) The two quantities are equal
D) The relationship cannot be determined from the information given


Explanation:

Join all three vertices with the centre (O) of the circle in triangle ABC

Ar. of triangle ABC = Ar. of AOB + Ar. of BOC + Ar. of AOC
\(\frac{1}{2}(6)(8) = \frac{1}{2}(6)(r) + \frac{1}{2}(8)(r) + \frac{1}{2}(10)(r) = 12r\)

Sir, I am bit confused in these two lines,

Ar. of triangle ABC = Ar. of AOB + Ar. of BOC + Ar. of AOC
\(\frac{1}{2}(6)(8) = \frac{1}{2}(6)(r) + \frac{1}{2}(8)(r) + \frac{1}{2}(10)(r) = 12r\)




Please explain, Thanks in Advance :heart

\(24 = 12r\)
\(r = 2\)

Whenever a circle is inscribed inside an equilateral triangle with each side \(a\), its radius is given by \(r = \frac{a}{2\sqrt{3}}\)

i.e. radius of circle inscribed in triangle DEF = \(\frac{4}{2\sqrt{3}}\)

Col. A: \(π(2)^2\)
Col. B: \(π(\frac{4}{2\sqrt{3}})^2\)

Col. A: \(4π\)
Col. B: \(\frac{4π}{3} = 1.33π\)

Hence, option A
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Re: Area of circle inscribed in a triangle [#permalink] New post   02 Jun 2021, 04:28
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kumarneupane4344 wrote:
KarunMendiratta wrote:
KarunMendiratta wrote:
Quantity A
Quantity B
Area of circle inscribed in a right triangle ABC with sides 6, 8 and 10
Area of circle inscribed in an equilateral triangle DEF with each side 4


A) Quantity A is greater
B) Quantity B is greater
C) The two quantities are equal
D) The relationship cannot be determined from the information given


Explanation:

Join all three vertices with the centre (O) of the circle in triangle ABC

Ar. of triangle ABC = Ar. of AOB + Ar. of BOC + Ar. of AOC
\(\frac{1}{2}(6)(8) = \frac{1}{2}(6)(r) + \frac{1}{2}(8)(r) + \frac{1}{2}(10)(r) = 12r\)

Sir, I am bit confused in these two lines,

Ar. of triangle ABC = Ar. of AOB + Ar. of BOC + Ar. of AOC
\(\frac{1}{2}(6)(8) = \frac{1}{2}(6)(r) + \frac{1}{2}(8)(r) + \frac{1}{2}(10)(r) = 12r\)




Please explain, Thanks in Advance :heart

\(24 = 12r\)
\(r = 2\)

Whenever a circle is inscribed inside an equilateral triangle with each side \(a\), its radius is given by \(r = \frac{a}{2\sqrt{3}}\)

i.e. radius of circle inscribed in triangle DEF = \(\frac{4}{2\sqrt{3}}\)

Col. A: \(π(2)^2\)
Col. B: \(π(\frac{4}{2\sqrt{3}})^2\)

Col. A: \(4π\)
Col. B: \(\frac{4π}{3} = 1.33π\)

Hence, option A


Attachment:
Ar. of triangle ABC.png
Ar. of triangle ABC.png [ 11.57 KiB | Viewed 14623 times ]
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Re: Area of circle inscribed in a triangle [#permalink] New post   02 Jun 2021, 04:39
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I got it. Thank you very much KarunMendiratta sir from my deeper heart :heart :heart
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Re: Area of circle inscribed in a triangle [#permalink] New post   03 Sep 2022, 18:36
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KarunMendiratta wrote:
KarunMendiratta wrote:
Quantity A
Quantity B
Area of circle inscribed in a right triangle ABC with sides 6, 8 and 10
Area of circle inscribed in an equilateral triangle DEF with each side 4


A) Quantity A is greater
B) Quantity B is greater
C) The two quantities are equal
D) The relationship cannot be determined from the information given


Explanation:

Join all three vertices with the centre (O) of the circle in triangle ABC

Ar. of triangle ABC = Ar. of AOB + Ar. of BOC + Ar. of AOC
\(\frac{1}{2}(6)(8) = \frac{1}{2}(6)(r) + \frac{1}{2}(8)(r) + \frac{1}{2}(10)(r) = 12r\)

\(24 = 12r\)
\(r = 2\)

Whenever a circle is inscribed inside an equilateral triangle with each side \(a\), its radius is given by \(r = \frac{a}{2\sqrt{3}}\)

i.e. radius of circle inscribed in triangle DEF = \(\frac{4}{2\sqrt{3}}\)

Col. A: \(π(2)^2\)
Col. B: \(π(\frac{4}{2\sqrt{3}})^2\)

Col. A: \(4π\)
Col. B: \(\frac{4π}{3} = 1.33π\)

Hence, option A

Another easy way for circle Radius is by the formulae

For right angled triangle inradius=(Hypotenuse-(sum of other two sides))/2
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Re: Area of circle inscribed in a triangle [#permalink] New post   30 Mar 2025, 04:50
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Not trigonometry is required here!

You can reduce the size of the triangle ABC to an equilateral triangle ABG of sides 6. The circle in this triangle must be smaller than the one in the bigger triangle ABC.

However, this new equilateral triangle ABG of sides 6 still definitely contains a bigger circle than the straight up smaller triangle DEF. Therefore, if the circle in ABC > the circle in ABG, and the circle in ABG > the circle in DEF; the circle in ABC must also > the circle in DEF
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Re: Area of circle inscribed in a triangle [#permalink] New post   25 Oct 2025, 06:17
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Step 1 – Quantity A
For a right triangle with sides a, b, and hypotenuse c, the radius r of the inscribed circle is given by:
r = (a + b - c) / 2

For sides a = 6, b = 8, and c = 10:
r = (6 + 8 - 10) / 2 = 4 / 2 = 2

Area of the inscribed circle:
A = πr2 = π(2)2 = 4π


Step 2 – Quantity B
The second triangle is equilateral with side length a = 4.
Height h of an equilateral triangle:
h = (√3 / 2)a = (√3 / 2) × 4 = 2√3
If you forget this formula, you can find h using the Pythagorean theorem- you can either solve for the height or realize that the triangle formed by the height is a 30-60-90 triangle.
The perpendicular from a vertex divides the triangle into two 30°–60°–90° right triangles. Each has half the base = 2 and hypotenuse = 4, so the height is 2√3.

The radius r of the inscribed circle in an equilateral triangle is:
r = h / 3 = (2√3) / 3
Area of this circle:
A = πr2 = π((2√3)/3)2 = π × (12/9) = 4π / 3



--> You can also easily get the radius(r) of the circle inscribed inside any equilateral triangle with side 'a', by using this formula--> r = a/(2√3)

Step 3 – Comparison
Quantity A = 4π, Quantity B = (4π) / 3
Therefore, Quantity A > Quantity B
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Re: Area of circle inscribed in a triangle [#permalink]
25 Oct 2025, 06:17
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