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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 8239 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   10 Feb 2024, 08:58
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