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The sides of equilateral ABD are tangent to a circle with c
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05 Jan 2021, 09:36
05 Jan 2021, 09:36
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90% (01:40) correct
9% (00:45) wrong based on 11 sessions
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The sides of equilateral \( \triangle ABD\) are tangent to a circle with center O. If \(AB = 12 \sqrt{3}\) and \(OD = 12\), the circumference of the circle is
A. \(6 \pi\)
B. \(12 \pi\)
C. \(6 \pi \sqrt{3}\)
D. \(12 \pi \sqrt{3}\)
E. \(24 \pi\)
Kudos for the right answer and explanation
Re: The sides of equilateral ABD are tangent to a circle with c
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06 Jan 2021, 00:14
06 Jan 2021, 00:14
1
The triangle is an equilateral triangle.
The formulas for equilateral triangles are : (a- side of the triangle)
The in-radius of the equilateral triangle OC= \(\frac{a}{2\sqrt{3}}\)
The circum-radius of the equilateral triangle OA-OB-OD= \(\frac{a}{\sqrt{3}}\)
Side is given as
\(AB = 12 \sqrt{3}\)
So OC = \(\frac{12\sqrt{3}\ }{ 2\sqrt{3}\}\)
OC = 6
Circumference = \(2 * pi * radius\) = \(2*pi*6\) = \(12pi\)
Answer B
The formulas for equilateral triangles are : (a- side of the triangle)
The in-radius of the equilateral triangle OC= \(\frac{a}{2\sqrt{3}}\)
The circum-radius of the equilateral triangle OA-OB-OD= \(\frac{a}{\sqrt{3}}\)
Side is given as
\(AB = 12 \sqrt{3}\)
So OC = \(\frac{12\sqrt{3}\ }{ 2\sqrt{3}\}\)
OC = 6
Circumference = \(2 * pi * radius\) = \(2*pi*6\) = \(12pi\)
Answer B
Re: The sides of equilateral ABD are tangent to a circle with c
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06 Jan 2021, 09:21
06 Jan 2021, 09:21
1
As ABD is an equilaeral and circle is inscribed in the equilateral,
OC : OD : CD = 1:2: sqrt(3)
OD is 12 , so OC is 6.
The circumference of the circle o is 2x pi x OC = 2 x pi x 6
=12pi
Answer : B
OC : OD : CD = 1:2: sqrt(3)
OD is 12 , so OC is 6.
The circumference of the circle o is 2x pi x OC = 2 x pi x 6
=12pi
Answer : B
