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18 Feb 2021, 01:15
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
83% (01:53) correct
17%
(02:37)
wrong
based on 86
sessions
History
Date
Time
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Not Attempted Yet
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. \(24 \pi\)
Attachment:
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18 Feb 2021, 01:26
Kudos
Bookmarks
Bunuel
Side of equilateral ∆ = 12√3
i.e Height of Equilateral ∆ i.e. \(AC= \frac{√3}{2}*Side = = \frac{√3}{2}*12√3 = 18\)
AO = OD = 12 (Given
AC = AO + OC
i.e 18 = 12+OC
i.e. OC = 6 = Radius of Incircle
Circumference \(= 2πr = 12π\)
Answer: Option B
18 Feb 2021, 01:46
Kudos
Bookmarks
OD = OA
Side of the triangle = a = 12√3
Height of the triangle = √3a/2
= 12√3*√3/2
= 18 = AC
OC = AC - OA
= 18 - 12
= 6 = r
Circumference = 2πr
= 2π*6
= 12π
Option B
Posted from my mobile device
Side of the triangle = a = 12√3
Height of the triangle = √3a/2
= 12√3*√3/2
= 18 = AC
OC = AC - OA
= 18 - 12
= 6 = r
Circumference = 2πr
= 2π*6
= 12π
Option B
Posted from my mobile device
