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In the figure above, an equilateral triangle ABC is inscribed in
[#permalink]
24 May 2021, 02:21
24 May 2021, 02:21
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71% (00:56) correct
28% (01:17) wrong based on 7 sessions
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In the figure above, an equilateral triangle ABC is inscribed in circle O..png [ 6.37 KiB | Viewed 2893 times ]
In the figure above, an equilateral triangle ABC is inscribed in circle O.
Quantity A |
Quantity B |
OD |
DE |
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
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
In the figure above, an equilateral triangle ABC is inscribed in
[#permalink]
26 May 2021, 07:48
26 May 2021, 07:48
1
Explanation:
Concept: Whenever an equilateral triangle is inscribed in a circle, the centroid and centre are at the same position.
i.e. OA = OB = OC = OE = radius (r) = \(\frac{2}{3}\) Height of the triangle
Now, AD = AO + OD
H = \(\frac{2}{3}\)H + OD
OD = \(\frac{1}{3}\)H
Also, DE = OE - OD
DE = \(\frac{2}{3}\)H - \(\frac{1}{3}\)H = \(\frac{1}{3}\)H
Col. A: \(\frac{1}{3}\)H
Col. B: \(\frac{1}{3}\)H
Hence, option C
Concept: Whenever an equilateral triangle is inscribed in a circle, the centroid and centre are at the same position.
i.e. OA = OB = OC = OE = radius (r) = \(\frac{2}{3}\) Height of the triangle
Now, AD = AO + OD
H = \(\frac{2}{3}\)H + OD
OD = \(\frac{1}{3}\)H
Also, DE = OE - OD
DE = \(\frac{2}{3}\)H - \(\frac{1}{3}\)H = \(\frac{1}{3}\)H
Col. A: \(\frac{1}{3}\)H
Col. B: \(\frac{1}{3}\)H
Hence, option C
gmatclubot
In the figure above, an equilateral triangle ABC is inscribed in [#permalink]
26 May 2021, 07:48
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