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Diameter BC 2 and AB 1
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27 Jul 2020, 08:19
27 Jul 2020, 08:19
Expert Reply
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Question Stats:
74% (00:28) correct
25% (00:49) wrong based on 35 sessions
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GRE The area of triangle ABC.jpg [ 15.78 KiB | Viewed 2050 times ]
Diameter \(BC=2\) and \( AB= 1\)
Quantity A |
Quantity B |
The area of triangle ABC |
\(\frac{\sqrt{3}}{2}\) |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Re: Diameter BC 2 and AB 1
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27 Jul 2020, 08:29
27 Jul 2020, 08:29
Expert Reply
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Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
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Joined: 25 Jul 2020
Posts: 38
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GRE 1: Q170 V157
Re: Diameter BC 2 and AB 1
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27 Jul 2020, 09:02
27 Jul 2020, 09:02
1
Apply Pythagoras' theorem: the height is 1, and the hypotenuse is 2. Then
\(a^2 + b^2 = c^2\)
\(a = \sqrt{c^2 - b^2}\)
\(a = \sqrt{4 - 1}\)
\(a = \sqrt{3}\)
Hence the area of the triangle is 1/2 ab = 1/2 (\(\sqrt{3}\)) -> answer is C.
\(a^2 + b^2 = c^2\)
\(a = \sqrt{c^2 - b^2}\)
\(a = \sqrt{4 - 1}\)
\(a = \sqrt{3}\)
Hence the area of the triangle is 1/2 ab = 1/2 (\(\sqrt{3}\)) -> answer is C.
