GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Diameter BC 2 and AB 1
[#permalink]
27 Jul 2020, 08:19
27 Jul 2020, 08:19
Expert Reply
00:00
Question Stats:
74% (00:28) correct
25% (00:49) wrong based on 35 sessions
Hide Show timer Statistics
Attachment:
GRE The area of triangle ABC.jpg [ 15.78 KiB | Viewed 2054 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
[#permalink]
27 Jul 2020, 08:29
27 Jul 2020, 08:29
Expert Reply
Post A Detailed Correct Solution For The Above Questions And Get A Kudos.
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
Intern
Joined: 25 Jul 2020
Posts: 38
Given Kudos: 0
GRE 1: Q170 V157
Re: Diameter BC 2 and AB 1
[#permalink]
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.
