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The area of ABC
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Updated on: 26 Dec 2018, 11:20
Updated on: 26 Dec 2018, 11:20
Expert Reply
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Question Stats:
66% (01:09) correct
33% (01:06) wrong based on 224 sessions
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Attachment:
#GREpracticequestion The area of ABC.jpg [ 18.79 KiB | Viewed 11867 times ]
Quantity A |
Quantity B |
The area of \(\triangle ABC\) |
The area of \(\triangle DEF\) |
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.
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Re: The area of ABC
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26 Apr 2017, 11:59
26 Apr 2017, 11:59
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Triangle ABC is a isosceles right-angled triangle.A = C = 45 degrees.Therefore, AB = BC = Sq.root(2)
Area of triangle ABC = 1/2 * Sq.root(2) * Sq.root(2)
= 1/2 * 2
= 1
In Triangle DEF , angle F= 30 degrees.
Sin 30 = 1/2 =DE/2
DE = 1
Also, Cos 30 = Sq.root(3)/2 = EF/2
EF = Sq.root(3) = 1.732
This implies area of triangle DEF = 1/2 * 1 * 1.732
= .866
Clearly,Area of ABC is greater than Area of DEF.
Area of triangle ABC = 1/2 * Sq.root(2) * Sq.root(2)
= 1/2 * 2
= 1
In Triangle DEF , angle F= 30 degrees.
Sin 30 = 1/2 =DE/2
DE = 1
Also, Cos 30 = Sq.root(3)/2 = EF/2
EF = Sq.root(3) = 1.732
This implies area of triangle DEF = 1/2 * 1 * 1.732
= .866
Clearly,Area of ABC is greater than Area of DEF.
General Discussion
Re: The area of ABC
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29 Sep 2018, 10:27
29 Sep 2018, 10:27
1
Right triangle 1:1:root{2} and 1:root{3}:2....the one with more symmetry will have the higher area.
