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.
Nov 17Live Online Classes with Top GRE Expert Rich Zwelling
08:00 PM EST
-10:00 PM EST
Get live instruction from a top-scoring, industry-recognized GRE teachers + 6 months of access to TTP Self-Study included! Tues/Thur Sept 7, 2026---Feb. 2, 2027, 1:00pm-3:00pm EST
Dec 14Live Online Classes with Top GRE Expert Michael Butville
08:00 PM EST
-10:00 PM EST
Get live instruction from a top-scoring, industry-recognized GRE teachers + 6 months of access to TTP Self-Study included! Mon/Wed Dec. 14, 2026---Feb. 22, 2027, 8:00pm-10:00pm EST
The area of the triangular region
[#permalink]
17 Jun 2018, 13:33
17 Jun 2018, 13:33
1
1
Expert Reply
00:00
Question Stats:
77% (00:39) correct
22% (00:53) wrong based on 76 sessions
Hide Show timer Statistics
Attachment:
triangle.jpg [ 6.88 KiB | Viewed 7536 times ]
Quantity A |
Quantity B |
The area of the triangular region |
25 |
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: The area of the triangular region
[#permalink]
18 Jun 2018, 20:55
18 Jun 2018, 20:55
2
From the figure we know that this is a 45-45-90 triangle.
In a 45-45-90 triangle sides are distributed with length that it is \(x-x-x\sqrt{2}\)
Given the side with length 10 is opposite to the 90 degree angle.
\(x\sqrt{2} = 10\)
Therefore, \(x = 10/\sqrt{2}\)
\(x = 2*5/\sqrt{2}\)
\(x = 5\sqrt{2}\)
Since this is an isosceles right triangle with height = base = x = \(5\sqrt{2}\)
we can get the area of a triangle as \(\frac{1}{2}* base * height\)
area = \(\frac{1}{2} * 5\sqrt{2}* 5\sqrt{2} = 25\)
In a 45-45-90 triangle sides are distributed with length that it is \(x-x-x\sqrt{2}\)
Given the side with length 10 is opposite to the 90 degree angle.
\(x\sqrt{2} = 10\)
Therefore, \(x = 10/\sqrt{2}\)
\(x = 2*5/\sqrt{2}\)
\(x = 5\sqrt{2}\)
Since this is an isosceles right triangle with height = base = x = \(5\sqrt{2}\)
we can get the area of a triangle as \(\frac{1}{2}* base * height\)
area = \(\frac{1}{2} * 5\sqrt{2}* 5\sqrt{2} = 25\)
Re: The area of the triangular region
[#permalink]
19 Jan 2019, 13:40
19 Jan 2019, 13:40
its 45 45 90
Now area is 1/2 * 10/√2 *10/√2
=25
Now area is 1/2 * 10/√2 *10/√2
=25
