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.
The area of the circle above, with center C, is
[#permalink]
12 Oct 2020, 07:33
12 Oct 2020, 07:33
2
Expert Reply
6
Bookmarks
00:00
Question Stats:
57% (01:52) correct
42% (01:43) wrong based on 142 sessions
Hide Show timer Statistics
Attachment:
GRE the area of the circle above.png [ 8.37 KiB | Viewed 6574 times ]
The area of the circle above, with center C, is \(36\pi\) , and \(x > 90\)
Quantity A |
Quantity B |
The area of the triangle |
18 |
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.
Kudos for the right answer and explanation
Re: The area of the circle above, with center C, is
[#permalink]
Updated on: 15 Oct 2020, 03:42
Updated on: 15 Oct 2020, 03:42
3
The area of any triangle with a x° angle in the domain of 90° < x < 180°, is less than the area of a triangle with x=90°. Having said that, we can calculate the area of a 90 degress triangle:
Area of the circle
\(Á_{circle} = \pi*r = 36\pi\)
\(r = 6\)
but, we know that the area of a triangle is given by:
\(Á_{triangle} = \frac{b*h}{2}\)
and, because we have a isoseles, the area is:
\(Á_{triangle} = \frac{r*r}{2} = \frac{6*6}{2}\)
\(Á_{triangle} = 18\)
but, x is MORE THAN 90°, therefore, we are completely sure that the area of any triangle in that domain must be less than 18.
Option B
Area of the circle
\(Á_{circle} = \pi*r = 36\pi\)
\(r = 6\)
but, we know that the area of a triangle is given by:
\(Á_{triangle} = \frac{b*h}{2}\)
and, because we have a isoseles, the area is:
\(Á_{triangle} = \frac{r*r}{2} = \frac{6*6}{2}\)
\(Á_{triangle} = 18\)
but, x is MORE THAN 90°, therefore, we are completely sure that the area of any triangle in that domain must be less than 18.
Option B
Re: The area of the circle above, with center C, is
[#permalink]
15 Oct 2020, 02:28
15 Oct 2020, 02:28
1
cote wrote:
The area of any triangle with a x° angle in the domain of 90° < x < 180°, is less than the area of a triangle with x=90°. Having said that, we can calculate the area of a 90 degress triangle:
Area of the circle
\(Á_{circle} = \pi*r = 36\pi\)
\(r = 6\)
but, we know that the area of a triangle is given by:
\(Á_{triangle} = \frac{b*h}{2}\)
and, because we have a isoseles, the area is:
\(Á_{triangle} = \frac{r*r}{2} = \frac{6*6}{2}\)
\(Á_{triangle} = 18\)
but, x is LESS THAN 90°, therefore, we are completely sure that the area of any triangle in that domain must be less than 18.
Option B
Area of the circle
\(Á_{circle} = \pi*r = 36\pi\)
\(r = 6\)
but, we know that the area of a triangle is given by:
\(Á_{triangle} = \frac{b*h}{2}\)
and, because we have a isoseles, the area is:
\(Á_{triangle} = \frac{r*r}{2} = \frac{6*6}{2}\)
\(Á_{triangle} = 18\)
but, x is LESS THAN 90°, therefore, we are completely sure that the area of any triangle in that domain must be less than 18.
Option B
it is mentioned in question that
x is > 90.,
therefore ans should be A
