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.
In the figure above, circle O is inscribed in equilateral tr
[#permalink]
29 Jul 2020, 10:54
29 Jul 2020, 10:54
1
Expert Reply
1
Bookmarks
00:00
Question Stats:
76% (02:49) correct
23% (02:55) wrong based on 42 sessions
Hide Show timer Statistics
Attachment:
GRE In the figure above, circle O.jpg [ 19.25 KiB | Viewed 4137 times ]
In the figure above, circle O is inscribed in equilateral triangle ABC. If the area of ABC is \(24 \sqrt{3}\), what is the area of circle O?
A. \(2 \pi \sqrt{3}\)
B. \(4 \pi\)
C. \(4 \pi \sqrt{3}\)
D. \(8 \pi\)
E. \(12 \pi\)
Kudos for the right answer and explanation
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
In the figure above, circle O is inscribed in equilateral tr
[#permalink]
29 Jul 2020, 11:34
29 Jul 2020, 11:34
2
Carcass wrote:
Attachment:
GRE In the figure above, circle O.jpg
In the figure above, circle O is inscribed in equilateral triangle ABC. If the area of ABC is \(24 \sqrt{3}\), what is the area of circle O?
A. \(2 \pi \sqrt{3}\)
B. \(4 \pi\)
C. \(4 \pi \sqrt{3}\)
D. \(8 \pi\)
E. \(12 \pi\)
If \(s\) the side length of an equilateral triangle, then the area of the equilateral triangle \(= \frac{s^2\sqrt{3}}{4}\)
Given: The area of ABC is \(24 \sqrt{3}\)
So we can write: \(\frac{s^2\sqrt{3}}{4} = 24 \sqrt{3}\)
Divide both sides of the equation by \(\sqrt{3}\) to get: \(\frac{s^2}{4} = 24\)
Multiply both sides of the equation by \(4\) to get: \(s^2 = 96\)
This means \(s = \sqrt{96} = 4\sqrt{6}\)
Since each point of tangency must divide each side into two equal lengths, we get the following
From here, draw a line from the center of the circle to the point of tangency, and draw a line from the center to one vertex as follows:
Notice that we have a right triangle, because one of the circle properties tells us that a tangent line is perpendicular to the radius at the point of tangency.
Since we have a special 30-60-90 right triangle, we can compare it to the base 30-60-90 right triangle
Since we have similar triangles, the ratios of corresponding sides will be equal.
We can write: 2√6/√3 = x/1
Simplify both sides to get 2√2 = x
This means the RADIUS = 2√2
Area of circle = πr² = π(2√2)² = π(2√2)(2√2) = 8π
Answer: D
RELATED VIDEO
General Discussion
Re: In the figure above, circle O is inscribed in equilateral tr
[#permalink]
29 Jul 2020, 11:26
29 Jul 2020, 11:26
1
Let length of red line be r and length of blue line be R (note that AO=OC) and let length of side be a.
Area of equilateral triangle = (a*a*√3)/4 = 24√3
=> a = 4√6
Now, Ares of triangle can also be written as (1/2)*(AD)*(BC) = 24√3
(1/2)*(AD)*(4√6) = 24√3
=> AD = 6√2
=> R + r = 6√2
=> R = 6√2 - r
=> R^2 = (6√2)^2 + r^2 - 2*r*6√2
=> R^2 = 72 + r^2 - 12r√2 ............. (1)
Now, in triangle ODC
OD^2 + DC^2 = OC^2
=> r^2 + (2√6)^2 = R^2
Now replace value of R^2 from equation (1)
=> r^2 + 24 = 72 + r^2 - 12r√2
=> 12r√2 = 72 - 24 = 48
=> r = 4/√2 = 2√2
=> area = pi*2√2*2√2
=> area = 8*pi
So, the answer is D.
Area of equilateral triangle = (a*a*√3)/4 = 24√3
=> a = 4√6
Now, Ares of triangle can also be written as (1/2)*(AD)*(BC) = 24√3
(1/2)*(AD)*(4√6) = 24√3
=> AD = 6√2
=> R + r = 6√2
=> R = 6√2 - r
=> R^2 = (6√2)^2 + r^2 - 2*r*6√2
=> R^2 = 72 + r^2 - 12r√2 ............. (1)
Now, in triangle ODC
OD^2 + DC^2 = OC^2
=> r^2 + (2√6)^2 = R^2
Now replace value of R^2 from equation (1)
=> r^2 + 24 = 72 + r^2 - 12r√2
=> 12r√2 = 72 - 24 = 48
=> r = 4/√2 = 2√2
=> area = pi*2√2*2√2
=> area = 8*pi
So, the answer is D.
Attachments
GRE In the figure above, circle O.jpg [ 12.98 KiB | Viewed 3997 times ]
