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In the figure above, circle O is inscribed in equilateral tr
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29 Jul 2020, 10:54
29 Jul 2020, 10:54
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GRE In the figure above, circle O.jpg [ 19.25 KiB | Viewed 4140 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. \(12 \pi\)
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In the figure above, circle O is inscribed in equilateral tr
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29 Jul 2020, 11:34
29 Jul 2020, 11:34
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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
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Re: In the figure above, circle O is inscribed in equilateral tr
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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.
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GRE In the figure above, circle O.jpg [ 12.98 KiB | Viewed 4000 times ]
