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A circle is inscribed in an equilateral triangle, Find area [#permalink]
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A circle is inscribed in an equilateral triangle, such that the two figures touch at exactly 3 points, one on each side of the triangle. Which of the following is closest to the percent of the area of the triangle that lies within the circle? A 20 B 45 C 60 D 55 E 77
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A circle is inscribed in an equilateral triangle, Find area [#permalink]
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PathFinder007 wrote: A circle is inscribed in an equilateral triangle, such that the two figures touch at exactly 3 points, one on each side of the triangle. Which of the following is closest to the percent of the area of the triangle that lies within the circle?
A 20 B 45 C 60 D 55 E 77 Assume this as an equilateral triangle with side as a (Yes I know, this isn't an equilateral triangle hence i asked you to assume, this is the only diagram I got from internet and I am too lazy to draw one myself) Attachment: File comment: Diagram
Su55k02_m10 (1).gif [ 9.13 KiB  Viewed 33416 times ]
so now area of an equilateral triangle = \(\frac{\sqrt3}{4} * a^2\) 1 and area of the triangle is also equal to Area of triangle AOC+Area of AOB + Area of BOC = \(\frac{1}{2} * a * r\)( r is height of individual triangle) 2 from equation 1 & 2 above \(\frac{\sqrt3}{4} * a^2 = 3 * \frac{1}{2} * a * r\) from here we can get the value of a i.e. \(a = 2\sqrt{3} * r\) 3 Now, In the question we need to find out \(\frac{Area of circle inscribed}{Area of the equilateral triangle}\) which is equal to \(\frac{{\pi * r^2}}{{3* 1/2 * (2 \sqrt3 r)^2}}\) substituting the value of a from equation 3 =\(\frac{\pi}{{3 * \sqrt3}}\) \(\approx \frac{3.14}{{3 * 1.72}}\) \(\approx \frac{3.14}{{3 * 1.72}}\) \(\approx \frac{2}{3} \approx 0.66\) Notice that we reduced the numerator by \(0.26\) so our answer is going to be a bit inflated. Looking at the answer choices, C is the closest. (Notice that there's nothing between 60 and 70 in the options so we can be a little imprecise in this case) Hence the solution is C
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Originally posted by Anamika2014 on 08 Sep 2014, 08:42.
Last edited by Anamika2014 on 09 Sep 2014, 10:27, edited 1 time in total.



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A circle is inscribed in an equilateral triangle, Find area [#permalink]
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PathFinder007 wrote: A circle is inscribed in an equilateral triangle, such that the two figures touch at exactly 3 points, one on each side of the triangle. Which of the following is closest to the percent of the area of the triangle that lies within the circle?
A 20 B 45 C 60 D 55 E 77 Refer diagram below: Attachment:
qSvMt.png [ 9.17 KiB  Viewed 33071 times ]
Let the dimension of equilateral Triangle = a Area of the triangle\(= \frac{\sqrt{3}}{4} a^2\) Radius of inscribed circle \(= \frac{\sqrt{3}}{6} a\) Area of inscribed circle \(= \frac{\pi a^2}{12}\) Percentage area \(= \frac{\frac{\pi a^2}{12}}{\frac{\sqrt{3} a^2}{4}} * 100 = \frac{\pi}{3\sqrt{3}} * 100 = 62.8%\) Answer = C
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Originally posted by PareshGmat on 08 Sep 2014, 21:26.
Last edited by PareshGmat on 09 Sep 2014, 19:38, edited 1 time in total.



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Re: A circle is inscribed in an equilateral triangle, Find area [#permalink]
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08 Sep 2014, 21:46
Anamika2014 wrote: PathFinder007 wrote: A circle is inscribed in an equilateral triangle, such that the two figures touch at exactly 3 points, one on each side of the triangle. Which of the following is closest to the percent of the area of the triangle that lies within the circle?
A 20 B 45 C 60 D 55 E 77 Assume this as an equilateral triangle with side as a (Yes I know, this isn't an equilateral triangle hence i asked you to assume, this is the only diagram I got from internet and I am too lazy to draw one myself) Attachment: Su55k02_m10 (1).gif so now area of an equilateral triangle = \(\frac{\sqrt3}{4} * a^2\) 1 and area of the triangle is also equal to Area of triangle AOC+Area of AOB + Area of BOC = \(\frac{1}{2} * a * r\)( r is height of individual triangle) 2 from equation 1 & 2 above \(\frac{\sqrt3}{4} * a^2 = 3 * \frac{1}{2} * a * r\) from here we can get the value of a i.e. \(a = 2\sqrt{3} * r\) 3 Now, In the question we need to find out \(\frac{Area of circle inscribed}{Area of the equilateral triangle}\) which is equal to \(\frac{{\pi * r^2}}{{3* 1/2 * (2 \sqrt3 r)^2}}\) substituting the value of a from equation 3 =\(\frac{\pi}{{3 * \sqrt3}}\) \approx \frac{3.14}{{3 * 1.72}}
\(\approx \frac{3.4}{{3 * 1.72}}[m]\approx \frac{2}{3} \approx 0.66\) Notice that we reduced the numerator by \(0.26\) so our answer is going to be a bit inflated. Looking at the answer choices, C is the closest. (Notice that there's nothing between 60 and 70 in the options so we can be a little imprecise in this case) Hence the solution is C It has to be 3.14 instead of 3.40
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Re: A circle is inscribed in an equilateral triangle, Find area [#permalink]
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08 Sep 2014, 22:20
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PathFinder007 wrote: A circle is inscribed in an equilateral triangle, such that the two figures touch at exactly 3 points, one on each side of the triangle. Which of the following is closest to the percent of the area of the triangle that lies within the circle?
A 20 B 45 C 60 D 55 E 77 PareshGmat's solution is crisp and perfect. Let me just add the explanation for this: Radius of inscribed circle =\(\frac{\sqrt{3}a}{6}\) We know that the altitude of the equilateral triangle will be \(\frac{\sqrt{3}a}{2}\) The altitude will also be the median and will pass through the center of the circle (since it is an equilateral triangle). We know that centroid divides the median in the ratio 2:1. The centroid will be the center of the circle since each median will pass through it due to symmetry. Hence the radius of the circle will be one third of the altitude. Radius = \(\frac{\sqrt{3}a}{2} * \frac{1}{3} = \frac{\sqrt{3}a}{6}\)
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Re: A circle is inscribed in an equilateral triangle, Find area [#permalink]
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03 Feb 2016, 09:56
PathFinder007 wrote: A circle is inscribed in an equilateral triangle, such that the two figures touch at exactly 3 points, one on each side of the triangle. Which of the following is closest to the percent of the area of the triangle that lies within the circle?
A 20 B 45 C 60 D 55 E 77 To answer this question we need (Area of circle/Area of Triangle)*100 If the circle is in Equilateral triangle then (1/3)*Height of equilateral Triangle = Radius of Circle i.e. Radius of Circle = \((1/3)*(\sqrt{3}/2)*side = Side/2\sqrt{3}\) (Area of circle/Area of Triangle)*100 = \((Pi Side^2/12)*100/ \sqrt{3}*Side^2/4 = pi*100/3\sqrt{3} = 60%\) Answer: Option C
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A circle is inscribed in an equilateral triangle, Find area [#permalink]
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PathFinder007 wrote: A circle is inscribed in an equilateral triangle, such that the two figures touch at exactly 3 points, one on each side of the triangle. Which of the following is closest to the percent of the area of the triangle that lies within the circle?
A 20 B 45 C 60 D 55 E 77 let r=both the radius of the inscribed circle and the base of each of the 6 identical 306090 right triangles subsumed by the equilateral triangle area of circle=⫪r^2 area of equilateral triangle=6*1/2*r*r√3=3r^2√3 ⫪r^2/3r^2√3=⫪/3√3≈.60=60% C
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Re: A circle is inscribed in an equilateral triangle, Find area [#permalink]
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14 Mar 2017, 23:44
PareshGmat wrote: PathFinder007 wrote: A circle is inscribed in an equilateral triangle, such that the two figures touch at exactly 3 points, one on each side of the triangle. Which of the following is closest to the percent of the area of the triangle that lies within the circle?
A 20 B 45 C 60 D 55 E 77 Refer diagram below: Attachment: qSvMt.png Let the dimension of equilateral Triangle = a Area of the triangle\(= \frac{\sqrt{3}}{4} a^2\) Radius of inscribed circle \(= \frac{\sqrt{3}}{6} a\) Area of inscribed circle \(= \frac{\pi a^2}{12}\) Percentage area \(= \frac{\frac{\pi a^2}{12}}{\frac{\sqrt{3} a^2}{4}} * 100 = \frac{\pi}{3\sqrt{3}} * 100 = 62.8%\) Answer = C How did you know to apply these formulas? Was there some kind of algebra you did to arrive at root3 divided by 4? I don't understand where you got root 3 divided 4  I know it works because I plugged it in and tried but don't understand what math I need to do to get there.



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Re: A circle is inscribed in an equilateral triangle, Find area [#permalink]
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Nunuboy1994 wrote: PareshGmat wrote: PathFinder007 wrote: A circle is inscribed in an equilateral triangle, such that the two figures touch at exactly 3 points, one on each side of the triangle. Which of the following is closest to the percent of the area of the triangle that lies within the circle?
A 20 B 45 C 60 D 55 E 77 Refer diagram below: Attachment: qSvMt.png Let the dimension of equilateral Triangle = a Area of the triangle\(= \frac{\sqrt{3}}{4} a^2\) Radius of inscribed circle \(= \frac{\sqrt{3}}{6} a\) Area of inscribed circle \(= \frac{\pi a^2}{12}\) Percentage area \(= \frac{\frac{\pi a^2}{12}}{\frac{\sqrt{3} a^2}{4}} * 100 = \frac{\pi}{3\sqrt{3}} * 100 = 62.8%\) Answer = C How did you know to apply these formulas? Was there some kind of algebra you did to arrive at root3 divided by 4? I don't understand where you got root 3 divided 4  I know it works because I plugged it in and tried but don't understand what math I need to do to get there. If the side of an equilateral triangle is s, The altitude of the equilateral triangle \(= \frac{\sqrt{3}}{2} * s\) (Draw the altitude of an equilateral triangle and then use pythagorean theorem on the right triangle obtained) The area of the equilateral triangle \(= (\frac{1}{2})*base*altitude = (\frac{1}{2})*s*\frac{\sqrt{3}}{2}s = \frac{\sqrt{3}}{4} s^2\)
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Re: A circle is inscribed in an equilateral triangle, Find area [#permalink]
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15 Mar 2017, 18:39
Okay the formulas appear to be more clear and logical now but my question is how do you divide a number by an imperfect square for example, in a equilateral triangle with sides of 2, I used the formula root divided by 4 times 2^2 to get the area. The formula needed to solve the problem as stated is area of circle/ area of triangle * 100 so now we have the denominator. Knowing that 1/3 of the height of equilateral triangle is the radius we can use the Pythagorean Theorem or other formula Karishma stated and arrive at root/ three. The area of the circle is then pi times root/3 squared. We can then simplify and arrive at pi times 1/3 (1/3 is the radius square in simplest form) divided by root 3. On my calculator this equals .60 though how would you make that calculation without a calculator?



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Re: A circle is inscribed in an equilateral triangle, Find area [#permalink]
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Hi, Since couple of replies already used equilateral triangle formula, to solve the question Let’s see now how to solve this question, with using the basic area of triangle formula, If the question asks for percentages or ratios always we can try some number out, Here say the side of the equilateral triangle is “6”, We have to find the area of the triangle and area of the circle inscribed, That is, (Area of the circle inscribed/Area of the triangle)*100 Area of the circle is (pi)* r^2 Area of the triangle is ½ * base * height Refer to the diagram, So, here area of the triangle, ½ * 6 * 3 (root 3) = 9 root 3, Also remember that, radius of the circle is always 1/3 rd the height of the equilateral triangle (This is because, Medians of the triangle intersects at 1:2 ratio). So here the radius would be, 1/3(3 root 3) = root 3 So the Area of the circle is, pi * (root 3)^2 = 3 * pi, We can approximate “pi” value as “3“, because answer choices are wide enough, So then area of the circle approximately is 9. Now, Area of the triangle, we can approximate as 9 root 3 = 9 * 1.7 = 15.3, So we can further approximate it as 15. So we can find the percentage now, (Area of the circle inscribed/Area of the triangle)*100 (9/15)* 100. This is 60%. So the answer is C. Hope this helps
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Re: A circle is inscribed in an equilateral triangle, Find area [#permalink]
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16 Mar 2017, 07:01
VeritasPrepKarishma wrote: PathFinder007 wrote: A circle is inscribed in an equilateral triangle, such that the two figures touch at exactly 3 points, one on each side of the triangle. Which of the following is closest to the percent of the area of the triangle that lies within the circle?
A 20 B 45 C 60 D 55 E 77 PareshGmat's solution is crisp and perfect. Let me just add the explanation for this: Radius of inscribed circle =\(\frac{\sqrt{3}a}{6}\) We know that the altitude of the equilateral triangle will be \(\frac{\sqrt{3}a}{2}\) The altitude will also be the median and will pass through the center of the circle (since it is an equilateral triangle). We know that centroid divides the median in the ratio 2:1. The centroid will be the center of the circle since each median will pass through it due to symmetry. Hence the radius of the circle will be one third of the altitude. Radius = \(\frac{\sqrt{3}a}{2} * \frac{1}{3} = \frac{\sqrt{3}a}{6}\) PareshGmat's solution is comprehensive ..Could you please give some similar problems Like it...I think this is harder than OG books questions.



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Re: A circle is inscribed in an equilateral triangle, Find area [#permalink]
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16 Mar 2017, 08:09
Nunuboy1994 wrote: Okay the formulas appear to be more clear and logical now but my question is how do you divide a number by an imperfect square for example, in a equilateral triangle with sides of 2, I used the formula root divided by 4 times 2^2 to get the area. The formula needed to solve the problem as stated is area of circle/ area of triangle * 100 so now we have the denominator. Knowing that 1/3 of the height of equilateral triangle is the radius we can use the Pythagorean Theorem or other formula Karishma stated and arrive at root/ three. The area of the circle is then pi times root/3 squared. We can then simplify and arrive at pi times 1/3 (1/3 is the radius square in simplest form) divided by root 3. On my calculator this equals .60 though how would you make that calculation without a calculator? You are asked for the closest value, so approximate. sqrt(2) = 1.4 sqrt(3) = 1.7 If needed, these values are likely to be given in the question anyway. Though most likely, options in actual GMAT questions will retain the irrational numbers. So you would probably see options in terms of sqrt(3).
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Re: A circle is inscribed in an equilateral triangle, Find area [#permalink]
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17 Mar 2017, 03:45
let the side of the triangle be a area = √3 a^2/4 median length = √3 a/2 radius = 1/3 *√3 a/ 2 = a/ 2√3
area of circle = pi *a/2√3 * a/2√3 = pi * a^2/ 12
therefore percentage = (pi a^2/12)/ √3 a^2/4
around 60%



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