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by diebeatsthegmat » Fri Feb 04, 2011 10:38 pm
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) 50%
(B) 55%
(C) 60%
(D) 65%
(E) 70%


can u guy please explain me the question? i dont understand it!
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by Night reader » Fri Feb 04, 2011 11:13 pm
diebeatsthegmat 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? <--this is our click
(A) 50%
(B) 55%
(C) 60%
(D) 65%
(E) 70%


can u guy please explain me the question? i dont understand it!
we need to find S(circle)/S(triangle)
since triangl (ABC) is equ-l we should take r=A*Sqrt(3)/6
height of traingl(ABC)=Sqrt(3)A/2, S(triangl-ABC)=Sqrt(3)A*A/4 OR ___ (A^2 *Sqrt(3))/4
S(circle)=pr^2 OR S(circle)=(p*3*A^2)/36
So (p*3*A^2)/36 : (A^2 *Sqrt(3))/4 = 3*p/9*Sqrt(3) =[p=3.14] 3.14/3*Sqrt(3)

percentage --> 3.14/Sqrt(3) *100=0.6042*100 OR 60%
Last edited by Night reader on Sat Feb 05, 2011 2:21 am, edited 1 time in total.
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by sanju09 » Sat Feb 05, 2011 12:43 am
diebeatsthegmat 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) 50%
(B) 55%
(C) 60%
(D) 65%
(E) 70%


can u guy please explain me the question? i dont understand it!
If p is each altitude of the equilateral triangle then its area is p^2/√3, and the in-radius of the triangle is p/3, and the area of the triangle that lies within the circle is the area of circle whose radius is p/3, which is π p^2/9.

The required percent is hence

= {(π p^2/9)/ (p^2/√3)} × 100

= {(3/9)/ (1/√3)} × 100 [for approximation we can take π = 3]

= 100 √3/3

= 173/3 [for approximation we can take √3 = 1.73]

≈ [spoiler]58%

C is the closest[/spoiler]
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by Night reader » Sat Feb 05, 2011 2:22 am
the correct answer should be 60%, I just revised my prior calc and found by plugging in p=3.14

so 58% closest answer is correct
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by GMATGuruNY » Sat Feb 05, 2011 4:37 am
diebeatsthegmat 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) 50%
(B) 55%
(C) 60%
(D) 65%
(E) 70%


can u guy please explain me the question? i dont understand it!
Image

Triangle:
b = 6
h = 3√3
A = 1/2 * 6 * 3√3 = 9√3 ≈ 15

Circle:
r = √3
A = π(√3)^2 = 3π ≈ 9

Circle/Triangle = 9/15 ≈ 60%.

The correct answer is C.
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