The figure below shows an equilateral triangle (∆ABC) tangent to circl

Question

The figure below shows an equilateral triangle (∆ABC) tangent to circle O at three points. If the perimeter of ∆ABC is 18, the area of circle O =

(A) \(2π\)

(B) \(\frac{5π}{2}\)

(C) \(2 \sqrt{2}π\)

(D) \(3π\)

(E) \(2 π\sqrt{3}\)

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2020-12-14_18-21-59.png [ 21.08 KiB | Viewed 2896 times ]

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Official Answer

D

Gknight5603 · Answer

area = inradius*semiperemeter

so inradius = √3 
area= 3π

Posted from my mobile device

CrackverbalGMAT · Answer

SOLUTION:

The in-radius of an equilateral triangle is √3 * (side) /6
(In-radius is 1/3 the length of an altitude, because each altitude is also a median of the triangle)

Perimeter of the given triangle =18 units
=>Each side of the triangle =18/3 = 6units [Equilateral triangle]
In radius = √3 * (6) / 6 =√3

Hence area = π r^2
=π *√3 * √3
=3π (OPTION D)

Hope this helps

Devmitra Sen (Math)

griffith7 · Answer

Less efficient way to solve (without any knowledge of in-radius formula):

equilateral  = (s^2√3)/4 = (36√3)/4 = 9√3

Then split equilateral triangle into six 30-60-90 triangles, in which the shortest leg of each are circle radii.

one of six inner 30-60-90 triangles  = (9√3)/6 = (3√3)/2

Since the base of this triangle is 3 (half of the side of the equilateral) and we now know the area, the height (=radius) can be solved for easily. Then simply solve for circle area.

LucienH · Answer

Perimeter = 18

Let S be a side of the triangle 
3S = 18 -> S = 6

the area of the triangle is 6^2*sqrt(3)/4
simplified as 9sqrt(3)

you can then split the triangle into 3 smaller equilateral triangles that both have their base as a side of the large triangle, and the two equal sides as radii of the circle

then the area of each mini triangle = 3sqrt(3)
Since the area = 1/2*base*height, you can solve for the radius:

3sqrt(3) = 1/2*6*radius
radius = sqrt(3)

Therefore, the area of the circle is 3Pi

The figure below shows an equilateral triangle (∆ABC) tangent to circl

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