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Intern  S
Joined: 19 Dec 2016
Posts: 45
Location: India
WE: Consulting (Computer Software)
A circle is inscribed in an equilateral triangle of side 24cm, touchin  [#permalink]

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Question Stats: 80% (02:29) correct 20% (02:32) wrong based on 43 sessions

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A circle is inscribed in an equilateral triangle of side 24cm, touching its sides. What is the area of the remaining portion of the triangle?

A. 144√3−48π cm2
B. 121√3−36π cm2
C. 144√3−36π cm2
D. 121√3−48π cm2
E. 133√3−48π cm2

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Originally posted by GTExl on 13 Jul 2017, 13:51.
Last edited by GTExl on 13 Jul 2017, 15:27, edited 1 time in total.
Manager  B
Joined: 24 Jun 2017
Posts: 117
Re: A circle is inscribed in an equilateral triangle of side 24cm, touchin  [#permalink]

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solved it but marked wrong answer:))

1. finding the altitude, altitude in equilateral triangle = median
so if ABC is the triangle and AB=BC=AC =24
then altitude = √(AB^2 - (AC/2)^2) = root(24^2-12^2)=12√3

2. then Area of equilateral triangle
1/2*AB*altitude= 1/2*24*12√3=144√3

3. then finding the radius of inscribed circle, there is a formula for that
r=AB/2√3=24/2√3

4. So area of the circle will be πr^2=π*(24/2√3)^2=48π

Then the area of the remaining portion of the triangle 144√3-48π
Intern  S
Joined: 19 Dec 2016
Posts: 45
Location: India
WE: Consulting (Computer Software)
Re: A circle is inscribed in an equilateral triangle of side 24cm, touchin  [#permalink]

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Attachment:
File comment: triangle img52.jpg [ 13.93 KiB | Viewed 2345 times ]

Let r = radius of the inscribed circle. Then,

Area of Δ ABC = Area of Δ OBC + Area of Δ OCA + area of Δ OAB

=(1/2 × r× BC)+(1/2 × r × CA)+(1/2 × r × AB)
=1/2 x r x(BC + CA + AB)
=1/2 x r x (24 +24 + 24)
=36r cm2 ----- 1st

Also Area of Δ ABC = 1/2 x BC x PA
BC= 24
and PA^2=(BA)^2 - (BP)^2 = 24^2 - 12^2
PA^2= 576 -144 =432
PA=√432 = 12√3

So Area of Δ ABC = 1/2 x 24 x 12√3 = 144√3 -----2nd

now putting 1st = 2nd
36r = 144√3
r=4√3
Area of a circle = π(4√3)^2 = 48π

area of the remaining portion of the triangle= 144√3 - 48π

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Intern  B
Joined: 22 Sep 2016
Posts: 6
Re: A circle is inscribed in an equilateral triangle of side 24cm, touchin  [#permalink]

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Area of triangle = (1/2)bh
b=24
h=(√3)*12 this can be found by splitting the triangle in half applying 30-60-90 rule
Area of triangle = 144√3

Area of circle (tricky part)
Place a dot in the center of the triangle and connect the dot to each vertex of the triangle. This will form 3 small isosceles triangles. Now split one triangle in half to form a smaller triangle(30-60-90 degrees!) Radius of the circle = shortest leg = 12/(√3).

Area of circle = πr^2
Area of circle = π (12/√3)^2 = π48

Area of triangle - area of circle = 144√3 - 48π

Sent from my LG-H910 using GMAT Club Forum mobile app Re: A circle is inscribed in an equilateral triangle of side 24cm, touchin   [#permalink] 13 Jul 2017, 15:05
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