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In the figure above an equilateral triangle is inscribed in a circle h
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07 Sep 2023, 01:35
07 Sep 2023, 01:35
Expert Reply
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81% (03:09) correct
18% (02:29) wrong based on 11 sessions
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In the figure above an equilateral triangle is inscribed in a circle having centre at O. If the area of the triangle is \(9 \sqrt{3}\) , what is the value of a?
A. \(2 \sqrt{3}
\)
B. 3
C. 2
D. -2
E. \(-2 \sqrt{3}\)
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Re: In the figure above an equilateral triangle is inscribed in a circle h
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14 Sep 2023, 13:14
14 Sep 2023, 13:14
1
side of eq. triangle = \sqrt{3} * (radius of circle)
area of triangle= 9\sqrt{3}
consider half triangle, i.e. 30-60-90
let, base = x
height = \sqrt{3} * x
area of half triangle = (9 * \sqrt{3}) / 2
1/2 * b * h = (9 * \sqrt{3}) / 2
b * h = (9 * \sqrt{3})
x * x / \sqrt{3} = 9 * \sqrt{3}
x * x = 9
x = 3
base of equilateral triangle is 6
from the first equation we get r = 2 * sqrt(3) which is option A.
area of triangle= 9\sqrt{3}
consider half triangle, i.e. 30-60-90
let, base = x
height = \sqrt{3} * x
area of half triangle = (9 * \sqrt{3}) / 2
1/2 * b * h = (9 * \sqrt{3}) / 2
b * h = (9 * \sqrt{3})
x * x / \sqrt{3} = 9 * \sqrt{3}
x * x = 9
x = 3
base of equilateral triangle is 6
from the first equation we get r = 2 * sqrt(3) which is option A.
Re: In the figure above an equilateral triangle is inscribed in a circle h
[#permalink]
14 Sep 2023, 15:28
14 Sep 2023, 15:28
1
Expert Reply
OE
Equilateral triangle inscribed in a circle, has got centre of the circle where all medians
intersect.
Medians of equilateral triangle, intersect at a point which divides every median in
the ratio 2:1.
Where part towards the vertex is 2 units and the part towards the side is 1 unit.
Area of equilateral triangle =\( \dfrac{\sqrt{3}}{4}\) side^2 \(=9 \sqrt{3} = \dfrac{\sqrt{3}}{4}\)
In the given equilateral triangle Median/ Altitude / Height of equilateral triangle = โ3/2 (๐ ๐๐๐) = โ3/2 (6)
Median = 3 โ3
If we split 3 โ3 , in the ratio 2:1.
It is 2 โ3 :1 โ3.
So, if we look at the figure part towards upper vertex is 2 โ3 units.
Hence a = 2 โ3 .
Ans. (A)
Equilateral triangle inscribed in a circle, has got centre of the circle where all medians
intersect.
Medians of equilateral triangle, intersect at a point which divides every median in
the ratio 2:1.
Where part towards the vertex is 2 units and the part towards the side is 1 unit.
Area of equilateral triangle =\( \dfrac{\sqrt{3}}{4}\) side^2 \(=9 \sqrt{3} = \dfrac{\sqrt{3}}{4}\)
In the given equilateral triangle Median/ Altitude / Height of equilateral triangle = โ3/2 (๐ ๐๐๐) = โ3/2 (6)
Median = 3 โ3
If we split 3 โ3 , in the ratio 2:1.
It is 2 โ3 :1 โ3.
So, if we look at the figure part towards upper vertex is 2 โ3 units.
Hence a = 2 โ3 .
Ans. (A)
