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05 Oct 2018, 00:46
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63% (02:16) correct
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An equilateral triangle that has an area of 16√3 is inscribed in a circle. What is the area of the circle?
(A) 3π
(B) 16π/3
(C) 64π/3
(D) 12√3*π
(E) 20√3*π
(A) 3π
(B) 16π/3
(C) 64π/3
(D) 12√3*π
(E) 20√3*π
05 Oct 2018, 01:38
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Bunuel
Area of equilateral triangle = \(16√3 = √3/4 a^2\)
Reducing, \(a = 8\)
Radius of Inscribed circle in a equilateral triangle,\(r = √3/6 a\)
\(r = 4/√3\)
Area of circle,\(A = πr^2\)
\(A = π*16/3\)
+1 for B
Area of an equilateral triangle is = √3/4a2 = 16√3
this gives a=8
We also know that half of an equilateral triangle is a 30-60-90 triangle with the longest side 'a', and sides of this triangle are in the ratio of x:x√3: 2x.
so a=2x
X=a/2
Hence, x√3=√3a/2 this is nothing but the median(altitude) of an equilateral triangle.
The important thing to remember is that 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 as each median will pass through it due to the symmetry.
the radius of the circle will be 2/3( median)= 2/3* √3a/2=√3a/3
a=8,this gives us radius= 8*√3/3
Area= πr2=π∗64/3
C:)
this gives a=8
We also know that half of an equilateral triangle is a 30-60-90 triangle with the longest side 'a', and sides of this triangle are in the ratio of x:x√3: 2x.
so a=2x
X=a/2
Hence, x√3=√3a/2 this is nothing but the median(altitude) of an equilateral triangle.
The important thing to remember is that 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 as each median will pass through it due to the symmetry.
the radius of the circle will be 2/3( median)= 2/3* √3a/2=√3a/3
a=8,this gives us radius= 8*√3/3
Area= πr2=π∗64/3
C:)
