
The ratio of the perimeter of an equilateral triangle having an altitude equal to the radius of a circle, to the perimeter of an equilateral triangle inscribed in the circle is:
A 1:2
B 1:3
C \[1:\sqrt 3\]
D \[\sqrt 3 :2\]
E 2:3
Answer
596.7k+ views
Hint: In this problem, first we need to find the sides of the equilateral triangles in terms of the radius of the circle, and hence find out the ratio of the perimeters. The sides of the equilateral triangles are equal, and the angle between any two sides is 60 degree. We will draw relationships between the radius of the circle and sides to find out the required.
Complete step-by-step answer:
Consider, the radius of the circle be\[r\], the side of the first equilateral triangle be \[{a_1}\] and the side of the equilateral triangle inscribed in the circle be \[{a_2}\].
The formula for the height \[h\] of the first equilateral triangle having side \[{a_1}\] is shown below.
\[h = \dfrac{{\sqrt 3 }}{2}{a_1}\]
Now, the height of the first equilateral triangle is equal to the radius of the circle, therefore,
\[\begin{gathered}
\,\,\,\,\,\,r = \dfrac{{\sqrt 3 }}{2}{a_1} \\
\Rightarrow {a_1} = \dfrac{2}{{\sqrt 3 }}r \\
\end{gathered}\]
Now, from the above figure,
\[\begin{gathered}
\,\,\,\,\,r\cos 30^\circ = \dfrac{{{a_2}}}{2} \\
\Rightarrow {a_2} = 2r\cos 30^\circ \\
\Rightarrow {a_2} = 2r\left( {\dfrac{{\sqrt 3 }}{2}} \right) \\
\Rightarrow {a_2} = \sqrt 3 r \\
\end{gathered}\]
Now, the ratio of the perimeter of the triangles having sides \[{a_1}\] and \[{a_2}\] is calculated as follows:
\[\begin{gathered}
\,\,\,\,\,\,\dfrac{{{\text{perimeter of triangle having side }}{a_1}}}{{{\text{perimeter of triangle having side }}{a_2}}} \\
\Rightarrow \dfrac{{3\left( {{a_1}} \right)}}{{3\left( {{a_2}} \right)}} \\
\Rightarrow \dfrac{{3\left( {\dfrac{2}{{\sqrt 3 }}r} \right)}}{{3\left( {\sqrt 3 r} \right)}} \\
\Rightarrow \dfrac{2}{3} \\
\Rightarrow 2:3 \\
\end{gathered}\]
Thus, the option (E) is the correct answer.
Note: The perimeter of the triangle is the sum of the length of the sides of the triangle. In an equilateral triangle, all the sides are equal, hence, the perimeter of the equilateral triangle is equal to 3 times its side.
Complete step-by-step answer:
Consider, the radius of the circle be\[r\], the side of the first equilateral triangle be \[{a_1}\] and the side of the equilateral triangle inscribed in the circle be \[{a_2}\].
The formula for the height \[h\] of the first equilateral triangle having side \[{a_1}\] is shown below.
\[h = \dfrac{{\sqrt 3 }}{2}{a_1}\]
Now, the height of the first equilateral triangle is equal to the radius of the circle, therefore,
\[\begin{gathered}
\,\,\,\,\,\,r = \dfrac{{\sqrt 3 }}{2}{a_1} \\
\Rightarrow {a_1} = \dfrac{2}{{\sqrt 3 }}r \\
\end{gathered}\]
Now, from the above figure,
\[\begin{gathered}
\,\,\,\,\,r\cos 30^\circ = \dfrac{{{a_2}}}{2} \\
\Rightarrow {a_2} = 2r\cos 30^\circ \\
\Rightarrow {a_2} = 2r\left( {\dfrac{{\sqrt 3 }}{2}} \right) \\
\Rightarrow {a_2} = \sqrt 3 r \\
\end{gathered}\]
Now, the ratio of the perimeter of the triangles having sides \[{a_1}\] and \[{a_2}\] is calculated as follows:
\[\begin{gathered}
\,\,\,\,\,\,\dfrac{{{\text{perimeter of triangle having side }}{a_1}}}{{{\text{perimeter of triangle having side }}{a_2}}} \\
\Rightarrow \dfrac{{3\left( {{a_1}} \right)}}{{3\left( {{a_2}} \right)}} \\
\Rightarrow \dfrac{{3\left( {\dfrac{2}{{\sqrt 3 }}r} \right)}}{{3\left( {\sqrt 3 r} \right)}} \\
\Rightarrow \dfrac{2}{3} \\
\Rightarrow 2:3 \\
\end{gathered}\]
Thus, the option (E) is the correct answer.
Note: The perimeter of the triangle is the sum of the length of the sides of the triangle. In an equilateral triangle, all the sides are equal, hence, the perimeter of the equilateral triangle is equal to 3 times its side.
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