
How do you find the area of the equilateral triangle inscribed in a circle?
Answer
411k+ views
Hint: We will consider a circle with a radius r and an equilateral triangle inscribed in it. We are going to use the sine rule to get the area of the triangle. Area of the triangle is the half of the product of base and height, so with the help of the rule we will obtain the value of base and height. Then the area of the triangle.
Complete step-by-step answer:
First let’s draw the given scenario,
This is the overall situation. Now we know that \[\vartriangle ABC\] is inscribed in the circle with radius r and center O. also that the given triangle is equilateral that has all angles measuring \[{60^ \circ }\].
Now the formula to calculate the area of the triangle is,
\[area\vartriangle = \dfrac{1}{2} \times base \times height\]
Now in the triangle above we will take AC as the base and BM as the height of the triangle. But we need to find the dimensions of the base and height.
For that we will take the help of sine rule that relates the lengths of the sides of the triangle with the sines of the angles opposite to the side. That is in the above case;
\[\dfrac{{AC}}{{\sin {{60}^ \circ }}} = \dfrac{{OA}}{{\sin {{30}^ \circ }}}\]
So the value of AC is,
\[AC = OA\dfrac{{\sin {{60}^ \circ }}}{{\sin {{30}^ \circ }}}\]
Now we know that,
\[\sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2}\& \sin {30^ \circ } = \dfrac{1}{2}\]
So substituting these values we get,
\[AC = OA\dfrac{{\dfrac{{\sqrt 3 }}{2}}}{{\dfrac{1}{2}}}\]
On calculating further,
\[AC = OA\sqrt 3 \]
\[AC = r\sqrt 3 \] this is the value of base in the above case.
Now we need to find the value of BM.
We can observe from the figure above that, \[BM = BO + OM\]
Here BO is clearly visible as r but for OM;
In \[\vartriangle OAM\],
\[\sin {30^ \circ } = \dfrac{{{\text{Opposite}}}}{{{\text{Hypotenuse}}}} = \dfrac{{OM}}{r}\]
So the value of OM is,
\[OM = r.\sin {30^ \circ }\]
Now putting both the values in the equation of BM,
\[BM = r + r.\sin {30^ \circ }\]
\[BM = r + \dfrac{r}{2}\]
Taking the LCM,
\[BM = \dfrac{3}{2}r\]
Now using the formula of area;
\[A\left( {\vartriangle ABC} \right) = \dfrac{1}{2} \times \left( {\sqrt 3 r} \right) \times \dfrac{3}{2}r\]
On calculating we get,
\[A\left( {\vartriangle ABC} \right) = \dfrac{{3\sqrt 3 }}{4}{r^2}\]
This is the answer.
So, the correct answer is “\[A\left( {\vartriangle ABC} \right) = \dfrac{{3\sqrt 3 }}{4}{r^2}\]”.
Note: Note that use of sine formula is the way to get the dimensions of the base and height. We use the sin formula since it is a case of a triangle and the triangle is an equilateral triangle. But it is applicable to any other triangle also.
If a, b and c are the three sides of the \[\vartriangle ABC\]
then we can say that \[\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}}\]
Complete step-by-step answer:
First let’s draw the given scenario,

This is the overall situation. Now we know that \[\vartriangle ABC\] is inscribed in the circle with radius r and center O. also that the given triangle is equilateral that has all angles measuring \[{60^ \circ }\].
Now the formula to calculate the area of the triangle is,
\[area\vartriangle = \dfrac{1}{2} \times base \times height\]
Now in the triangle above we will take AC as the base and BM as the height of the triangle. But we need to find the dimensions of the base and height.
For that we will take the help of sine rule that relates the lengths of the sides of the triangle with the sines of the angles opposite to the side. That is in the above case;
\[\dfrac{{AC}}{{\sin {{60}^ \circ }}} = \dfrac{{OA}}{{\sin {{30}^ \circ }}}\]
So the value of AC is,
\[AC = OA\dfrac{{\sin {{60}^ \circ }}}{{\sin {{30}^ \circ }}}\]
Now we know that,
\[\sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2}\& \sin {30^ \circ } = \dfrac{1}{2}\]
So substituting these values we get,
\[AC = OA\dfrac{{\dfrac{{\sqrt 3 }}{2}}}{{\dfrac{1}{2}}}\]
On calculating further,
\[AC = OA\sqrt 3 \]
\[AC = r\sqrt 3 \] this is the value of base in the above case.
Now we need to find the value of BM.
We can observe from the figure above that, \[BM = BO + OM\]
Here BO is clearly visible as r but for OM;
In \[\vartriangle OAM\],
\[\sin {30^ \circ } = \dfrac{{{\text{Opposite}}}}{{{\text{Hypotenuse}}}} = \dfrac{{OM}}{r}\]
So the value of OM is,
\[OM = r.\sin {30^ \circ }\]
Now putting both the values in the equation of BM,
\[BM = r + r.\sin {30^ \circ }\]
\[BM = r + \dfrac{r}{2}\]
Taking the LCM,
\[BM = \dfrac{3}{2}r\]
Now using the formula of area;
\[A\left( {\vartriangle ABC} \right) = \dfrac{1}{2} \times \left( {\sqrt 3 r} \right) \times \dfrac{3}{2}r\]
On calculating we get,
\[A\left( {\vartriangle ABC} \right) = \dfrac{{3\sqrt 3 }}{4}{r^2}\]
This is the answer.
So, the correct answer is “\[A\left( {\vartriangle ABC} \right) = \dfrac{{3\sqrt 3 }}{4}{r^2}\]”.
Note: Note that use of sine formula is the way to get the dimensions of the base and height. We use the sin formula since it is a case of a triangle and the triangle is an equilateral triangle. But it is applicable to any other triangle also.
If a, b and c are the three sides of the \[\vartriangle ABC\]

then we can say that \[\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}}\]
Recently Updated Pages
Master Class 11 Accountancy: Engaging Questions & Answers for Success

Express the following as a fraction and simplify a class 7 maths CBSE

The length and width of a rectangle are in ratio of class 7 maths CBSE

The ratio of the income to the expenditure of a family class 7 maths CBSE

How do you write 025 million in scientific notatio class 7 maths CBSE

How do you convert 295 meters per second to kilometers class 7 maths CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

Write down 5 differences between Ntype and Ptype s class 11 physics CBSE
