Answer

Verified

381.3k+ 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 \[\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}}\]

Recently Updated Pages

what is the correct chronological order of the following class 10 social science CBSE

Which of the following was not the actual cause for class 10 social science CBSE

Which of the following statements is not correct A class 10 social science CBSE

Which of the following leaders was not present in the class 10 social science CBSE

Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE

Which one of the following places is not covered by class 10 social science CBSE

Trending doubts

Harsha Charita was written by A Kalidasa B Vishakhadatta class 7 social science CBSE

Which are the Top 10 Largest Countries of the World?

Banabhatta wrote Harshavardhanas biography What is class 6 social science CBSE

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

How do you graph the function fx 4x class 9 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

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

Why is there a time difference of about 5 hours between class 10 social science CBSE