Radius of circum circle of an …

This question is asked by the student of class 9th. Therefore, we cannot use Trigonometry to solve it.

Here is the another solution.

Let the side of equilateral triangle be 'a' cm.

Since in an equilateral triangle, the altitude is perpendicular to the opposite side and bisects it.

Also, in an equilateral triangle ABC,

3 x Area of triangle OBC = Area of triangle ABC

=>          3 x 1/2 x Base x Height = 1/2 x Base x Height

=>          3 x 1/2 x a x OM = 1/2 x a x AM

=>          3OM = AM

=>          3OM = 20 + OM

=>          2OM = 20

=>          OM = 10 cm

Now, in triangle OBC, using Pythagoras Theorem,

BM² = OB² - OM²

=>          BM^{<font size="2">2</font>} = 20^{<font size="2">2</font>} - 10^{<font size="2">2</font>}

=>          BM^{<font size="2">2</font>} = 400 - 100 = 300

=>          BM = 10(3)^1/2

Therefore, the side of equilateral triangle ABC, BC = 2 x BM = 20(3)^1/2

Ans.

Here we know that

{tex}\angle {/tex}OBM = 30

So In right angle triangle OMB,

cos 30 = {tex}{BM\over 20}{/tex}{tex}=> {\sqrt 3\over 2}={ BM\over 20}{/tex}

BM= {tex}10\sqrt3{/tex}cm

Now we know that perpendicular from the centre divides the chord in two equal part so

BC= 2BM = {tex}20\sqrt 3 cm{/tex}