An equilateral triangle is inscribed into a circle with radius 10cm. What is the area of the triangle?

$As shown in the figure,$

$△ABC is an equilateral triangle inscribed in a circle centered at O.$

$Here, the radius of the circle is 10cm.$

$As the triangle is an equilateral, so the value of each angle is 60° and half of it is 30°.$

$Here, OD \perp BC and OD line divides BC equally.$

$Now, from △BOD,$

$Cos30°= \frac{Base}{Hypotenuse}= \frac{BD}{OB} =\frac{BD}{10}$

$=\frac{\sqrt{3}}{2}=\frac{BD}{10}$

$\therefore BD = 5\sqrt{3}$

$\therefore One Arm of the triangle is = BD + CD = 5\sqrt{3} + 5\sqrt{3} = 10\sqrt{3} cm$

$Now, we know that,$

$Area of a equilateral triangle is \frac{\sqrt{3}}{4}{\left(Arm\right)}^{2}c{m}^2$

$=\frac{\sqrt{3}}{4}{\left(10\sqrt{3}\right)}^2 c{m}^2 =\frac{\sqrt{3}}{4}(100\times 3)c{m}^2=75\sqrt{3} c{m}^2 \left(answer\right).$