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What is the maximum area of a triangle that can be inscribed in a circle of radius a?
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Verified Answer
The correct answer is:
$\frac{3 \sqrt{3} a^{2}}{4}$
For area of triangle to be maximum, it should be equilateral triangle.
Area of $\Delta \mathrm{OAB}=\frac{1}{2} \mathrm{ab} \sin \theta$
$=\frac{1}{2} \cdot$ a $a \cdot \sin \theta$
$=\frac{1}{2} \mathrm{a}^{2} \cdot \sin 120^{\circ}$
$=\frac{\sqrt{3}}{4} \mathrm{a}^{2}$
Area of triangle $\mathrm{OAB}, \mathrm{OBC}, \mathrm{OAC}$
$=$ Area of triangle $\mathrm{ABC}$
$=3 \times \frac{\sqrt{3}}{4} a^{2}=\frac{3 \sqrt{3}}{4} \mathrm{a}^{2}$
Area of $\Delta \mathrm{OAB}=\frac{1}{2} \mathrm{ab} \sin \theta$
$=\frac{1}{2} \cdot$ a $a \cdot \sin \theta$
$=\frac{1}{2} \mathrm{a}^{2} \cdot \sin 120^{\circ}$
$=\frac{\sqrt{3}}{4} \mathrm{a}^{2}$
Area of triangle $\mathrm{OAB}, \mathrm{OBC}, \mathrm{OAC}$
$=$ Area of triangle $\mathrm{ABC}$
$=3 \times \frac{\sqrt{3}}{4} a^{2}=\frac{3 \sqrt{3}}{4} \mathrm{a}^{2}$
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