In the given figure, $ABC$ is an equilateral triangle which is inscribed inside a circle and whose radius is $r$. Which of the following is the area of the triangle?

### Concept & Area of a Triangle The area of a triangle is given by the standard formula: $$\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}$$ Here, we must express the base $BC$ and the height $AE$ (where $E$ is the foot of the perpendicular from $A$ to $BC$ passing through center $D$) in terms of the circle's radius $r$ and the given segment $DE$. ### Step-by-Step Solution 1. **Determine the Height:** The line segment passing from vertex $A$ through center $D$ to the base $BC$ at point $E$ is the height of the equilateral triangle. $AD$ is the radius $r$. Thus, the total height is $AE = AD + DE = r + DE$. 2. **Determine the Base:** In the right-angled triangle $\triangle BDE$, the hypotenuse $BD$ is also a radius $r$. Using the Pythagorean theorem: $BE^2 + DE^2 = BD^2$ $BE^2 = r^2 - DE^2 \implies BE = \sqrt{r^2 - DE^2}$ The full base $BC$ is twice $BE$, so $BC = 2\sqrt{r^2 - DE^2}$. 3. **Calculate the Area:** $\text{Area} = \frac{1}{2} \times BC \times AE$ $\text{Area} = \frac{1}{2} \times \left(2\sqrt{r^2 - DE^2}\right) \times (r + DE)$ $\text{Area} = \sqrt{(r - DE)(r + DE)} \times (r + DE)$ $\text{Area} = (r - DE)^{\frac{1}{2}} (r + DE)^{\frac{1}{2}} (r + DE)^1 = (r - DE)^{\frac{1}{2}} (r + DE)^{\frac{3}{2}}$ ### Exam Strategy & Shortcut Recognize that $\sqrt{r^2 - DE^2}$ factors into $\sqrt{r-DE}\sqrt{r+DE}$. When you multiply this by the height $(r+DE)$, the $(r+DE)$ terms combine their exponents ($1/2 + 1 = 3/2$), immediately leading to option (d). ### Common Pitfall Failing to recognize that $AD$ and $BD$ are both radii ($r$) of the circumcircle. Students often get stuck looking for the side length of the equilateral triangle using trigonometry, which makes the algebra much messier than simply using the Pythagorean theorem on $\triangle BDE$. ### Final Answer Therefore, the correct answer is **$(r - DE)^{\frac{1}{2}} (r + DE)^{\frac{3}{2}}$**.