Subject: General Aptitude · Chapter: Area · Exam: 2005 · Marks: · Difficulty:
In the given figure, $A B C$ 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? $(r+D E)^{\frac{1}{2}}(r-D E)^{\frac{3}{2}}$
(b)$(r-D E)^{\frac{1}{2}}(r+D E)^{2}$
(c)$(r-D E)^{2}(r+D E)^{2}$
(d)$(r-D E)^{\frac{1}{2}}(r+D E)^{\frac{3}{2}}$
Answer
Answer (as printed): D
Explanation
We have: $A E \perp B C$ and $A D=B D=C D=r$. $A E=A D+D E=r+D E$. In $\triangle B D C$, $$\begin{aligned} & B E=\sqrt{(B D)^{2}-(D E)^{2}}=\sqrt{r^{2}-(D E)^{2}}=\sqrt{(r-D E)(r+D E)} . \\ & \therefore \quad \text { Area of the triangle }=\frac{1}{2} \times B C \times A E \\ & =\frac{1}{2} \times 2 B E \times A E=B E \times A E=\sqrt{(r-D E)(r+D E)}(r+D E) \\ & =(r-D E)^{\frac{1}{2}}(r+D E)^{\frac{3}{2}} . \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.