Question

In a special racing event, the person who enclosed the maximum area would be the winner and would get ₹ 100 every square metre of area covered by him/her. Jonsson, who successfully completed the race and was the eventual winner, enclosed the area shown in the figure below. What is the prize money won?
\textit{Note: The arc from C to D makes a complete semi-circle. Given: } $AB=3$ m, $BC=10$ m, $CD=BE=2$ m. \begin{center} \includegraphics[width=0.7\textwidth]{01.jpeg} \end{center}

• A. ₹ 2914
• B. ₹ 2457
• C. ₹ 2614
• D. ₹ 2500

Hint:

When a composite region is given, split it into basic shapes (rectangle/triangle/circle). Keep an eye on which dimension is a diameter vs. radius, and confirm the right angle for triangle area as $\tfrac{1}{2}\times \text{base}\times \text{height}$.

Answer 1

The Correct Option is B

Step 1: Decompose the enclosed shape.
It consists of three parts: (i) a rectangle of length $BC=10$ m and height $BE=2$ m,
(ii) a right triangle with base $AB=3$ m and height $BE=2$ m (right angle at $B$),
(iii) a semicircle with diameter $CD=2$ m $\Rightarrow$ radius $r=1$ m attached on the left. Step 2: Compute individual areas.
Rectangle: $A_{\text{rect}} = BC \times BE = 10 \times 2 = 20\ \text{m}^2$.
Triangle: $A_{\text{tri}} = \tfrac{1}{2}\times AB \times BE = \tfrac{1}{2}\times 3 \times 2 = 3\ \text{m}^2$.
Semicircle: $A_{\text{semi}} = \tfrac{1}{2}\pi r^2 = \tfrac{1}{2}\pi(1)^2 = \dfrac{\pi}{2}\ \text{m}^2$. Step 3: Total enclosed area and prize.
Total area $A = 20 + 3 + \dfrac{\pi}{2} = 23 + \dfrac{\pi}{2} \approx 23 + 1.5708 = 24.5708\ \text{m}^2$.
Prize money $= 100 \times A \approx 100 \times 24.5708 = \boxed{₹ 2457\ \text{(approx.)}}$.