Given a circle with circumference $44 ext{ cm}$, where $BC$ is the diameter and $D$ is the midpoint of $AC$. Compare the following quantities:

- Quantity I: Area of the shaded region
- Quantity II: $7\pi ext{ cm}^2$

![Screenshot 2025-09-06 21.56.55.png](https://res.cloudinary.com/aptidude/image/upload/v1757176088/aptidude_questions/sdgsfygietobkbzzqq6e.png)

Choose the correct relation between Quantity I and Quantity II:

Question

Given a circle with circumference $44 	ext{ cm}$, where $BC$ is the diameter and $D$ is the midpoint of $AC$. Compare the following quantities:

- Quantity I: Area of the shaded region
- Quantity II: $7\pi 	ext{ cm}^2$

![Screenshot 2025-09-06 21.56.55.png](https://res.cloudinary.com/aptidude/image/upload/v1757176088/aptidude_questions/sdgsfygietobkbzzqq6e.png)

Choose the correct relation between Quantity I and Quantity II:

Accepted Answer

**Solution:**

1. Given the circumference of the circle is $44 	ext{ cm}$.

2. Calculate the radius $r$ of the circle using the formula:
   $$	ext{Circumference} = 2\pi r$$
   $$44 = 2 	imes \pi 	imes r$$
   $$r = \frac{44}{2\pi} = \frac{44}{2 	imes 3.14} = 7 	ext{ cm}$$

3. Since $BC$ is the diameter, its length is $2r = 14 	ext{ cm}$.

4. $D$ is the midpoint of $AC$, so $AD = DC = \frac{AC}{2}$.

5. The shaded region's area (Quantity I) is given by:
   $$	ext{Area of semicircle} - 	ext{Area of triangle}$$

- Area of semicircle with radius $7$ cm:
     $$\frac{1}{2} \pi r^2 = \frac{1}{2} 	imes \pi 	imes 7^2 = \frac{1}{2} 	imes \pi 	imes 49 = 24.5 \pi 	ext{ cm}^2$$

- Area of triangle formed with base $14$ cm and height $7$ cm (since $D$ is midpoint):
     $$\frac{1}{2} 	imes 14 	imes 7 = 49 	ext{ cm}^2$$

6. Area of shaded region:
   $$24.5 \pi - 49$$
   Approximating $\pi$ as $3.14$:
   $$24.5 	imes 3.14 = 76.93$$
   $$76.93 - 49 = 27.93 	ext{ cm}^2$$

7. Quantity II is:
   $$7 \pi = 7 	imes 3.14 = 21.98 	ext{ cm}^2$$

8. Comparing the two:
   $$27.93 > 21.98$$

**Answer:** Quantity I > Quantity II (Option a)

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