An apple pie with radius *R* cm is initially cut into *X* identical pieces, each having an area of $0.77 ext{ cm}^2$. Later, it was discovered that half of the pie was rotten. The remaining half was then cut into $(X - 3)$ pieces, each piece having an area of $0.616 ext{ cm}^2$.

Find the circumference of the original pie.

Question

An apple pie with radius *R* cm is initially cut into *X* identical pieces, each having an area of $0.77 	ext{ cm}^2$. Later, it was discovered that half of the pie was rotten. The remaining half was then cut into $(X - 3)$ pieces, each piece having an area of $0.616 	ext{ cm}^2$.

Find the circumference of the original pie.

Accepted Answer

**Solution:**

1. Let the radius of the original pie be *R* cm.

2. The total area of the pie is given by:
   $$\pi R^2$$

3. Initially, the pie is cut into *X* pieces, each with area $0.77 	ext{ cm}^2$. So,
   $$X 	imes 0.77 = \pi R^2$$

4. It is found that half of the pie is rotten, so the remaining half has area:
   $$\frac{\pi R^2}{2}$$

5. This remaining half is cut into $(X - 3)$ pieces, each with area $0.616 	ext{ cm}^2$. Hence,
   $$(X - 3) 	imes 0.616 = \frac{\pi R^2}{2}$$

6. From step 3, express $X$:
   $$X = \frac{\pi R^2}{0.77}$$

7. Substitute $X$ into step 5:
   $$\left( \frac{\pi R^2}{0.77} - 3 ight) 	imes 0.616 = \frac{\pi R^2}{2}$$

8. Multiply out:
   $$0.616 	imes \frac{\pi R^2}{0.77} - 3 	imes 0.616 = \frac{\pi R^2}{2}$$

9. Simplify the first term:
   $$\frac{0.616}{0.77} \pi R^2 - 1.848 = \frac{\pi R^2}{2}$$

10. Calculate $\frac{0.616}{0.77} \approx 0.8$, so:
    $$0.8 \pi R^2 - 1.848 = 0.5 \pi R^2$$

11. Rearranging:
    $$0.8 \pi R^2 - 0.5 \pi R^2 = 1.848$$

12. Simplify:
    $$0.3 \pi R^2 = 1.848$$

13. Solve for $R^2$:
    $$R^2 = \frac{1.848}{0.3 \pi} = \frac{1.848}{0.3 	imes 3.1416} \approx \frac{1.848}{0.94248} \approx 1.96$$

14. Therefore,
    $$R = \sqrt{1.96} = 1.4 	ext{ cm}$$

15. Calculate the circumference:
    $$2 \pi R = 2 	imes 3.1416 	imes 1.4 = 8.8 	ext{ cm}$$

**Answer:** None of the given options match the circumference $8.8 	ext{ cm}$.

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