d. 18cm
Q6. A cover, which covers the water bottle fully, is ordered for a water bottle with a
height of 15cm. The radius is one third of the height. Find the area of the cover ordered.
a. 630 cm square
b. 628.5 cm square
c. 600 cm square
d. None of the above
Q7. Find the slant height of a cone with a height of 5cm and radius 12cm.
a. 13m
b. 11cm
c. 13cm
d. 16cm
Q8. A rectangular field needs to be fenced. The length of the field is 30m. The cost of
fencing per meter is 33.7 rupees. The length and breadth are in the ratio 5:13. Find the
area of the field along with the cost of fencing.
a. Cost : 7279 rupees, area : 2000 square meters
b. Cost : 7305 rupees, 2340 square meters
c. Cost : 7279 rupees, 2340 square meters
d. None of the above

Let's solve the questions step by step.

### Q6: Find the area of the cover ordered.
Given:
- Height of the bottle = 15 cm
- Radius of the bottle = \(\frac{1}{3}\) of height = \(\frac{15}{3} = 5 \, \text{cm}\)

The cover will fully cover the bottle, which implies it's like a cylinder including the lateral surface area and the area of the two circular ends (top and bottom).

The total surface area \(A\) of the cover for the cylinder can be found using the formula:
\[
A = 2\pi r h + 2\pi r^2
\]
Where:
- \(r = 5 \, \text{cm}\)
- \(h = 15 \, \text{cm}\)
- \(\pi \approx 3.14\)

First, calculate the lateral surface area (curved surface area):
\[
2\pi r h = 2 \times 3.14 \times 5 \times 15 = 471 \, \text{cm}^2
\]

Next, calculate the area of the top and bottom circles:
\[
2\pi r^2 = 2 \times 3.14 \times 5^2 = 157 \, \text{cm}^2
\]

Total surface area:
\[
A = 471 + 157 = 628 \, \text{cm}^2
\]
Thus, the area of the cover ordered is approximately **628.5 cm²**.

**Answer**: \( \text{b. 628.5 cm}^2 \)

---

### Q7: Find the slant height of the cone.
Given:
- Height of the cone \( h = 5 \, \text{cm} \)
- Radius \( r = 12 \, \text{cm} \)

The slant height \( l \) of the cone can be found using the Pythagorean theorem:
\[
l = \sqrt{r^2 + h^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \, \text{cm}
\]

**Answer**: \( \text{c. 13 cm} \)

---

### Q8: Find the area of the field and the cost of fencing.
Given:
- Length of the field \( l = 30 \, \text{m} \)
- Length and breadth ratio = 5:13

Let the breadth be \( b \). Since the ratio of length to breadth is 5:13:
\[
\frac{l}{b} = \frac{5}{13}
\]
So, breadth \( b = \frac{13}{5} \times 30 = 78 \, \text{m} \).

The area \( A \) of the field:
\[
A = l \times b = 30 \times 78 = 2340 \, \text{m}^2
\]

To find the cost of fencing, calculate the perimeter \( P \) of the field:
\[
P = 2(l + b) = 2(30 + 78) = 2 \times 108 = 216 \, \text{m}
\]

Cost of fencing per meter = 33.7 rupees.
Total cost of fencing:
\[
\text{Cost} = 216 \times 33.7 = 7279.2 \, \text{rupees}
\]

**Answer**: \( \text{c. Cost: 7279 rupees, Area: 2340 square meters} \)

---

### Do you want details or have any questions?

### Related Questions:
1. What is the formula for the surface area of a cylinder?
2. How do you calculate the slant height of a cone with different dimensions?
3. What is the formula for the area of a circle?
4. How do you find the perimeter of a rectangular field?
5. What is the relationship between the length and breadth in terms of their ratio?
6. How do you convert the cost of fencing from rupees per meter to total cost?
7. Can the formula for the surface area of a cylinder be applied to other shapes?
8. How would the area change if the radius of the bottle was doubled?

**Tip:** Always double-check your calculations, especially with units and conversions, to avoid small errors that can change the outcome significantly.

Explore math problems involving the surface area of a cylinder for a bottle cover, calculating the slant height of a cone, and determining the area and cost of fencing a rectangular field. Get detailed step-by-step solutions and learn key geometric formulas. Suitable for high school students.

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