Question 4:

The swimming pool has a trapezoidal cross-section. To calculate the volume:

1. Area of Trapezoidal Cross-section:

   • The formula for the area of a trapezoid is: $A = \frac{1}{2}(b_1 + b_2)h$ where $b_1 = 5$ m, $b_2 = 1$ m, and $h = 3$ m.

     Substituting: $A = \frac{1}{2}(5 + 1)(3) = \frac{1}{2}(6)(3) = 9 \, \text{m}^2$

2. Volume of the Swimming Pool:

   • Multiply the area by the length ($20$ m): $V = A \times \text{length} = 9 \times 20 = 180 \, \text{m}^3$

3. Convert to Liters:

   • Since $1 \, \text{m}^3 = 1000 \, \text{liters}$: $V = 180 \times 1000 = 180,000 \, \text{liters}$

Final Answer for Question 4:

The pool requires 180,000 liters of water.

Question 5:

The surface area of a sphere is given as $100 \, \text{cm}^2$. The formulas are:

1. Surface Area Formula: $A = 4\pi r^2$ Solving for the radius $r$: $r^2 = \frac{A}{4\pi} = \frac{100}{4\pi} \approx \frac{100}{12.566} \approx 7.9577$ $r \approx \sqrt{7.9577} \approx 2.8 \, \text{cm}$

2. Volume of the Sphere: The formula for volume is: $V = \frac{4}{3}\pi r^3$ Substituting $r \approx 2.8$: $V = \frac{4}{3}\pi (2.8)^3 \approx \frac{4}{3}\pi (21.952) \approx \frac{4}{3}(68.986) \approx 91.9 \, \text{cm}^3$

Final Answer for Question 5:

(a) Radius: 2.8 cm
(b) Volume: 91.9 cm³

Would you like further details or explanations?

Related Questions:

1. How do you calculate the area of a trapezoid for other shapes?
2. How can you generalize the formula for converting cubic meters to liters?
3. What are alternative methods to calculate the surface area of a sphere?
4. How does the radius affect the volume of a sphere geometrically?
5. How can we apply these calculations in real-life problems like tank filling or designing?

Tip: Always double-check unit conversions to ensure consistent measurements!