Calculating Volume and Area for Geometric Shapes
Let's solve each part of this question one at a time.
a) To calculate the volume of a pyramid with a square base, we can use the formula:
\[ V = \frac{1}{3} \times \text{base area} \times \text{height} \]
In this case, the base is a square with sides of 5 cm, so the base area (A) is:
\[ A = \text{side} \times \text{side} = 5\, \text{cm} \times 5\, \text{cm} = 25\, \text{cm}^2 \]
The height (h) of the pyramid is given as 10 cm. Now we can calculate the volume:
\[ V = \frac{1}{3} \times 25\, \text{cm}^2 \times 10\, \text{cm} \]
\[ V = \frac{1}{3} \times 250\, \text{cm}^3 \]
\[ V = 83.33\, \text{cm}^3 \]
To convert cubic centimeters to liters, we use the fact that 1 liter equals 1000 cubic centimeters:
\[ V = 83.33\, \text{cm}^3 \times \frac{1\, \text{L}}{1000\, \text{cm}^3} = 0.08333\, \text{L} \]
Rounded to two decimal places, the volume in liters is:
\[ V = 0.08\, \text{L} \]
b) For the second part, we have a rectangular plate that is 500 mm by 300 mm, and four corners are rounded to form sectors each with a radius of 25 mm. To determine the final area of the plate, we need to find the area of the rectangle and then subtract the areas of the four sectors.
First, calculate the area of the rectangle:
\[ A_{\text{rectangle}} = \text{length} \times \text{width} = 500\, \text{mm} \times 300\, \text{mm} = 150,000\, \text{mm}^2 \]
Each corner sector is a quarter of a circle with a radius of 25 mm. The area of a full circle is \( \pi r^2 \). A quarter of this area is \( \frac{1}{4} \pi r^2 \) for each sector:
\[ A_{\text{sector}} = \frac{1}{4} \pi (25\, \text{mm})^2 \]
\[ A_{\text{sector}} = \frac{1}{4} \pi \times 625\, \text{mm}^2 \]
\[ A_{\text{sector}} = 156.25 \pi\, \text{mm}^2 \]
Since there are four of these sectors, the total area to be subtracted is:
\[ 4 \times 156.25 \pi\, \text{mm}^2 = 625 \pi\, \text{mm}^2 \]
Now, subtract the total sectors' area from the rectangle's area:
\[ A_{\text{final}} = 150,000\, \text{mm}^2 - 625 \pi\, \text{mm}^2 \]
Assuming \( \pi \approx 3.14159 \):
\[ A_{\text{final}} = 150,000\, \text{mm}^2 - 1963.495 \, \text{mm}^2 \]
\[ A_{\text{final}} \approx 148,036.505\, \text{mm}^2 \]
The final area of the plate, after rounding to the nearest millimeter squared, is approximately 148,037 mm².
