Example Question - area of rectangle
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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.
Calculating Shaded Area of Figure
To find the shaded area of the given figure, you need to calculate the area of the large rectangle and then subtract the area of the smaller, unshaded rectangle that is inside the larger rectangle. The area of a rectangle is found using the formula: Area = length × width 1. Calculate the area of the large rectangle: Length = 17 cm Width = 11 cm Area (large rectangle) = 17 cm × 11 cm = 187 cm² 2. Calculate the area of the smaller rectangle: From the given lengths in the image, you can deduce that the length of the smaller (inner) rectangle would be 17 cm - 3 cm - 3 cm (since there are two 3 cm lengths subtracted from the total width), and for the width 11 cm - 3 cm - 3 cm (two 3 cm lengths subtracted from the total length). Length (small rectangle) = 17 cm - 6 cm = 11 cm Width (small rectangle) = 11 cm - 6 cm = 5 cm Area (small rectangle) = 11 cm × 5 cm = 55 cm² 3. Calculate the shaded area by subtracting the area of the small rectangle from the area of the large rectangle: Shaded area = Area (large rectangle) - Area (small rectangle) Shaded area = 187 cm² - 55 cm² = 132 cm² Therefore, the shaded area of the figure is 132 cm².
Calculating the Area of a Rectangle
The image shows a rectangle, and it seems you might want to calculate its area. The formula to calculate the area of a rectangle is: Area = length × width However, the image does not provide a direct question to answer. Assuming you are asked to calculate the area, based on the lengths provided in the image: Length = 6 cm Width = 4 cm So you would compute the area as follows: Area = 6 cm × 4 cm = 24 cm² Hence, the area of the rectangle is 24 square centimeters. If there's another specific question regarding the image, please provide the question, and I'll be glad to help you with it.
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