Problem 52
Question
A stained glass window is to be placed in a house. The window consists of a rectangle, 6 feet high by 3 feet wide, with a semicircle at the top. Approximately how many feet of stripping, to the nearest tenth of a foot, will be needed to frame the window?
Step-by-Step Solution
Verified Answer
Approximately 19.7 feet of stripping will be needed to frame the window.
1Step 1: Identify Dimensions of the Rectangle
The rectangle is 6 feet high and 3 feet wide. The perimeter of a rectangle is given by \(2(length + width)\) but since the rectangular part of the window doesn't include the bottom side of the strip, the calculation will be \(2(height) + width\). Plugging the values in gives \(2(6) + 3 = 15\) feet.
2Step 2: Calculate Semicircle's Dimensions
The diameter of the semicircle is the width of the rectangle which is 3 feet. The radius therefore is half of the diameter, \(3/2 = 1.5\) feet.
3Step 3: Calculate Circumference of Semicircle
The circumference of a circle is \(2 \pi radius\). Because the window has a semicircle, the calculation will be \(\pi radius\). Plugging in the values gives \(\pi(1.5) = 4.712\) feet.
4Step 4: Add the Rectangular and Semicircular Perimeters
Add the total length of the rectangle and semicircle to get the amount of stripping needed. So, \(15 + 4.712 = 19.712\) feet.
5Step 5: Approximation
Round the final answer to the nearest tenth, giving a total of 19.7 feet of stripping needed to frame the window.
Key Concepts
Perimeter CalculationCircumference of a SemicircleGeometric Problem Solving
Perimeter Calculation
Understanding the principles of perimeter calculation is an essential part of mathematics, especially in geometry. The perimeter is the distance around the boundary of a two-dimensional shape.
When dealing with a rectangle, for instance, the formula for calculating the perimeter is quite straightforward—the sum of all its sides. However, if one side is not contributing to the boundary (such as in our stained glass window example, where the bottom doesn't need stripping because the semicircle is there), we adjust our calculation accordingly. In this case, the formula becomes twice the height plus the width.
When provided with the dimensions, like a 6 feet by 3 feet rectangle, the calculation is simple: multiply the height by 2 and add the width. This gives us an important part of the overall perimeter needed for materials, such as the stripping for our window's frame. It's essential to consider any unique aspects of the shape in question to accurately calculate the needed materials.
When dealing with a rectangle, for instance, the formula for calculating the perimeter is quite straightforward—the sum of all its sides. However, if one side is not contributing to the boundary (such as in our stained glass window example, where the bottom doesn't need stripping because the semicircle is there), we adjust our calculation accordingly. In this case, the formula becomes twice the height plus the width.
When provided with the dimensions, like a 6 feet by 3 feet rectangle, the calculation is simple: multiply the height by 2 and add the width. This gives us an important part of the overall perimeter needed for materials, such as the stripping for our window's frame. It's essential to consider any unique aspects of the shape in question to accurately calculate the needed materials.
Circumference of a Semicircle
A circle's circumference, or perimeter, is known as the distance around it, calculated using the familiar formula, \(2\text{π}r\), where \(r\) is the radius of the circle. However, when dealing with a semicircle, we must use half of this value, since a semicircle is, by definition, half of a circle. Therefore, the circumference of a semicircle, not including the diameter, is given by \(\text{π}r\).
In the context of our stained glass window, where the width of the rectangle forms the diameter of the semicircle, we first calculate the radius by dividing the diameter by 2 and then apply our semicircle circumference formula. This allows us to understand the exact length of stripping needed for the curved part of our window frame. Given the radius, we can promptly ascertain this length and contribute to the total measurement needed for our project.
In the context of our stained glass window, where the width of the rectangle forms the diameter of the semicircle, we first calculate the radius by dividing the diameter by 2 and then apply our semicircle circumference formula. This allows us to understand the exact length of stripping needed for the curved part of our window frame. Given the radius, we can promptly ascertain this length and contribute to the total measurement needed for our project.
Geometric Problem Solving
Geometric problem-solving challenges us to apply mathematical formulas and reasoning to find solutions to real-world problems, like determining the amount of material needed to frame a stained glass window. This exercise doesn't just test our ability to perform calculations—it also requires logical thinking to address the problem's specific conditions.
Here, combining the steps of calculating the perimeter of the rectangular part and the circumference of the semicircle is key. Such a process may seem straightforward, but it requires an understanding of how to dissect the problem into smaller, manageable parts.
With all the individual lengths calculated, the penultimate step is summing them up to determine the total amount of stripping required. This exercise in geometric problem-solving demonstrates how mathematics can be a valuable tool in practical applications, fostering a mindset that can dissect and approach various challenges efficiently and accurately.
Here, combining the steps of calculating the perimeter of the rectangular part and the circumference of the semicircle is key. Such a process may seem straightforward, but it requires an understanding of how to dissect the problem into smaller, manageable parts.
With all the individual lengths calculated, the penultimate step is summing them up to determine the total amount of stripping required. This exercise in geometric problem-solving demonstrates how mathematics can be a valuable tool in practical applications, fostering a mindset that can dissect and approach various challenges efficiently and accurately.
Other exercises in this chapter
Problem 52
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