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If you have 550 ft of fencing what are the areas of the different rectangles you could enclose with the fencing?
The largest area rectangle that can be enclosed by 550 feet of Fencing is a square, which would have sides of 550/4 = 137.5 feet and therefore an area of 1.89 X 104 square feet, to the justified number of significant digits. There is no theoretical lower limit greater than zero on the area of a rectangle that would satisfy the stated conditions. As a possible practical limit, consider a rectangle with two sides each of length 273 feet and the other two sides 1 foot each. This would have an area of only 273 square feet. If the short sides could be made only 0.1 foot in length, the longer side lengths would be 274.9 feet and the area only 27.49 square feet.
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If you have 36 feet of fencingwhat are the areas of the different rectangles you could enclose with the fencing?
To determine the areas of different rectangles that can be enclosed with 36 feet of fencing, we can express the perimeter as ( P = 2(l + w) = 36 ), where ( l ) is the length and ( w ) is the width. This simplifies to ( l + w = 18 ). The area ( A ) of the rectangle can be expressed as ( A = l \times w ). By substituting ( w = 18 - l ) into the area formula, we find that the area varies depending on the values of ( l ) and ( w ), reaching a maximum area of 81 square feet when ( l = w = 9 ) (a square). Other combinations of length and width will yield varying areas, all less than or equal to 81 square feet.
What is true about the sum of the length and the width for any rectangles with the same perimeterbut different areas?
For rectangles with the same perimeter, the sum of the length and width is constant, as it is directly related to the perimeter formula (P = 2(length + width)). However, even though they share the same perimeter, rectangles can have different areas depending on the specific values of length and width. This means that while the sum of length and width remains unchanged, the individual dimensions can vary to produce different areas.
When you use the Distributive Property to find areas of rectangles why does it make sense to divide the rectangles so you get groups of 10?
rddffdg
How are the areas of similar rectangles related to the scale factor?
The areas are proportional to the square of the scale factor.
Find 4 rectangles with same area with different perimeter?
10cm by 10cm (perimeter=40cm), 5cm by 20cm (perimeter=50cm), 50cm by 2cm (perimeter=104cm), 100cm by 1cm (perimeter=202cm). All of these rectangles' areas are 100cm2
If you have 36 feet of fencingwhat are the areas of the different rectangles you could enclose with the fencing?
To determine the areas of different rectangles that can be enclosed with 36 feet of fencing, we can express the perimeter as ( P = 2(l + w) = 36 ), where ( l ) is the length and ( w ) is the width. This simplifies to ( l + w = 18 ). The area ( A ) of the rectangle can be expressed as ( A = l \times w ). By substituting ( w = 18 - l ) into the area formula, we find that the area varies depending on the values of ( l ) and ( w ), reaching a maximum area of 81 square feet when ( l = w = 9 ) (a square). Other combinations of length and width will yield varying areas, all less than or equal to 81 square feet.
Type of fencing that enabled farmers to enclose land on the treeless plains?
Barbed wire was a type of fencing that enabled farmers to enclose land on the treeless plains. It was cost-effective and easy to install, allowing for the effective enclosure of large areas of land.
What is true about the sum of the length and the width for any rectangles with the same perimeterbut different areas?
For rectangles with the same perimeter, the sum of the length and width is constant, as it is directly related to the perimeter formula (P = 2(length + width)). However, even though they share the same perimeter, rectangles can have different areas depending on the specific values of length and width. This means that while the sum of length and width remains unchanged, the individual dimensions can vary to produce different areas.
Why could two students obtain different values for the calculated areas of the same rectangle?
because it was estimation, the lengths were different and the rectangles are not the same
How are rectangles related to the distributive property?
Rectangles are related to the distributive property because you can divide a rectangle into smaller rectangles. The sum of the areas of the smaller rectangles will equal the area of the larger rectangle.
How do you find the areas of the rectangles?
multiply the length with the breadth.
Are the areas all the same size?
No, areas can vary in size based on their dimensions. Different geometric shapes, such as squares, rectangles, circles, and triangles, have different formulas to calculate their areas. Additionally, irregular shapes will have unique methods to determine their areas.
Can rectangles with the same perimeter have different areas?
Yes. Say there are two rectangles, both with perimeter of 20. One of the rectangles is a 2 by 8 rectangle. The area of this rectangle is 2 x 8 which is 16. The other rectangle is a 4 by 6 rectangle. It has an area of 4 x 6 which is 24.
When you use the Distributive Property to find areas of rectangles why does it make sense to divide the rectangles so you get groups of 10?
rddffdg
How are the areas of similar rectangles related to the scale factor?
The areas are proportional to the square of the scale factor.
Find 4 rectangles with same area with different perimeter?
10cm by 10cm (perimeter=40cm), 5cm by 20cm (perimeter=50cm), 50cm by 2cm (perimeter=104cm), 100cm by 1cm (perimeter=202cm). All of these rectangles' areas are 100cm2
Why should you learn about how to calculate the perimeter and area of an object?
Knowing how to calculate areas is useful when ordering carpets, other floor coverings or quantities of turf to create a lawn. Calculating perimeters is required to order the correct length of fencing or hedging plants to enclose a garden or other area.
