Dimensions above-ground,rectangular pool are 25 feet long, 18 feet wide and 6 feet deep.a.How much water is needed to fill the pool?
(25*18*6 ft^3)(7.48052gal/ft^3)=20197.4 gal
The garden area is x*y.The walkway extends 2 feet on each side, or 4 feet more width (x+4) and 4 feet more depth (y+4).The size of the total space, garden plus walkway, is (x+4) by (y+4).The walkway area is the total area minus the garden area:Walkway area = Total area - Garden area= (x+4)(y+4) - xy= xy + 4x + 4y + 16 - xy= 4x + 4y + 16e.g.: small garden of 2 feet x 3 feet. Area = 6 sqft.Total with walkway is 6 feet x 7 feet. Total area = 42 sqft.Walkway area = 4*2 + 4*3 +16 = 8 + 12 + 16 = 36 sqftcheck: 42 sqft total - 6 sqft garden = 36 sqft walkways. Check!Two side walkways are 2 ft wide by 2 ft deep = 2 * 2*2 = 8 sqftTwo end walkways are 2 ft deep by 3 ft wide = 2 * 2*3 = 12 sqftFour corners are 2 ft square = 4*2*2 = 16 sqfttotal = 36 sqft. Check!
A rectangular pool is 7 feet wide. It is 3 times as long as it is long. What is the dimensions of the pool?
If it's 3 times as long as it is long, it sounds like it's ever expanding and will never stop growing. If you meant 3 times as long as it is wide, it's 21' long since 3 X 7 =21.
The steps to solve this are:The perimeter of a rectangle is given by p=2(l+w) where p is the perimeter, l is the length and w is the width. We are told the length of the field is 10 times the width so l=10w. Using this in the formula above we get:2090=2(10w+w)22w=2090w=95 ftl=950 ft
There will be two answers depending on whether it is the US gallon which is 231 cu.in. or the Imperial gallon (UK) which is 277.42 cu.in.With US measure, 400 gallons = 400*231 = 92400 cu.in.With UK measure, 400 gallons = 400*277.42 = 110968 cu.in.Let the width be x in. depth = 2x and length = 4x. The volume = 8x^3 cu.in.With US measure: 8x^3 = 92400 or x = 22.60445 in. So the width is 22.6 in, depth is 45.2089 in and length is 90.4178 in.With UK measure: 8x^3 = 110968 or x = 24.0271 in. So the width is 24.0271 in, depth is 48.0543 in and length is 96.10867 in.
The area of a rectangle is the product of two adjacent sides. Let the smaller side be equal to x, then [math] 180 = x \cdot 5x [/math] or [math] x^2 = 36 [/math]. Since sides of rectangles are positive we have x = 6. Te larger side would be 5x = 30. The perimeter is therefore 6 + 6 + 30 + 30 = 72.
The rectangular ground is 40 m x 30 m. A 3 m wide path is parallel to the length and the width of the ground. What is the area of the path.The area of the path = [40+30]*3 - 3*3 = 70*3–9 = 210–9 = 201 sq m.
Width=xLength=4xperimeter= 2(x+4x)=10x10x=600x=60Dimension= 240*60 (L*B)
Find the dimensions of the original field. A rectangular field is 4 times as long as it is wide. ?
First we need to define our variables: w = original width and (since it is 4 times as long) 4w = original length. If the length is decreased by 10 feet, then the new length is 4w-10. If the width is increased by 2 feet, then the new width is w + 2. Perimeter means the distance around the field, so that would mean add up the 4 sides (two new lengths and two new widths) So, (4w-10) + (4w-10) + (w+2) + (w+2) = 80 Simplify: 8w - 20 + 2w + 4 = 80 10w - 16 = 80 Now solve. Add 16 to both sides: 10w = 96 Divide by 10: w = 9.6 Remember what w stood for? It was the original width. 4w was the original length, so 4(9.6) = 38.4. The dimensions are 38.4 feet long and 9.6 feet wide. Hope this helped.
A rectangular reflecting pool in a park is 20 ft. wide and 30 ft. long......math question?
Suppose that the width of the strip of grass is x. If you draw this out, adding the strip of grass all the way around the rectangle will add 2x to both of the length and width of the rectangle. So, the combined area of the pool and the strip of grass is: (20 + 2x)(30 + 2x). The area enclosed by the pool is just (20)(30) = 600, so the area of the just the strip of grass is: (20 + 2x)(30 + 2x) - 600. We want this to equal 336 ft^2, so we have the equation: (20 + 2x)(30 + 2x) - 600 = 336 ==> 4x^2 + 100x + 600 - 600 = 336, by FOILing ==> 4x^2 + 100x - 336 = 0, by setting the right side equal to zero ==> 4(x + 28)(x - 3) = 0, by factoring the left side ==> x = -28 and x = 3, by the zero-product property. Since we the strip of grass cannot have a negative width, x = 3 is the only reasonable solution. Therefore, the strip of grass should be 3 feet wide. I hope this helps!