perimeter of a rectangle


A beach volleyball court is 10 yards wide and 17 yards long. The rope used for the boundary line cos

A beach volleyball court is 10 yards wide and 17 yards long. The rope used for the boundary line costs $2.00 per yard. How much would it cost to buy a new boundary line for the court? [U]Approach:[/U] [LIST] [*]A volleyball court is shaped as a rectangle. [*]And the boundary line runs on the perimeter of the rectangle. [*]So we want the perimeter of the rectangle [/LIST] Using our [URL='https://www.mathcelebrity.com/rectangle.php?l=17&w=10&a=&p=&pl=Calculate+Rectangle']rectangle calculator with length = 17 and width = 10[/URL], we have: P = [B]54[/B]

A child's bedroom is rectangular in shape with dimensions 17 feet by 15 feet. How many feet of wallp

A child's bedroom is rectangular in shape with dimensions 17 feet by 15 feet. How many feet of wallpaper border are needed to wrap around the entire room? A rectangle has an Perimeter (P) of: P = 2l + 2w We're given l = 17 and w = 15. So we have: P = 2(17) + 2(15) P = 34 + 30 P = [B]64[/B]

A farmer has 165 feet of fencing material in which to enclose a rectangular grazing area. He wants t

A farmer has 165 feet of fencing material in which to enclose a rectangular grazing area. He wants the length x to be greater than 50 feet and the width y to be no more than 20 feet. Write a system to represent this situation. Perimeter of a rectangle: P = 2l + 2w We have P = 165 and l = x --> x>50 and width y <= 20. Plug these into the perimeter formula [B]165 = 2x + 2y where x > 50 and y <= 20[/B]

A garden has a length that is three times its width. If the width is n feet and fencing cost $8 per

A garden has a length that is three times its width. If the width is n feet and fencing cost $8 per foot, what is the cost of the fencing for the garden? Garden is a rectangle which has Perimeter P of: P = 2l + 2w l = 3w P = 2(3w) + 2w P = 6w + 2w P = 8w Width w = n, so we have: P = 8n Cost = 8n * 8 = [B]64n dollars[/B]

A parallelogram has a perimeter of 54 centimeters. Two of the sides are each 17 centimeters long. Wh

A parallelogram has a perimeter of 54 centimeters. Two of the sides are each 17 centimeters long. What is the length of each of the other two sides? A parallelogram is a rectangle bent on it's side. So we have the perimeter formula P below: P = 2l + 2w We're given w = 17 and P = 54. So we plug this into the formula for perimeter: 2l + 2(17) = 54 2l + 34 = 54 Using our [URL='https://www.mathcelebrity.com/1unk.php?num=2l%2B34%3D54&pl=Solve']equation calculator[/URL], we get [B]l = 10[/B].

A rectangle has a length that is 8.5 times its width. IF the width is n, what is the perimeter of th

A rectangle has a length that is 8.5 times its width. IF the width is n, what is the perimeter of the rectangle. w = n l = 8.5n P = 2(8.5n) + 2n P = 17n + 2n P = [B]19n[/B]

A RECTANGLE HAS A PERIMETER OF 196 CENTIMETERS. IF THE LENGTH IS 6 TIMES ITS WIDTH FIND TH DIMENSION

A RECTANGLE HAS A PERIMETER OF 196 CENTIMETERS. IF THE LENGTH IS 6 TIMES ITS WIDTH FIND TH DIMENSIONS OF THE RECTANGLE? Whoa... stop screaming with those capital letters! But I digress... The perimeter of a rectangle is: P = 2l + 2w We're given two equations: [LIST=1] [*]P = 196 [*]l = 6w [/LIST] Plug these into the perimeter formula: 2(6w) + 2w = 196 12w + 2w = 196 [URL='https://www.mathcelebrity.com/1unk.php?num=12w%2B2w%3D196&pl=Solve']Plugging this equation into our search engine[/URL], we get: [B]w = 14[/B] Now we put w = 14 into equation (2) above: l = 6(14) [B]l = 84 [/B] So our length (l), width (w) of the rectangle is (l, w) = [B](84, 14) [/B] Let's check our work by plugging this into the perimeter formula: 2(84) + 2(14) ? 196 168 + 28 ? 196 196 = 196 <-- checks out

a rectangle has an area of 238 cm 2 and a perimeter of 62 cm. What are its dimensions?

a rectangle has an area of 238 cm 2 and a perimeter of 62 cm. What are its dimensions? We know the rectangle has the following formulas: Area = lw Perimeter = 2l + 2w Given an area of 238 and a perimeter of 62, we have: [LIST=1] [*]lw = 238 [*]2(l + w) = 62 [/LIST] Divide each side of (1) by w: l = 238/w Substitute this into (2): 2(238/w + w) = 62 Divide each side by 2: 238/w + w = 31 Multiply each side by w: 238w/w + w^2 = 31w Simplify: 238 + w^2 = 31w Subtract 31w from each side: w^2 - 31w + 238 = 0 We have a quadratic. So we run this through our [URL='https://www.mathcelebrity.com/quadratic.php?num=w%5E2-31w%2B238%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic equation calculator[/URL] and we get: w = (14, 17) We take the lower amount as our width and the higher amount as our length: [B]w = 14 l = 17 [/B] Check our work for Area: 14(17) = 238 <-- Check Check our work for Perimeter: 2(17 + 14) ? 62 2(31) ? 62 62 = 62 <-- Check

A rectangle is cut in half to create two squares that each has an area of 25. What is the perimeter

[SIZE=4]A rectangle is cut in half to create two squares that each has an area of 25. What is the perimeter of the original rectangle? A. 20 B. 25 C. 30 D. 50[/SIZE] [SIZE=4]Area of a square: A = s^2 We're given A = 25: s^2= 25 s = 5 This means the rectangle width is 5. The rectangle length is 2(5) = 10. Perimeter of a rectangle: P = 2l + 2w P = 2(10) + 2(5) P = 20 + 10 P = [B]30 (choice C)[/B] [B][/B] [B][MEDIA=youtube]tKpS1gQY68o[/MEDIA][/B][/SIZE]

A rectangle shaped parking lot is to have a perimeter of 506 yards. If the width must be 100 yards b

A rectangle shaped parking lot is to have a perimeter of 506 yards. If the width must be 100 yards because of a building code, what will the length need to be? Perimeter of a rectangle (P) with length (l) and width (w) is: 2l + 2w = P We're given P = 506 and w = 100. We plug this in to the perimeter formula and get: 2l + 2(100) = 506 To solve this equation for l, we [URL='https://www.mathcelebrity.com/1unk.php?num=2l%2B2%28100%29%3D506&pl=Solve']type it in our search engine[/URL] and we get: l = [B]153[/B]

A rectangular field is to be enclosed with 1120 feet of fencing. If the length of the field is 40 fe

A rectangular field is to be enclosed with 1120 feet of fencing. If the length of the field is 40 feet longer than the width, then how wide is the field? We're given: [LIST=1] [*]l = w + 40 [/LIST] And we know the perimeter of a rectangle is: P = 2l + 2w Substitute (1) into this formula as well as the given perimeter of 1120: 2(w + 40) + 2w = 1120 Multiply through and simplify: 2w + 80 + 2w = 1120 Group like terms: 4w + 80 = 1120 [URL='https://www.mathcelebrity.com/1unk.php?num=4w%2B80%3D1120&pl=Solve']Typing this equation into our search engine[/URL], we get: [B]w = 260[/B]

A rectangular football pitch has its length equal to twice its width and a perimeter of 360m. Find i

A rectangular football pitch has its length equal to twice its width and a perimeter of 360m. Find its length and width. The area of a rectangle (A) is: A = lw --> where l is the length and w is the width We're given l = 2w, so we substitute this into the Area equation: A = (2w)w A = 2w^2 We're given the area of the pitch is 360, so we set: 2w^2 = 360 We [URL='https://www.mathcelebrity.com/1unk.php?num=2w%5E2%3D360&pl=Solve']type this equation into our search engine[/URL], follow the links, and get: w = [B]6*sqrt(5) [/B] Now we take this, and substitute it into this equation: 6*sqrt(5)l = 360 Dividing each side by 6*sqrt(5), we get: l = [B]60/sqrt(5)[/B]

A rectangular house is 68 yards wide and 112 yards long. What is its perimeter?

A rectangular house is 68 yards wide and 112 yards long. What is its perimeter? The perimeter of a rectangle is: P = 2l + 2w Plugging in our length of 112 and our width of 68, we get: P = 2(112) + 2(68) P = 224 + 136 P = [B]360[/B]

A rectangular parking lot has a perimeter of 152 yards. If the length of the parking lot is 12 yards

A rectangular parking lot has a perimeter of 152 yards. If the length of the parking lot is 12 yards greater than the width. What is the width of the parking lot? The perimeter of a rectangle is: 2l + 2w = P. We're given 2 equations: [LIST=1] [*]2l + 2w = 152 [*]l = w + 12 [/LIST] Substitute equation (2) into equation (1) for l: 2(w + 12) + 2w = 152 2w + 24 + 2w = 152 Combine like terms: 4w + 24 = 152 To solve for w, we [URL='https://www.mathcelebrity.com/1unk.php?num=4w%2B24%3D152&pl=Solve']type this equation into our search engine[/URL] and we get: w =[B] 32[/B]

A rectangular piece of paper has the dimensions of 10 inches by 7 inches.What is the perimeter of th

A rectangular piece of paper has the dimensions of 10 inches by 7 inches.What is the perimeter of the piece of paper Using our [URL='https://www.mathcelebrity.com/rectangle.php?l=10&w=7&a=&p=&pl=Calculate+Rectangle']rectangle calculator[/URL], we get perimeter P: P = [B]34[/B]

A rectangular room is 3 times as long as it is wide, and its perimeter is 56 meters

A rectangular room is 3 times as long as it is wide, and its perimeter is 56 meters Given l = length and w = width, The perimeter of a rectangle is 2l + 2w, we have: [LIST=1] [*]l = 3w [*]2l + 2w = 56 [/LIST] Substitute equation (1) into equation (2) for l: 2(3w) + 2w = 56 6w + 2w = 56 To solve this equation for w, we [URL='https://www.mathcelebrity.com/1unk.php?num=6w%2B2w%3D56&pl=Solve']type it in our math engine[/URL] and we get: w = [B]7 [/B] To solve for l, we substitute w = 7 into equation (1): l = 3(7) l = [B]21[/B]

A rectangular room is 3 times as long as it is wide, and its perimeter is 56 meters.

A rectangular room is 3 times as long as it is wide, and its perimeter is 56 meters. We're given the following: [LIST] [*]l = 3w [/LIST] We know the Perimeter (P) of a rectangle is: P = 2l + 2w Substituting l = 3w and P = 56 into this equation, we get: 2(3w) + 2w = 56 Multiplying through, we get: 6w + 2w = 56 (6 +2)w = 56 8w = 56 [URL='https://www.mathcelebrity.com/1unk.php?num=8w%3D56&pl=Solve']Typing this equation into our search engine[/URL], we get: [B]w = 7[/B] Substitute w = 7 into l = 3w, we get: l = 3(7) [B]l = 21[/B]

A rectangular room is 3 times as long as it is wide, and its perimeter is 56 meters. Find the dimens

A rectangular room is 3 times as long as it is wide, and its perimeter is 56 meters. Find the dimensions of the room. We're given two items: [LIST] [*]l = 3w [*]P = 56 [/LIST] We know the perimeter of a rectangle is: 2l + 2w = P We plug in the given values l = 3w and P = 56 to get: 2(3w) + 2w = 56 6w + 2w = 56 To solve for w, we [URL='https://www.mathcelebrity.com/1unk.php?num=6w%2B2w%3D56&pl=Solve']plug this equation into our search engine[/URL] and we get: w = [B]7 [/B] To solve for l, we plug in w = 7 that we just found into the given equation l = 3w: l = 3(7) l = [B]21 [/B] So our dimensions length (l) and width (w) are: (l, w) = [B](21, 7)[/B]

A rectangular room is 3 times as long as it is wide, and its perimeter is 56 meters. Find the dimens

A rectangular room is 3 times as long as it is wide, and its perimeter is 56 meters. Find the dimension of the room. We're given: l = 3w The Perimeter (P) of a rectangle is: P = 2l + 2w With P = 56, we have: [LIST=1] [*]l = 3w [*]2l + 2w = 56 [/LIST] Substitute equation (1) into equation (2) for l: 2(3w) + 2w = 56 6w + 2w = 56 To solve this equation for w, we [URL='https://www.mathcelebrity.com/1unk.php?num=6w%2B2w%3D56&pl=Solve']type it in our search engine[/URL] and we get: w = [B]7 [/B] Now we plug w = 7 into equation (1) above to solve for l: l = 3(7) l = [B]21[/B]

A rectangular room is 3 times as long as it is wide, and its perimeter is 64 meters. Find the dimens

A rectangular room is 3 times as long as it is wide, and its perimeter is 64 meters. Find the dimension of the room. We're given: [LIST] [*]l = 3w [*]P = 64 [/LIST] We also know the perimeter of a rectangle is: 2l + 2w = P We plugin l = 3w and P = 64 into the perimeter equation: 2(3w) + 2w = 64 Multiply through to remove the parentheses: 6w + 2w = 64 To solve this equation for w, we type it in our search engine and we get: [B]w = 8[/B] To solve for l, we plug w = 8 into the l = 3w equation above: l = 3(8) [B]l = 24[/B]

A rectangular room is 4 times as long as it is wide, and its perimeter is 80 meters. Find the dimens

A rectangular room is 4 times as long as it is wide, and its perimeter is 80 meters. Find the dimension of the room The perimeter of a rectangle is P = 2l + 2w. We're given two equations: [LIST=1] [*]l = 4w [*]2l + 2w = 80. <-- Since perimeter is 80 [/LIST] Plug equation (1) into equation (2) for l: 2(4w) + 2w = 80 8w + 2w = 80 [URL='https://www.mathcelebrity.com/1unk.php?num=8w%2B2w%3D80&pl=Solve']Plugging this equation into our search engine[/URL], we get: w = [B]10[/B] To get l, we plug w = 10 into equation (1): l = 4(10) l = [B]40[/B]

A yard is 33.21 meters long and 17.6 meters wide. What length of fence must be purchased to enclose

A yard is 33.21 meters long and 17.6 meters wide. What length of fence must be purchased to enclose the entire yard? The yard is a rectangle. The perimeter of a rectangle is: P = 2l + 2w where l is the length and w is the width. Evaluating, using our [URL='https://www.mathcelebrity.com/rectangle.php?l=33.21&w=17.6&a=&p=&pl=Calculate+Rectangle']rectangle calculator[/URL], we get P = [B]101.62[/B]

Allan built an additional room onto his house. The length of the room is 3 times the width. The peri

Allan built an additional room onto his house. The length of the room is 3 times the width. The perimeter of the room is 60 feet. What is the length of the room A room is a rectangle. We know the perimeter of a rectangle is: P = 2l + 2w We're given two equations: [LIST=1] [*]l = 3w [*]P = 60 [/LIST] Plug (1) and (2) into our rectangle perimeter formula: 2(3w) + w = 60 6w + w = 60 [URL='https://www.mathcelebrity.com/1unk.php?num=6w%2Bw%3D60&pl=Solve']Type this equation into our search engine[/URL] to solve for w: w = 8.5714 Now plug w = 8.5714 into equation 1 to solve for l: l = 3(8.5714) l = [B]25.7142[/B]

If 800 feet of fencing is available, find the maximum area that can be enclosed.

If 800 feet of fencing is available, find the maximum area that can be enclosed. Perimeter of a rectangle is: 2l + 2w = P However, we're given one side (length) is bordered by the river and the fence length is 800, so we have: So we have l + 2w = 800 Rearranging in terms of l, we have: l = 800 - 2w The Area of a rectangle is: A = lw Plug in the value for l in the perimeter into this: A = (800 - 2w)w A = 800w - 2w^2 Take the [URL='https://www.mathcelebrity.com/dfii.php?term1=800w+-+2w%5E2&fpt=0&ptarget1=0&ptarget2=0&itarget=0%2C1&starget=0%2C1&nsimp=8&pl=1st+Derivative']first derivative[/URL]: A' = 800 - 4w Now set this equal to 0 for maximum points: 4w = 800 [URL='https://www.mathcelebrity.com/1unk.php?num=4w%3D800&pl=Solve']Typing this equation into the search engine[/URL], we get: w = 200 Now plug this into our perimeter equation: l = 800 - 2(200) l = 800 - 400 l = 400 The maximum area to be enclosed is; A = lw A = 400(200) A = [B]80,000 square feet[/B]

If the perimeter of a rectangular field is 120 feet and the length of one side is 25 feet, how wide

If the perimeter of a rectangular field is 120 feet and the length of one side is 25 feet, how wide must the field be? The perimeter of a rectangle P, is denoted as: P = 2l + 2w We're given l = 25, and P = 120, so we have 2(25) + 2w = 120 Simplify: 2w + 50 = 120 [URL='https://www.mathcelebrity.com/1unk.php?num=2w%2B50%3D120&pl=Solve']Typing this equation into our search engine[/URL], we get: [B]w = 35[/B]

If the perimeter of a rectangular sign is 44cm and the width is 2cm shorter than half the length, th

If the perimeter of a rectangular sign is 44cm and the width is 2cm shorter than half the length, then what are the length and width? The perimeter (P) of a rectangle is: 2l + 2w = P We're given P = 44, so we substitute this into the rectangle perimeter equation: 2l + 2w = 44 We're also given w = 0.5l - 2. Substitute the into the Perimeter equation: 2l + 2(0.5l - 2) = 44 Multiply through and simplify: 2l + l - 4 = 44 Combine like terms: 3l - 4 = 44 [URL='https://www.mathcelebrity.com/1unk.php?num=3l-4%3D44&pl=Solve']Type this equation into the search engine[/URL], and we get: [B]l = 16[/B] Substitute this back into the equation w = 0.5l - 2 w = 0.5(16) - 2 w = 8 - 2 [B]w = 6[/B]

Perimeter of a rectangle is 372 yards. If the length is 99 yards, what is the width?

Perimeter of a rectangle is 372 yards. If the length is 99 yards, what is the width? The perimeter P of a rectangle with length l and width w is: 2l + 2w = P We're given P = 372 and l = 99, so we have: 2(99) + 2w = 372 2w + 198 = 372 [SIZE=5][B]Step 1: Group constants:[/B][/SIZE] We need to group our constants 198 and 372. To do that, we subtract 198 from both sides 2w + 198 - 198 = 372 - 198 [SIZE=5][B]Step 2: Cancel 198 on the left side:[/B][/SIZE] 2w = 174 [SIZE=5][B]Step 3: Divide each side of the equation by 2[/B][/SIZE] 2w/2 = 174/2 w = [B]87[/B]

Rectangles and Parallelograms

Free Rectangles and Parallelograms Calculator - Solve for Area, Perimeter, length, and width of a rectangle or parallelogram and also calculates the diagonal length as well as the circumradius and inradius.

The largest American flag ever flown had a perimeter of 1,520 feet and a length of 505 feet. Find th

The largest American flag ever flown had a perimeter of 1,520 feet and a length of 505 feet. Find the width of the flag. for a rectangle, the Perimeter P is given by: P = 2l + 2w P[URL='https://www.mathcelebrity.com/rectangle.php?l=505&w=&a=&p=1520&pl=Calculate+Rectangle']lugging in our numbers for Perimeter and width into our rectangle calculator[/URL], we get: l =[B] 255[/B]

The length of a rectangle is 6 less than twice the width. If the perimeter is 60 inches, what are th

The length of a rectangle is 6 less than twice the width. If the perimeter is 60 inches, what are the dimensions? Set up 2 equations given P = 2l + 2w: [LIST=1] [*]l = 2w - 6 [*]2l + 2w = 60 [/LIST] Substitute (1) into (2) for l: 2(2w - 6) + 2w = 60 4w - 12 + 2w = 60 6w - 12 = 60 To solve for w, [URL='https://www.mathcelebrity.com/1unk.php?num=6w-12%3D60&pl=Solve']type this into our math solver [/URL]and we get: w = [B]12 [/B] To solve for l, substitute w = 12 into (1) l = 2(12) - 6 l = 24 - 6 l = [B]18[/B]

The length of a rectangle is equal to triple the width. Find the length of the rectangle if the peri

The length of a rectangle is equal to triple the width. Find the length of the rectangle if the perimeter is 80 inches. The perimeter (P) of a rectangle is: 2l + 2w = P We're given two equations: [LIST=1] [*]l = 3w [*]2l + 2w = 80 [/LIST] We substitute equation 1 into equation 2 for l: 2(3w) + 2w = 80 6w + 2w = 80 To solve this equation for w, we [URL='https://www.mathcelebrity.com/1unk.php?num=6w%2B2w%3D80&pl=Solve']type it in our search engine[/URL] and we get: w = 10 To solve for the length (l), we substitute w = 10 into equation 1 above: l = 3(10) l = [B]30[/B]

The length of a rectangle is three times its width.If the perimeter is 80 feet, what are the dimensi

The length of a rectangle is three times its width.If the perimeter is 80 feet, what are the dimensions? We're given 2 equations: [LIST=1] [*]l = 3w [*]P = 80 = 2l + 2w = 80 [/LIST] Substitute (1) into (2) for l: 2(3w) + 2w = 80 6w + 2w = 80 8w = 80 Divide each side by 8: 8w/8 = 80/8 w = [B]10 [/B] Substitute w = 10 into (1) l = 3(10) l = [B]30[/B]

The length of a rectangular building is 6 feet less than 3 times the width. The perimeter is 120 fee

The length of a rectangular building is 6 feet less than 3 times the width. The perimeter is 120 feet. Find the width and length of the building. Using our [URL='http://www.mathcelebrity.com/rectangle-word-problems.php?t1=perimeter&v1=120&t2=length&v2=6&op=less&v3=3&t4=times&t5=width&pl=Calculate']rectangular word problem calculator[/URL], we have: [LIST] [*][B]l = 43.5[/B] [*][B]w = 16.5[/B] [/LIST]

The length of a rectangular building is 6 feet less than 3 times the width. The perimeter is 120 fee

The length of a rectangular building is 6 feet less than 3 times the width. The perimeter is 120 feet. Find the width and length of the building. Using our [URL='http://www.mathcelebrity.com/rectangle-word-problems.php?t1=perimeter&v1=120&t2=length&v2=6&op=less&v3=3&t4=times&t5=width&pl=Calculate']rectangle word problem calculator[/URL], we get: [LIST] [*][B]w = 16.5[/B] [*][B]l = 43.5[/B] [/LIST]

the length of a rectangular map is 15 inches and the perimeter is 50 inches. Find the width

The length of a rectangular map is 15 inches and the perimeter is 50 inches. Find the width. Using our r[URL='http://www.mathcelebrity.com/rectangle.php?l=3&w=&a=&p=50&pl=Calculate+Rectangle']ectangle solver[/URL], we get [B]w = 10[/B].

The length of Sallys garden is 4 meters greater than 3 times the width. The perimeter of her garden

The length of Sallys garden is 4 meters greater than 3 times the width. The perimeter of her garden is 72 meters. Find the dimensions of Sallys garden. Gardens have a rectangle shape. Perimeter of a rectangle is 2l + 2w. We're given: [LIST=1] [*]l = 3w + 4 [I](3 times the width Plus 4 since greater means add)[/I] [*]2l + 2w = 72 [/LIST] We substitute equation (1) into equation (2) for l: 2(3w + 4) + 2w = 72 Multiply through and simplify: 6w + 8 + 2w = 72 (6 +2)w + 8 = 72 8w + 8 = 72 To solve this equation for w, we [URL='https://www.mathcelebrity.com/1unk.php?num=8w%2B8%3D72&pl=Solve']type it in our search engine[/URL] and we get: w = [B]8 [/B] To solve for l, we substitute w = 8 above into Equation (1): l = 3(8) + 4 l = 24 + 4 l = [B]28[/B]

The length of Sallys garden is 4 meters greater than 3 times the width. The perimeter of her garden

The length of Sallys garden is 4 meters greater than 3 times the width. The perimeter of her garden is 72 meters A garden is a rectangle, which has perimeter P of: P = 2l + 2w With P = 72, we have: 2l + 2w = 72 We're also given: l = 3w + 4 We substitute this into the perimeter equation for l: 2(3w + 4) + 2w = 72 6w + 8 + 2w = 72 To solve this equation for w, we t[URL='https://www.mathcelebrity.com/1unk.php?num=6w%2B8%2B2w%3D72&pl=Solve']ype it in our search engine[/URL] and we get: w =[B] 8[/B] Now, to solve for l, we substitute w = 8 into our length equation above: l = 3(8) + 4 l = 24 + 4 l = [B]28[/B]

The length of the flag is 2 cm less than 7 times the width. The perimeter is 60cm. Find the length a

The length of the flag is 2 cm less than 7 times the width. The perimeter is 60cm. Find the length and width. A flag is a rectangle shape. So we have the following equations Since P = 2l + 2w, we have 2l + 2w = 60 l = 7w - 2 Substitute Equation 1 into Equation 2: 2(7w -2) + 2w = 60 14w - 4 + 2w = 60 16w - 4 = 60 Add 4 to each side 16w = 64 Divide each side by 16 to isolate w w = 4 Which means l = 7(4) - 2 = 28 - 2 = 26

The perimeter of a bedroom door is 28 feet. It is 4 feet wide. How tall is it?

The perimeter of a bedroom door is 28 feet. It is 4 feet wide. How tall is it? Using our[URL='https://www.mathcelebrity.com/rectangle.php?l=&w=4&a=&p=28&pl=Calculate+Rectangle'] rectangle calculator[/URL], we get: l = [B]10[/B]

The perimeter of a college basketball court is 102 meters and the length is twice as long as the wid

The perimeter of a college basketball court is 102 meters and the length is twice as long as the width. What are the length and width? A basketball court is a rectangle. The perimeter P is: P = 2l + 2w We're also given l = 2w and P = 102. Plug these into the perimeter formula: 2(2w) + 2w = 102 4w + 2w = 102 6w = 102 [URL='https://www.mathcelebrity.com/1unk.php?num=6w%3D102&pl=Solve']Typing this equation into our calculator[/URL], we get: [B]w = 17[/B] Plug this into the l = 2w formula, we get: l = 2(17) [B]l = 34[/B]

The perimeter of a poster is 20 feet. The poster is 6 feet tall. How wide is it?

The perimeter of a poster is 20 feet. The poster is 6 feet tall. How wide is it? [U]Assumptions and givens:[/U] [LIST] [*]The poster has a rectangle shape [*]l = 6 [*]P = 20 [*]The perimeter of a rectangle (P) is: 2l + 2w = P [/LIST] Plugging in our l and P values, we get: 2(6) + 2w = 20 Multiplying through and simplifying, we get: 12 + 2w = 20 To solve for w, we [URL='https://www.mathcelebrity.com/1unk.php?num=12%2B2w%3D20&pl=Solve']type this equation into our search engine [/URL]and we get: w = [B]4[/B]

The perimeter of a rectangle is 400 meters. The length is 15 meters less than 4 times the width. Fin

The perimeter of a rectangle is 400 meters. The length is 15 meters less than 4 times the width. Find the length and the width of the rectangle. l = 4w - 15 Perimeter = 2l + 2w Substitute, we get: 400 = 2(4w - 15) + 2w 400 = 8w - 30 + 2w 10w - 30 = 400 Add 30 to each side 10w = 370 Divide each side by 10 to isolate w w = 37 Plug that back into our original equation to find l l = 4(37) - 15 l = 148 - 15 l = 133 So we have (l, w) = (37, 133)

The perimeter of a rectangle parking lot is 340 m. If the length of the parking lot is 97 m, what is

The perimeter of a rectangle parking lot is 340 m. If the length of the parking lot is 97 m, what is its width? The formula for a rectangles perimeter P, is: P = 2l + 2w where l is the length and w is the width. Plugging in our P = 340 and l = 97, we have: 2(97) + 2w = 340 Multiply through, we get: 2w + 194 = 340 [URL='https://www.mathcelebrity.com/1unk.php?num=2w%2B194%3D340&pl=Solve']Type this equation into our search engine[/URL], we get: [B]w = 73[/B]

The perimeter of a rectangular backyard is 162 feet. It is 52 feet long. How wide is it?

The perimeter of a rectangular backyard is 162 feet. It is 52 feet long. How wide is it? We [URL='https://www.mathcelebrity.com/rectangle.php?l=52&w=&a=&p=162&pl=Calculate+Rectangle']use our rectangle solver to solve for w[/URL]. We get: [B]w = 29[/B]

The perimeter of a rectangular field is 250 yards. If the length of the field is 69 yards, what is

The perimeter of a rectangular field is 250 yards. If the length of the field is 69 yards, what is its width? Set up the rectangle perimeter equation: P = 2l + 2w For l = 69 and P = 250, we have: 250= 2(69) + 2w 250 = 138 + 2w Using our [URL='http://www.mathcelebrity.com/1unk.php?num=2w%2B138%3D250&pl=Solve']equation solver[/URL], we get: [B]w = 56 [/B]

The perimeter of a rectangular field is 300m. If the width of the field is 59m, what is its length

The perimeter of a rectangular field is 300m. If the width of the field is 59m, what is its length? Set up the perimeter (P) of a rectangle equation given length (l) and width (w): 2l + 2w = P We're given P = 300 and w = 59. Plug these into the perimeter equation: 2l + 2(59) = 300 2l + 118 = 300 [URL='https://www.mathcelebrity.com/1unk.php?num=2l%2B118%3D300&pl=Solve']Typing this equation into our search engine[/URL], we get: [B]l = 91[/B]

The perimeter of a rectangular notecard is 16 inches. The notecard is 5 inches wide. How tall is it?

The perimeter of a rectangular notecard is 16 inches. The notecard is 5 inches wide. How tall is it? Perimeter of a rectangle P is: P = 2l + 2w We have: 2l + 2w = 16 We are given w = 5, so we have: 2l + 2(5) = 16 2l + 10 = 16 [URL='https://www.mathcelebrity.com/1unk.php?num=2l%2B10%3D16&pl=Solve']Plugging this into our equation calculator[/URL], we get [B]l = 3[/B].

The perimeter of a rectangular outdoor patio is 54 ft. The length is 3 ft greater than the width. Wh

The perimeter of a rectangular outdoor patio is 54 ft. The length is 3 ft greater than the width. What are the dimensions of the patio? Perimeter of a rectangle is: P = 2l + 2w We're given l = w + 3 and P = 54. So plug this into our perimeter formula: 54= 2(w + 3) + 2w 54 = 2w + 6 + 2w Combine like terms: 4w + 6 = 54 [URL='https://www.mathcelebrity.com/1unk.php?num=4w%2B6%3D54&pl=Solve']Typing this equation into our search engine[/URL], we get: [B]w = 12[/B] Plug this into our l = w + 3 formula: l = 12 + 3 [B]l = 15[/B]

The perimeter of a rectangular parking lot is 258 meters. If the length of the parking lot is 71, wh

The perimeter of a rectangular parking lot is 258 meters. If the length of the parking lot is 71, what is its width? The perimeter for a rectangle (P) is given as: 2l + 2w = P We're given P = 258 and l = 71. Plug these values in: 2(71) + 2w = 258 142 + 2w = 258 [URL='https://www.mathcelebrity.com/1unk.php?num=142%2B2w%3D258&pl=Solve']Typing this equation into our search engine[/URL], we get: [B]w = 58[/B]

The perimeter of a rectangular shelf is 60 inches. The shelf is 7 inches deep. How wide is it?

The perimeter of a rectangular shelf is 60 inches. The shelf is 7 inches deep. How wide is it? The perimeter for a rectangle is given below: P = 2l + 2w We're given l = 7 and P = 60. Plug this into the perimeter formula: 60 = 2(7) + 2w 60 = 14 + 2w Rewritten, it's 2w + 14 = 60. [URL='https://www.mathcelebrity.com/1unk.php?num=2w%2B14%3D60&pl=Solve']Typing this equation into our search engine[/URL], we get [B]w = 23[/B].

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