length of a rectangle

A bag of fertilizer covers 300 square feet of lawn. Find how many bags of fertilizer should be purchased to cover a rectangular lawn 290 feet by 150 feet. The area of a rectangle is length * width, so we have: A = 290 * 150 A = 43,500 sq ft. Now, to find the number of bags needed for a 300 square feet per bag of fertilizer, we have: Bags Needed = Total Square Feet of Lawn / Square Feet covered per bag Bags Needed = 43,500 / 300 Bags Needed = [B]145[/B]

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 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 flower bed is to be 3 m longer than it is wide. The flower bed will an area of 108 m2 . What will its dimensions be? A flower bed has a rectangle shape, so the area is: A = lw We are given l = w + 3 Plugging in our numbers given to us, we have: 108 = w(w + 3) w^2 + 3w = 108 Subtract 108 from each side: w^2 + 3w - 108 = 0 [URL='https://www.mathcelebrity.com/quadratic.php?num=w%5E2%2B3w-108%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']Type this problem into our search engine[/URL], and we get: w = (9, -12) Since length cannot be negative, w = 9. And l = 9 + 3 --> l = 12 So we have [B](l, w) = (12, 9)[/B] Checking our work, we have: A = (12)9 A = 108 <-- Match!

A framed print measures 80cm by 65cm. The frame is 5cm wide. Find the area of the unframed print. We subtract 5 cm from the length and the width to account for the frame: Unframed Length: 80 - 5 = 75 Unframed Width: 65 - 5 = 60 Area of the unframed rectangle is: A = lw A = 75(60) A = [B]4,500 sq cm[/B]

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. 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 pool is 5 meters wide and 21 meter long what is the area of the pool? A pool is a rectangle. So the area for a rectangle is: A = lw [I]where l is the length and w is the width.[/I] [URL='https://www.mathcelebrity.com/rectangle.php?l=21&w=5&a=&p=&pl=Calculate+Rectangle']Plugging in our width of 5 and length of 21 to our rectangle calculator[/URL], we get: A = [B]105 m^2[/B]

a rectangle has a length of x-7 and a width of x + 5. Write an expression that represents the area of the rectangle in terms of x. Area of a rectangle (A) with length(l) and width (w) is expressed as follows: A = lw Plugging in our values given above, we have: [B]A = (x - 7)(x + 5)[/B]

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 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? 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

[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 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 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 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 hotel room is 4 yards by 5 yards. The owner of the hotel wants to recarpet the room with carpet that costs $76.00 per square yard. How much will it cost to recarpet the room? $ The area of a rectangle is length * width, so we have: A = 5 yards * 4 yards A = 20 square yards Total cost = Cost per square yard * total square yards Total Cost = $76 * 20 Total Cost = [B]$1520[/B]

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 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 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. 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 standard volleyball court has an area of 1800ft. The length is 60. What is the width of the volleyball court Plugging [URL='https://www.mathcelebrity.com/rectangle.php?l=60&w=&a=1800&p=&pl=Calculate+Rectangle']this into our rectangle calculator[/URL] and we get: w = [B]30[/B]

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 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]

area of a rectangle Let l be the length and w be the width of a rectangle. The Area (A) is: A = [B]lw[/B]

[B]I need 36 m of fencing for my rectangular garden. I plan to build a 2m tall fence around the garden. The width of the garden is 6 m shorter than twice the length of the garden. How many square meters of space do I have in this garden? List the answer being sought (words) ______Need_________________________ What is this answer related to the rectangle?_Have_________________________ List one piece of extraneous information____Need_________________________ List two formulas that will be needed_______Have_________________________ Write the equation for width_____________Have_________________________ Write the equation needed to solve this problem____Need____________________[/B]

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 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, 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? 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]

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 dimensions of a rectangle are 30 cm and 18 cm. When its length decreased by x cm and its width is increased by x cm, its area is increased by 35 sq. cm. a. Express the new length and the new width in terms of x. b. Express the new area of the rectangle in terms of x. c. Find the value of x. Calculate the current area. Using our [URL='https://www.mathcelebrity.com/rectangle.php?l=30&w=18&a=&p=&pl=Calculate+Rectangle']rectangle calculator with length = 30 and width = 18[/URL], we get: A = 540 a) Decrease length by x and increase width by x, and we get: [LIST] [*]length = [B]30 - x[/B] [*]width = [B]18 + x[/B] [/LIST] b) Our new area using the lw = A formula is: (30 - x)(18 + x) = 540 + 35 Multiplying through and simplifying, we get: 540 - 18x + 30x - x^2 = 575 [B]-x^2 + 12x + 540 = 575[/B] c) We have a quadratic equation. To solve this, [URL='https://www.mathcelebrity.com/quadratic.php?num=-x%5E2%2B12x%2B540%3D575&pl=Solve+Quadratic+Equation&hintnum=+0']we type it in our search engine, choose solve[/URL], and we get: [B]x = 5 or x = 7[/B] Trying x = 5, we get: A = (30 - 5)(18 + 5) A = 25 * 23 A = 575 Now let's try x = 7: A = (30 - 7)(18 + 7) A = 23 * 25 A = 575 They both check out. So we can have

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 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 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 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 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 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. 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 a wooden frame is 1 foot longer than its width and its area is equal to 12ft The frame is a rectangle. The area of a rectangle is A = lw. So were given: [LIST=1] [*]l = w + 1 [*]lw = 12 [/LIST] Substitute equation (1) into equation (2) for l: (w + 1) * w = 12 Multiply through and simplify: w^2 + w = 12 We have a quadratic equation. To solve for w, we type this equation into our search engine and we get two solutions: w = 3 w = -4 Since width cannot be negative, we choose the positive result and have: w = [B]3[/B] To solve for length, we plug w = 3 into equation (1) above and get: l = 3 + 1 l = [B]4[/B]

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 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 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 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 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 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 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? 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 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, 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 width of a rectangle is fixed at 4cm. For what lengths will the area be less than 86 cm^2 The Area (A) of a rectangle is given by: A = lw With an area of [I]less than[/I] 86 and a width of 4, we have the following inequality: 4l < 86 To solve for l, we [URL='https://www.mathcelebrity.com/interval-notation-calculator.php?num=4l%3C86&pl=Show+Interval+Notation']type this inequality into our search engine[/URL] and we get: [B]l < 21.5[/B]

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