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 chalkboard is 3 feet tall and 4 feet long. What is its perimeter A chalkboard is a rectangle. So the perimeter is: 2l + 2w Using [URL='https://www.mathcelebrity.com/rectangle.php?l=4&w=3&a=&p=&pl=Calculate+Rectangle']our rectangle calculator[/URL], we get: P = [B]14[/B]

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 diving board is 10 feet long and 1 foot wide. What is its area? A diving board is a rectangle. And the area of a rectangle is: A = lw Plugging in our numbers, we get: A = 10(1) A = [B]10 sq feet[/B]

A family room measures 15.6 feet long and 18.4 feet wide. What is the area of the room? The room is rectangular. So our area A = lw. Using our [URL='https://www.mathcelebrity.com/rectangle.php?l=15.6&w=18.4&a=&p=&pl=Calculate+Rectangle']rectangle calculator[/URL], we get: A = [B]287.04 square feet[/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 playing card is 7 centimeters wide and 10 centimeters tall. What is its area? A playing card has a rectangle shape, so the area is l x w. A = l x w A = 10 cm x 7 cm A =[B] 70 cm^2[/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 postcard is 4 inches tall and 5 inches wide. What is its area? A postcard is a rectangle. The area is 4 x 5 = [B]20 square inches[/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 field is twice as long as it is wide. If the perimeter is 360 what are the dimensions? We are given or know the following about the rectangle [LIST] [*]l = 2w [*]P = 2l + 2w [*]Since P = 360, we have 2l + 2w = 360 [/LIST] Since l = 2w, we have 2l + (l) = 360 3l = 360 Divide by 3, we get [B]l = 120[/B] Which means w = 120/2 [B]w = 60[/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 garden is 5 ft longer than it is wide. Its area is 546 ft2. What are its dimensions? [LIST=1] [*]Area of a rectangle is lw. lw = 546ft^2 [*]We know that l = w + 5. [/LIST] Substitute (2) into (1) (w + 5)w = 546 w^2 + 5w = 546 Subtract 546 from each side w^2 + 5w - 546 = 0 Using the positive root in our [URL='http://www.mathcelebrity.com/quadratic.php?num=w%5E2%2B5w-546%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic calculator[/URL], we get [B]w = 21[/B]. This means l = 21 + 5. [B]l = 26[/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 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 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. 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 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 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 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 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 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 train ticket is 8 centimeters tall and 10 centimeters long. What is its area? The ticket is a rectangle. The area is: A = lw Plugging in our numbers, we get: A = (8)(10) A = 80

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]

All squares are rectangles and all rectangles are parallelograms, therefore all squares are parallelograms. Is this true? [B]Yes.[/B] This is similar to A implies B and B implies C so A implies C also known as transitive property

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]

Anne wants to make a platform that is 7 feet wide and 10 feet long. If she uses boards that measure 6 inches wide by 2 feet long, how many boards will she need to complete the job? Area of platform which is a rectangle: A = lw A = 10 * 7 A = 70 Area of boards which are rectangles: A = lw A = 2 * 6 A = 12 We divide our platform area by our board area to get the number of boards needed: Boards needed = Platform Area / Board Area Boards needed = 70/12 Boards needed = 5.83333 We round up if we want full boards to be [B]6[/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]

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Given a description of a 2D shape, it will return the shape (Circle, rectangle, square, triangle, etc.)

Find the area of a rectangle with length 8 cm and width 5 cm. A = l x w A = 8cm * 5cm A = [B]40cm^2 [URL='https://www.mathcelebrity.com/rectangle.php?l=8&w=5&a=&p=&pl=Calculate+Rectangle']Also, you can use this calculation:[/URL][/B]

How do I find dimensions of a rectangle when it has been expanded?

[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]

[B]List the answer being sought (words) ______Area of the garden What is this answer related to the rectangle?_Have_________________________ List one piece of extraneous information____2m tall fence List two formulas that will be needed_______P = 36. P = 2l + 2w Write the equation for width_____________w = 2l - 6 Write the equation needed to solve this problem A = lw, P = 2l + 2w[/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 all A's are B's, then all B's are A's. Is this true? [B]No.[/B] Example: All dogs are mammals, but not all mammals are dogs. All squares are rectangles, but not all rectangles are squares.

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]

rectangle abcd prove: triangle adc is congruent to triangle bcd 1. Given: ABCD is a rectangle 2. AB = CD since opposite sides of rectangle are congruent 3. BC = AD since opposite sides of rectangle are congruent 4. AC = AC by the Reflexive Property of Equality 5. triangle ADC = triangle CBA by the Side-Side-Side (SSS) Property

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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.

Draw this rectangle and you'll see we have a pythagorean theorem equation. a^2 + b^2= c^2 b = 8 and c= 10 a^2 + 8^2 = 10^2 a^2 + 64 = 100 Subtract 64 from each side: a^2 = 36 a= 6 Therefore, perimeter P is: P = 2l + 2w P = 2(6) + 2(8) P = 12 + 16 P = [B]28[/B] [MEDIA=youtube]8lcpRet3r18[/MEDIA]

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 12ft² 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 Sally’s 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 Sally’s garden is 4 meters greater than 3 times the width. The perimeter of her garden is 72 meters. Find the dimensions of Sally’s 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 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? 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 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? [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. 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 it’s 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? 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 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 it’s 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? 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. 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 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].

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]