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Tutors Answer Your Questions about Rectangles (FREE)
Question 241315: The dimensions of a room are 21 feet by 6 feet. If carpet costs $16.25 per square
yard, how much will it cost to carpet the room?
Answer by checkley77(7072)   (Show Source):
Question 241177: the width of a rectangle is fixed at 11 cm. What lengths will make the perimeter greater than 96 cm?
Answer by rfer(2688) (Show Source):
Question 241075: the length of a rectangle is 5 meters less than twice its width.If the perimeter of the rectangle is 38 meters, then what is the value of the length(in meters)
Answer by checkley77(7072) (Show Source):
You can put this solution on YOUR website!L=2W-5
2L+2W=38
2(2W-5)+2W=38
4W-10+2W=38
6W=38+10
6W=48
W=48/6
W=8 ANS. FOR THE WIDTH.
L=2*8-5
L=16-5
L=11 ANS. FOR THE LENGTH.
PROOF:
2*11+2*8=38
22+16=38
38=38
Question 240948: A rectangular sign must have an area of 22 square yards. Its length must be 4 yards more than its width. Find the dimensions of the sign.
Answer by solver91311(5072)  (Show Source):
Question 240927: Find the perimeter of a rectangular picture if the width is 7/9 yd and the length is 5/9 yd. Show work.
Answer by checkley77(7072) (Show Source):
Question 240860: how do you find the fill in of a rectangle if the length is 5(1/4) and the width is 9(2/5)
Answer by checkley77(7072) (Show Source):
You can put this solution on YOUR website!Don't know what the "fill" is?
5.25*9.4=49.35 is the [area] of the rectangle.
2*5.25+2*9.4=10.5+18.8=29.3 is the [perimeter].
5.25^2+9.4^2=x^2
27.5625+88.36=x^2
115.9225=x^2
x=sqrt115.9225
x=10.7667 is the [diagonal].
Question 240741: If the dimensions of the larger rectangle are 9 by 18, what are the dimensions of the smaller rectangle?
Answer by rapaljer(3625) (Show Source):
You can put this solution on YOUR website!Is there a diagram or picture that we are not seeing here?
Dr. Robert J. Rapalje, Retired
Seminole State College of Florida
Altamonte Springs Campus
Question 240711: Rectangle perimeter is 240 yards. Length equals width plus 16 yards. What are the dimensions? What is the area?
Answer by unlockmath(121)  (Show Source):
You can put this solution on YOUR website!Hello,
Let's first show the width by using x. No w we can show the length by using x+16. We know the perimeter so we can form and equation.
x+x+(x+16)+(x+16)=240 yds Combine like terms gives us:
4x+32=240 Subtract 32 from both sides gives us:
4x=208 Divide 4 into both sides gives us:
x=52 OK, now we know:
width is 52 yds Length is 68 yds.
I'll let you figure out the area.
RJ Toftness
www.math-unlock.com
Question 240569: if thw area of the rectangle is 30 square inches what is the perimeter and value?
Answer by rapaljer(3625) (Show Source):
You can put this solution on YOUR website!Not enough information is given to solve this one. The area could be 30 square inches because it is 5 by 6 or it could be 3 by 10, or many different possiblities.
Dr. Robert J. Rapalje, Retired
Seminole State College of Florida
Altamonte Springs Campus
Question 240486: The width of a rectangular painting is tw thirds of its length. If the perimeter is 60 inches then what are the length and width?
Answer by checkley77(7072) (Show Source):
You can put this solution on YOUR website!W=2/3L
2W+2L=60
2(2L/3)+2L=60
4L/3+2L=60
(4L+3*2L)/3=60
(4L+6L)/3=60
10L/3=60
10L=3*60
10L=180
L=180/10
L=18 ANS.
W=2/3*18
W=36/3=12 ANS.
PROOF:
2*12+2*18=60
24+36=60
60=60
Question 240546: faye has 20 feet of fencing to make a regtangular pen for her dog.what is the largest area that she can fence in?
Answer by solver91311(5072) (Show Source):
Question 240425: a rectangle has a perimeter of 78 in. its length is 6 in more than twice its length.
Answer by checkley77(7072) (Show Source):
You can put this solution on YOUR website!L=2W+6
2L+2W=78
2(2W+6)+2W=78
4W+12+2W=78
6W=78-12
6W=66
W=66/6
W=11 ANS.
L=2*11+6
L=22+6
L=28 ANS.
PROOF:
2*28+2*11=78
56+22=78
78=78
Question 240399: The width of a rectangle is 2 m less than half the length. The perimeter is 44 m.Find the dimensions.
Answer by checkley77(7072) (Show Source):
You can put this solution on YOUR website!W=L/2-2
2L+2W=44
2L+2(L/2-2)=44
2L+L-4=44
3L=44+4
3L=48
L=48/3
L=16 ANS. FOR THE LENGTH.
W=16/2-2
W=8-2
L=6 ANS. FOR THE WIDTH.
PROOF:
2*16+2*6=44
32+12=44
44=44
Question 240233: ABCD is a rectangle. Each of the longer sides is 3 cm shorter than twice the shorter side. The perimeter of the rectangle is 24 cm. Find the length of the longer side
Answer by unlockmath(121) (Show Source):
You can put this solution on YOUR website!Hello,
Let's note down the information we do have first. Lets show the shorter side
with x and the longer side by 2x-3. Ok, with this we can write an equation as:
x+x+(2x-3)+(2x-3)=24 Combine like terms.
6x-6=24 Add 6 to both sides.
6x=30 Divide both sides by 6
x=5 So 5 cm is the length of the shorter sides. Plug in 5 to (2x-3) gives us:
7 cm for the longer sides.
RJ Toftness
www.math-unlock.com
Question 240049: Find the dimensions of a rectangular garden which encloses 48 square feet of land if the length is 3 times the width
Answer by checkley77(7072) (Show Source):
Question 240052: A painting covers 44 square inches one side of the painting is 3 inches more than twice the other side find the dimensions of the painting
Found 2 solutions by stanbon, checkley77:
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!A painting covers 44 square inches one side of the painting is 3 inches more than twice the other side find the dimensions of the painting
-----------------------------
side = "x"
side = 2x+3
--------------------
Equation:
x(2x+3) = 44
2x^2 + 3x - 44 = 0
---
x = [-3 +- sqrt(9 -4*2*-44)]/2
---
x = [-3 +- sqrt(361)]/2
---
Positive solution:
x = [-3 + 19]/2
x = 8 cm (one of the sides)
2x+3 = 19 (the other side)
=================================
Cheers,
Stan H.
Answer by checkley77(7072) (Show Source):
Question 239957: the length of a rectangle is 3 meters longer than the width. if the area is 28 square meters, what are the rectanles dimensions?
Answer by checkley77(7072) (Show Source):
Question 239827: the length of a rectangle is 8in. more than its width. The perimeter of the rectangle is 54 cm. What are the width and length of the rectangle
Answer by checkley77(7072) (Show Source):
You can put this solution on YOUR website!L=W+8
2L+2W=54
2(W+8)+2W=54
2W+16+2W=54
4W=54-16
4W=38
W=38/4
W=9.5 IN. IS THE WIDTH.
L=9.5+8
L=17.5 IN.
PROOF:
2*17.5+2*9.5=54
35+19=54
54=54
Question 239828: the width of a rectangle is one half its length. The perimeter of the rectangle is 54 cm. What are the width and length of the rectangle?
Answer by checkley77(7072) (Show Source):
You can put this solution on YOUR website!W=L/2
2L+2W=54
2L+2*L/2=54
2L+L=54
3L=54
L=54/3
L=18 ANS. FOR THE LENGTH.
W=18/2=9 ANS. FOR THE LENGTH.
PROOF:
2*9+2*18=54
18+36=54
54=54
Question 239304: i have a math problem that says my yard is 20 by 30 feet and i want to put a flower bed that boarders the yard. if the width is constant and i have 187ft^2 of flowers what is the width of the flower bed
Answer by edjones(3299)  (Show Source):
You can put this solution on YOUR website!Let x=how much you must increase the length of each side to have the area of the flower bed be 187'
(x+30)(x+20)-(30*20)=187 new area-old area=187
x^2+50x+600-600=187
x^2+50x-187=0
x=3.5' Quadratic formula below.
x/2 = 3.5/2 = 1.75' width of the flower bed is 1/2 of the increase in length of each side.
.
Ed
.
| Solved by pluggable solver: SOLVE quadratic equation with variable |
Quadratic equation  (in our case  ) has the following solutons:
For these solutions to exist, the discriminant  should not be a negative number.
First, we need to compute the discriminant :  .
Discriminant d=3248 is greater than zero. That means that there are two solutions:  .
Quadratic expression  can be factored:
Again, the answer is: 3.49561369755001, -53.49561369755.
Here's your graph:
 | |
Question 239058: How do I find the length and width of a rectangle with a perimeter of 82 inches and a diagonal of 29 inches?
Answer by Theo(675)  (Show Source):
You can put this solution on YOUR website!P = 2L + 2W = 82
D = 29
Diagonal forms right triangle with sides of the rectangle.
This means that:
L^2 + w^2 = D^2 which means that:
L^2 + W^2 = 29^2
Since 2L + 2W = 82, this means that:
2L = 82-2W which means that:
L = 41 - W
L^2 + W^2 = 29^2 becomes:
(41-W)^2 + W^2 = 29^2 which becomes:
41^2 - 82W + W^2 + W^2 = 29^2 which becomes:
41^2 - 82W + W^2 + W^2 - 29^2 = 0
combine like terms and simplify to get:
2W^2 - 82W + 840 = 0
divide both side by 2 to get:
W^2 - 41W + 420 = 0
This factors out to:
(W-20) * (W - 21) = 0
This makes W = 20 or W = 21
If W = 20, then L = 21
If W = 21, then L = 20
Your answers are:
L = 20 and W = 21
or:
L = 21 and L = 20
To confirm, plug these values into the original equations and see if the equations are true.
Assume L = 20 and W = 21.
L^2 + W^2 = 29^2 becomes 20^2 + 21^2 = 29^2 which becomes 841 = 841 which is true.
Making L = 21 and W = 20 will yield the same answer.
2L + 2W = 82 becomes:
40 + 42 = 82 which is also true.
The values shown look good.
Answers are:
L = 20 and W = 21
or:
L = 21 and W = 20
Question 238965: The length of a rectangle is five times its width. if the perimeter of the rectangle is 72 cm, find its area?
Thank you
Answer by nerdybill(2448)  (Show Source):
You can put this solution on YOUR website!The length of a rectangle is five times its width. if the perimeter of the rectangle is 72 cm, find its area?
.
Let W = width
then
5W = length (from:"length of a rectangle is five times its width")
.
Since we know the perimeter:
2(W + 5W) = 72
(W + 5W) = 36
6W = 36
W = 6 cm (width)
.
Length:
5W = 5(6) = 30 cm (length)
.
Area:
Width * Length
6 * 30 = 180 square cm
Question 237934: The length of a rectangular garden is 3 yd. more than twice its width. The perimeter of the garden is 36 yd. What are the width and length of the garden?
Answer by College Student(217)  (Show Source):
You can put this solution on YOUR website!width = w
lenght = l = 2w+3
Perimeter = 2w+2l=36
.
 <--- this is your width
.
Since lenght is 2w+3, we can say:
.
Check:
Since there are two widths and two lenghts to form a perimeter of a rectangle, we add: 5+5+13+13=36
Question 238808: the perimeter of a rectangle is 62m. if the lenght is 4m more than width find the lenght and the width
Answer by checkley77(7072) (Show Source):
You can put this solution on YOUR website!L=W+4
2L=2W=62
2(w+4)+2W=62
2W+8+2W=62
4W=62-8
4W=54
W=54/4
W=13.5 ANS.
L=13.5+4=17.5 ANS. FOR THE LENGTH.
PROOF:
2*17.5+2*13.5=62
35+27=62
62=62
Question 238752: If the area of a rectangle is x^2+7x+10, what are the dimensions of said rectangle
Answer by unlockmath(121) (Show Source):
You can put this solution on YOUR website!Hello,
First let's lay the very foundation of the area of a rectangle which is base times the height, right?
OK, with the following x^2+7x+10 we can factor it into:
(x+5)(x+2) This gives us the length of both sides.
Length is x+5
and width is x+2
RJ Toftness
www.math-unlock.com
Question 238717: ABCD is a rectangle, and its length is twice its width. The perimeter of the rectangle is 60 units. P, Q, R, and S are the midpoints of each side. What is the area of the quadrilateral PQRS?
Answer by JimboP1977(76)  (Show Source):
You can put this solution on YOUR website!The old saying that a picture says a thousand words is no more appropriate than here!
First thing to do is draw the rectangle out with the mid points.
A______P______B
|**************|
|**************|
S**************Q
|**************|
|**************|
D______R______C
Then let's put down what we know about this rectangle:
AB = 2AD (1)
2AB+2AD = 60 (2)
Subsituting in AB from (1) into (2) gives
2AB+AB = 60
3AB = 60
So we know AB = 20 units
So as we know P is halfway between A and B we know that AP = 10 units
S is halfway between AD so AS must be 5 units
We know that all the corner triangles APS, BPQ, CQR, DRS are equal in size so
The total area of the rectangle ABCD minus APS, BPQ, CQR and DRS must equal the area of the remaining quadrilateral PQRS.
Area of APS = 5*10*1/2 = 25 units
So the four triangles must have an area of 4*25 = 100
Rectangles area = AB*AD = 20*10 = 200
200 - 100 = 100 units
So PQRS = 100 units.
Does this make sense? I would suggest you draw it out for youself and go through each step to make sure you understand it.
James
Question 238418: A square has 3 cm in length.
What is the perimeter if the square?
Answer by Fombitz(2113)  (Show Source):
Question 238377: Let's say you have a rectangle and you would like to find the area of it in square units. The width is 7 units but the length us (2x-5) units. What do you do to find the area? It is a TAKS question and I would really appreciate your help, thank you.
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!Let's say you have a rectangle and you would like to find the area of it in square units. The width is 7 units but the length us (2x-5) units. What do you do to find the area?
Area = width*length
Area = 7*(2x-5)
Area = 14x-35 sq.units
============================
Cheers,
Stan H.
Question 238288: The length of a rectangle is 4 centimeters more than the width. the measure of the diagonal is 10 centimeters. Find the dimensions of the the rectangle
I would appreciate help =]
Answer by checkley77(7072) (Show Source):
You can put this solution on YOUR website!L=W+4
L^2+W^2=10^2
(W+4)^2+W^2=100
W^2+8W+16+W^2=100
2W^2+8W+16-100=0
2W^2+8W-84=0
2(W^2+4W-42)=0
W=(-4+-SQRT[4^2-4*1*-42])/2*1
W=(-4+-SQRT[16+168])/2
W=(-4+-SQRT184)/2
W=(-4+-13.565)/2
W=(-4+13.565)/2
W=9.565/2
W=4.78 ANS. FOR THE WIDTH
L=4.78+4=8.78 ANS. FOR THE LENGTH.
PROOF:
4.78^2+8.78^2=100
22.85+77.09=100
100~100
Question 238219: a rectangle has a length of 10 centimeters and an are of 40 square centimeters. what is the width of the rectangle?
Answer by EmKid(1) (Show Source):
You can put this solution on YOUR website!Area of a rectangle = length x width
So to find a side of a rectangle when we know one side and the area, we reverse this formula to say area / (divided by) length = width
40 / 10 = 4, so the width is 4.
Question 238221: A rancher wants to use 400 feet of fencing to enclose a rectangular area or 7500 feet squared. What dementions should the rectangle have?
Answer by JimboP1977(76) (Show Source):
You can put this solution on YOUR website!x is the length of one side of the rectangle
y is the width of one side of the rectangle
x*y = 7500
2x+2y = 400
x+y = 200
y = 200 - x (1)
Therefore
x*y = x(200-x) = 7500
200x-x^2 = 7500
Divide both sides by -1
x^2-200x = -7500
Complete the square
(x-100)^2 - 10000 = -7500
x = 2500^0.5 + 100
x = 150
y= 7500/150
so dimensions are 150 feet by 50 feet
Question 237933: The length of a rectangle is 8 in. more than its width. The perimeter of the rectangle is 24 in. What are the width and length of the rectangle?
Answer by unlockmath(121) (Show Source):
You can put this solution on YOUR website!The length of a rectangle is 8 in. more than its width. The perimeter of the rectangle is 24 in. What are the width and length of the rectangle?
Hello,
Let's determine what we know. Set up the width as x and the length as x+8 Hence,
we can set up our equation as:
x+x+(x+8)+(x+8)=24 combine like terms
4x+16=24 Subtract 16 from both sides.
4x=8 Divide both sides by 4
x=2
Therefore the width is 2 in. and the length is 10 in.
RJ Toftness
www.math-unlock.com
Question 237790: The shorter leg of a triangle is 15 meters .The hypotenus is 5 meters longer than the longer leg.Find the length of the longer leg ?
Answer by solver91311(5072) (Show Source):
Question 237603: A garden, measuring 10m x 16m, is enclosed by a side walk of width 1m .the area of the side walk, in m^2 is
a) 8 b) 28 c) 56 d) 208 e) 216
Found 2 solutions by nyc_function, checkley77:
Answer by nyc_function(260)  (Show Source):
You can put this solution on YOUR website!Here is the set up leading to a quadratic equation and then you can do the math.
160 = (10 + 2m)(16 + 2m)
Can you solve for m now?
Answer by checkley77(7072) (Show Source):
Question 237579: a square piece of paper is folded in half to form a rectangle . this rectangle has a perimeter of 24 m . the area of the original square in cm^2,was
a)12 b) 16 c) 32 d) 48 e) 64
Answer by Theo(675) (Show Source):
You can put this solution on YOUR website!let s = the side of the square.
area of the square is s^2
perimeter of the square is 4*s
divide the square in half and you have 2 sides of s and 2 sides of .5s
total number of sides of the half is 3s
perimeter of the half is 24.
s = 24/3 = 8
s^2 = 64
area of the square is 64 square cm.
Question 237390: The perimeter of a rectangular garden is 40 ft. The width is 2 ft more than one half the length. Find the length and width.
Answer by College Student(217) (Show Source):
You can put this solution on YOUR website!The Perimeter of a triangle is the sum of its 4 sides. You can express it as:  ...or... 
The problems tells us the width is 2ft more than half the lenght, therefore:
Now you can use the perimeter equation, but instead of using L and W, we will use L and the equivalency of W. This will help us figure out the value of L.
Since P=40 (given)
When you multiply 2 by 1/2L, the fraction dissapears, but remember to also multiply it by the next digit, which happens to also be 2. Thus...
Now figure out W:
Since 
Question 237100: QUESTION ONE
The perimeter of a rectangle parking lot is 146 meters. Find its dimensions if the length is 7 meters less than 4 times thw width
QUESTION TWO
The lenght of the second side of the triangle is 2 inches less tha the length of the first side. The length of the third side is 12 inches more than the length of the first side. The perimter of the triangle is 73 inches. Find the length of each side of the triangle.
THESE ARE LET STATMENTS!
Answer by rfer(2688) (Show Source):
Question 236948: A rectangle is twice as long as it is wide and has a diagonal of length 25 cm. Find its dimensions.
Found 2 solutions by checkley77, edjones:
Answer by checkley77(7072) (Show Source):
You can put this solution on YOUR website!x^2+(2x)^2=25^2
X^2+4X^2=625
5x^2=625
x^2=625/5
x^2=125
x=sqrt125
x=11.18 cm. is the width.
2*11.18=22.36 cm. is the length.
Proof:
11.18^2+22.36^2=625
124.996+499.97=625
625~625
Answer by edjones(3299) (Show Source):
Question 236905: How to work a problem where their is a rectangle with the length of 20 and the width of 14; and the shaded region of the border is 2ft wide?
Answer by nyc_function(260) (Show Source):
You can put this solution on YOUR website!This question is incomplete.
Given the information, what are you looking for?
Are you looking for the area of the shaded region?
Are you looking for something else?
===================================
I got your reply.
I will set it up and you can finish.
Area of shaded region = area of outside rectangle minus area of inside rectangle
Area of shaded region = (18)(24) - (20)(14)
Can you do it now?
Question 236867: ABCD is a quadrilateral. Angle A = 40 degrees, and angle B = 150 degrees.
Could ABCD be a trapezoid?
Answer by edjones(3299) (Show Source):
Question 236805: The length of a rectangle is 10 m less than twice the width of the rectangle. The area is 2800 m2. Find the length and width of the rectangle
Answer by checkley77(7072) (Show Source):
You can put this solution on YOUR website!L=2W-10
AREA=LW
2,800=(2W-10)W
2,800=2W^2-10W
2W^2-10W-2,800=0
2(W^2-5W-1,400)=0
2(W-40)(W+30)=0
W-40=0
W=40 ANS. FOR THE WIDTH.
L=2*40-10
L=80-10
L=70 ANS. FOR THE LENGTH.
pROOF:
2,800=40*70
2,800=2,800
Question 236772: the perimeter of a rectangular garden is 40 ft. The width is 2 ft more than one half the length. find the length and the width.
Answer by nyc_function(260) (Show Source):
You can put this solution on YOUR website!Here is the set up:
We use P = 2L + 2W
P = 40
length = x
width = (x/2) + 2
Here is your equation:
40 = 2(x) + 2[(x/2) + 2]
Can you finich now?
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