Tutors Answer Your Questions about Rectangles (FREE)
Question 140787: The diagonal of a rectangle platform is square root of 20 cm. If both dimensions are increased by 2 cm, then the resulting perimeter would be 5/3 of the original perimeter. Find the dimensions of the rectangle.
Click here to see answer by Fombitz(32388)  |
Question 141753: I am in need of a geometry refresher, so, whoever is kind enough to answer this, please explain as to how you derived your answer. What is the length of each diagonal of a rectangular box with length 53 cm, width 48 cm, and height 70 cm?
Click here to see answer by scott8148(6628)  |
Question 142712: This seemed simple enough but I keep screwing it up. I've had good help from this site and would really appreciate it again.
"Dana buys a peice of carpet that measures 20 sq. yds.. Will she be able to completely cover a rectengular floor that measures 12.5 ft. by 16.5 ft?"
Click here to see answer by edjones(8007)  |
Question 142712: This seemed simple enough but I keep screwing it up. I've had good help from this site and would really appreciate it again.
"Dana buys a peice of carpet that measures 20 sq. yds.. Will she be able to completely cover a rectengular floor that measures 12.5 ft. by 16.5 ft?"
Click here to see answer by scott8148(6628)  |
Question 143174: the area of a rectangular parcel of land is 720 sq. meters. the length of the land is 4 meters less than twice the width.
a. write an equation than can be used to find the dimensions of the land.
(i think i have this- 720=(2w-4)w)
b.solve the equation in part a. to find the dimensions of the land
Note: i already have the answer(l=36,w=20), i just don't know how to solve the equation)
Click here to see answer by jim_thompson5910(35256) |
Question 143174: the area of a rectangular parcel of land is 720 sq. meters. the length of the land is 4 meters less than twice the width.
a. write an equation than can be used to find the dimensions of the land.
(i think i have this- 720=(2w-4)w)
b.solve the equation in part a. to find the dimensions of the land
Note: i already have the answer(l=36,w=20), i just don't know how to solve the equation)
Click here to see answer by rapaljer(4671)  |
Question 143192: a farmer has 60m of chicken wire to enclose a rectangular yard for his chickens. find the dimensions that will give maximum area.
write equations to express the perimeter and the area in terms of the length and width.
Click here to see answer by edjones(8007)  |
Question 143297: I need help in solving the following. Thank you.
A rectangular garden has dimensions of 23ft by 11ft. A gravel path of equal width is to be built around the garden. How wide can the path be if there is enough gravel for 240 square feet. I was trying to solve this by using the formula (23-2x)(11-2x) = 240 but I cant find the solution to work. Any help would be appreciated to point me in the right direction. Thank you.
Click here to see answer by checkley77(12844) |
Question 143290: The area of a rectangular parcel of land is 720 square meters. The length of the land is 4 meters less than twice the width.
A. Write an equation that can be used to find the dimensions of the land.
B. Solve the equation written in part A to find the dimensions of the land.
Click here to see answer by checkley77(12844) |
Question 143841: PROBLEM:DIMENSIONS OF A PARKING LOT:
A SHOPPING CENTER HAS A RECTANGULAR AREA OF 40,000 YD^2 ENCLOSED ON THREE SIDES FOR A PARKING LOT. THE LENGTH IS 200 YD MORE THAN TWICE THE WIDTH. WHAT ARE THE DIMENSIONS OF THE LOT?
** IN THIS PRBLEM I ALWAYS SEEM TO KEEP GETTING A NEGATIVE ANSWER... I DO NOT UNDERSTAND THE PROPER WAY OF EVEN SOLVING THIS PROBLEM.. PLEASE HELP BY SHOWING ME HOW YOU WOULD WORK THIS THURALLY OUT.. STEP BY STEP, THANKS SO MUCH :)
Click here to see answer by checkley77(12844) |
Question 144235: You have 172 feet of fencing to enclose a rectangular region. Find the dimensions of the rectangle that maximize the enclosed area.
possible answers: D, 43 by 43 is the max.
Solution: The area the product of 2 sides. Since it's a rectangle, 2 adjacent sides will use 1/2 of the 172 feet, or 86 feet.
The area = L x (86-L), or 
Set the 1st derivative = 0.
86-2L=0 (Yes, it's a calculus problem any time you want max or minimum.)
86 = 2L
L = 43.
A) 45ft by 41ft B) 86ft by 21.5ft
C) 86ft by 86ft D) 43ft by 43ft
Click here to see answer by Alan3354(69443)  |
Question 144272: Please can you help me? Thank you.
Do you think that it would be easier or more difficult to visualize solids as compared to objects in one or two-dimensional planes? Are there obvious examples that you can come up with of solids that appear or are being used in our daily lives and the environment that surrounds us?
Click here to see answer by vleith(2983) |
Question 145162: Suppose the length of a rectangle is 3 inches longer than the width and that the perimeter of the rectangle is 78. Which I figured that out the length of the rectangle is 21 inches and the width is 18 inches. The next question asks "Set up and equation involving only W, the width of the rectangel?" I am not sure what that is asking???
Click here to see answer by shahid(44) |
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