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Algebra..Please Help...Thanks?
Algebra..Please Help...Thanks?
How do I find the answer to questions like these:
The perimeter of a rectangle is 18 feet, and the area of the rectangle is 20 square feet. What is the length of the rectangle?
The length of a rectangular floor is 2 feet more than its width. The area of the floor is 168 square feet. Kim wants to use a rug in the middle of the room and leave a 2 foot border of the floor visible on all sides. What should the width (the shorter side) of the rug be?
The perimeter of a rectangle is 18 feet, and the area of the rectangle is 20 square feet. What is the width of the rectangle?
All answers have to be in feet..
Thank you,
Buckeye
Answers:
William B: If 2L +2W =18
then 2(L+W) =18, L+W=9
W=9-L
L(9-L) = 20
9L -L^2=20
L^2 -9L +20=0
(L-4)(L-5) =0
length is 5, width is 4
2008-02-29 08:32:19
then 2(L+W) =18, L+W=9
W=9-L
L(9-L) = 20
9L -L^2=20
L^2 -9L +20=0
(L-4)(L-5) =0
length is 5, width is 4
2008-02-29 08:32:19
Crystal Mandie: 1. The length is 6 feet, and the height is 3 feet.
2.
2008-02-29 08:32:43
2.
2008-02-29 08:32:43
potlyfe: P= 2L + 2W (length and width)
A = L * W
in your case (1):
2L + 2W = 18
L * W = 20
therefore;
W = 20/L
so
2L + 2(20/L) = 18
2L + 40/L = 18
(2L^2) + 40 = 18L (multiply through by L)
(2L^2) - 18L + 40 = 0
you solve the quadratic and get 'L' which is the length to be 5ft hence the width is 4ft. this is the general method for solving all these kind of problems.
2008-02-29 08:33:17
A = L * W
in your case (1):
2L + 2W = 18
L * W = 20
therefore;
W = 20/L
so
2L + 2(20/L) = 18
2L + 40/L = 18
(2L^2) + 40 = 18L (multiply through by L)
(2L^2) - 18L + 40 = 0
you solve the quadratic and get 'L' which is the length to be 5ft hence the width is 4ft. this is the general method for solving all these kind of problems.
2008-02-29 08:33:17
Chosen Answer
Hilary M: Just off the top of my head the first one is 5 feet by 4 feet.
5+5+4+4=18 ft
5*4=20 feet
divide 18 by two first because you know that 2 sides, a height and a width of it, can't be more than 9 feet total... then find combinations of heights and widths that not only equal 9 feet (one height plus one width) but when multiplied equal 20.
for the first and the second concerning the 18 ft perimeter and the 20 sq feet area...the width is usually the shorter of the two and the length the longer....so length would be 5 ft and width would be 4
for the rug problem draw a picture. use variables. w represents width...length is w +2 because it is 2 feet longer than the width (outside area)...for the rug area the measurements would be width w-4, length would be (w+2)-4 or essentially w-2...that should get you started if you follow the rules your instructor gave you for formulas for area and for perimeter... Good Luck!
2008-02-29 08:37:35
Hilary M: Just off the top of my head the first one is 5 feet by 4 feet.
5+5+4+4=18 ft
5*4=20 feet
divide 18 by two first because you know that 2 sides, a height and a width of it, can't be more than 9 feet total... then find combinations of heights and widths that not only equal 9 feet (one height plus one width) but when multiplied equal 20.
for the first and the second concerning the 18 ft perimeter and the 20 sq feet area...the width is usually the shorter of the two and the length the longer....so length would be 5 ft and width would be 4
for the rug problem draw a picture. use variables. w represents width...length is w +2 because it is 2 feet longer than the width (outside area)...for the rug area the measurements would be width w-4, length would be (w+2)-4 or essentially w-2...that should get you started if you follow the rules your instructor gave you for formulas for area and for perimeter... Good Luck!
2008-02-29 08:37:35