1. The length of a rectangle exceeds its width by 4 feet. if the width is double
ID: 3099144 • Letter: 1
Question
1. The length of a rectangle exceeds its width by 4 feet. if the width is doubled and the length is decreased by 2 feet, a new rectangle is formed whose perimeter is 8 feet more than the perimeter of the original rectangle. Find the dimensions of the original rectangle. Find the dimensions of the original rectangle.
2. A side of a square is 10 meters longer than the side of an equilateral triangle. The perimeter of the square is 3 times the perimeter of the triangle. Find the length of each side of the triangle.
3. The length of each side of a hexagon is 4 inches less than the length of a side of a square. The perimeter of the hexagon is equal to the perimeter of the square. Find the length of a side of the hexagon and the length of a side of the square.
Explanation / Answer
1. The length of a rectangle exceeds its width by 4 feet. if the width is doubled and the length is decreased by 2 feet, a new rectangle is formed whose perimeter is 8 feet more than the perimeter of the original rectangle. Find the dimensions of the original rectangle. Find the dimensions of the original rectangle.
perimeter of a rectangle = 2 * length + 2* width P = 2*l + 2*w
The length of a rectangle exceeds its width by 4 feet. this translates to l1 = w1 + 4
substitute this into the perimeter equation
P1 = 2*(w1+4) + 2*w1 = 2w1 + 8 + 2w1 = 4*w1 + 8
if the width is doubled and the length is decreased by 2 feet, a new rectangle is formed whose perimeter is 8 feet more than the perimeter of the original rectangle.
a new rectangle is formed whose perimeter is 8 feet more than the perimeter of the original rectangle
P2 = P1 + 8
solve for P1
P1 = P2 - 8
if the width is doubled and the length is decreased by 2 feet,
w2 = w1*2 l2 = l1 - 2
P2 = 2*w2 + 2*l2 = 2*(w1*2) + 2*(l1-2) = 4*w1 + 2*l1 - 4
plug in l1 = w1 + 4 into this equation
P2 = 4*w1 + 2*(w1 + 4) - 4 = 4*w1 + 2*w1 + 8 - 4 = 6*w1 - 4
Plug this P2 into the equation P1 = P2 - 8
P1 = 6*w1 - 4 - 8 = 6*w1 - 12
Set this P1 equal to the P1 = 4*w1 + 8 found above and solve for w1
4*w1 + 8 = 6*w1 - 12
8 = 2*w1 - 12
20 = 2*w1
10 = w1
Now plug back into the perimeter equations that are in terms of w1
P1 = 4*w1 + 8 = 40 + 8 = 48
P2 = 6*w1 - 4 = 60 - 4 = 56
2. A side of a square is 10 meters longer than the side of an equilateral triangle. The perimeter of the square is 3 times the perimeter of the triangle. Find the length of each side of the triangle.
equilateral triangle has all 3 sides equal, square has all 4 sides equal
Psq = 4*lsq Ptr = 3*ltr
A side of a square is 10 meters longer than the side of an equilateral triangle.
lsq = ltr + 10
plug into Psq above
Psq = 4*(ltr+10) = 4*ltr + 40
The perimeter of the square is 3 times the perimeter of the triangle.
Psq = 3*Ptr
Plug in Ptr from above
Psq = 3*(3*ltr) = 9*ltr
now set the Psq's equal to each other and solve for ltr
Psq = 9*ltr
Psq = 4*ltr + 40
9*ltr = 4*ltr + 40
5*ltr = 40
ltr = 8
3. The length of each side of a hexagon is 4 inches less than the length of a side of a square. The perimeter of the hexagon is equal to the perimeter of the square. Find the length of a side of the hexagon and the length of a side of the square.
lsq = length of side of square
lhx = length of side of hexagon
Psq = perimeter of square
Phx = perimeter of hexagon
The length of each side of a hexagon is 4 inches less than the length of a side of a square.
lsq = lhx + 4
The perimeter of the hexagon is equal to the perimeter of the square.
Psq = Phx
Psq = 4*lsq
Phx = 6*lhx
4*lsq = 6*lhx
plug in lsq = lhx + 4 from above and solve for lhx
4*lsq = 6*lhx
4*(lhx+4) = 6*lhx
4*lhx + 16 = 6*lhx
16 = 2*lhx
8 = lhx
from above
lsq = lhx + 4
plug in the now known lhx
lsq = 8 + 4 = 12
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