SOLUTION: I'm having trouble with these two problems. The width of a rectangle is two thirds of its length,and its area is 216 square meters. Find the dimensions. The width and lengt

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Question 415139: I'm having trouble with these two problems.
The width of a rectangle is two thirds of its length,and its area is 216 square meters. Find the dimensions.

The width and length of a rectangle are consecutive odd integers.If the length is increased by five feet,the area of the resulting rectangle is 60 square feet.
Find the dimensions and the area of the original rectangle.

Found 2 solutions by ewatrrr, Theo:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!

Hi
The width of a rectangle is two thirds of its length
Let x and 2/3x represent the length and width
Question states***
x(2/3x)= 216m^2 |CHECKING our Answer***18m*12m = 216m^2
Solving for x
2/3x^2 = 216
x^2 = 324 x = 18m (tossing out negative solution for length)

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
First Problem:
L = length
W = width

width is 2/3 of the length.

this means that W = 2/3 * L

area of the rectangle is 216 square meters.

area of a rectangle is equal to L * W

L*W = 216

since W = 2/3 * L, then substitute for W to get:

L * (2/3 * L) = 216

simplify to get:

2/3 * L^2 = 216

multiply both sides of the equation by 3/2 to get:

L^2 = 216 * 3/2 = 324

take square root of both sides of this equation to get:

L = +/- 18

Since L can't be negative, then L = 18

W = 2/3 * L = 2/3 * 18 = 12

L = 18
W = 12

12 * 18 = 216

first problem solved.

problem number 2:

The width and length of a rectangle are consecutive odd integers.If the length is increased by five feet,the area of the resulting rectangle is 60 square feet.
Find the dimensions and the area of the original rectangle.

L = length of rectangle
W = width of rectangle

length and width are consecutive odd integers.

Let L = W + 2

This means that, if W is an odd integer, then L is the next consecutive odd integer.

if length is increased by 5 feet, the area of the rectangle becomes 60 square feet.

area of rectangle is L*W

if you increase L by 5, then the formula becomes:

(L+5)*W = 60

Since L = W + 2, you can substitute in this equation to get:

(W+7)*W = 60

Simplify to get:

W^2 + 7W = 60

subtract 60 from both sides of this equation to get:

W^2 + 7W - 60 = 0

factor this quadratic equation to get:

(W+12) * (W-5) = 0

This makes W = -12 or W = 5

W can't be negative, so the only viable answer is W = 5

Since L is equal to W + 2, you get L = 7

The original dimensions of the rectangles are L = 7 and W = 5

The original area of the rectangle is 7*5 = 35

Add 5 to the length and you get 12*5 = 60, satisfying the requirements of the problem.