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the perimeter of a rectangle is 160ft. One fourth the length is the same as twice the width. Find the dimensions of the rectangle.

User Denville
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2 Answers

2 votes

Final answer:

To find the dimensions of the rectangle, set up equations using the given information. Solve the equations simultaneously to find the values of L and W. Substituting the values back into the first equation, we find the dimensions of the rectangle are 64ft by 8ft.

Step-by-step explanation:

To solve this problem, we can set up equations using the given information. Let's assume the length of the rectangle is 'L' and the width is 'W'. We are given that one fourth of the length is equal to twice the width, so we can write the equation as: L/4 = 2W.

The perimeter of a rectangle is given by the formula P = 2L + 2W. Substituting the given perimeter of 160ft into the equation, we get: 2L + 2W = 160.

Now, we can solve these two equations simultaneously to find the dimensions of the rectangle. From the first equation, we can solve for L in terms of W: L = 8W. Substituting this value into the second equation, we get: 2(8W) + 2W = 160. Simplifying, we get 16W + 2W = 160, which gives us 18W = 160. Dividing both sides by 18, we find that W = 8. Substituting this value back into the first equation, we can solve for L: L = 8W = 8(8) = 64.

Therefore, the dimensions of the rectangle are 64ft by 8ft.

User Ewiinnnnn
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8.5k points
4 votes

Answer:

w = 80/9 ft.

L = 640/9 ft.

Step-by-step explanation:

L = length w = width

2w = L/4 so 8w = L

160 = 2L + 2w

160 = 2(8w) + 2w

160 = 16w+2w

160 = 18w

80/9 = w

User Martey
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7.7k points