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The length of a rectangle is 1 m longer than 3 times its width. The area is 80 m2. Find the dimensions of this rectangle

User Fredericka
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Answer: Let's use the given information to set up two equations, one for the length and one for the area, in terms of the width.

Let x be the width of the rectangle, in meters. Then, according to the problem, the length of the rectangle is 1 meter longer than 3 times the width, or 3x + 1.

The area of a rectangle is given by the formula A = length × width, so we can write:

A = (3x + 1) x

We know that the area is 80 m², so we can substitute this value and solve for x:

80 = (3x + 1) x

Expanding the right side gives:

80 = 3x² + x

Rearranging and setting equal to zero:

3x² + x - 80 = 0

This is a quadratic equation that we can solve using the quadratic formula:

x = (-b ± sqrt(b² - 4ac)) / 2a

where a = 3, b = 1, and c = -80. Plugging in these values, we get:

x = (-1 ± sqrt(1² - 4(3)(-80))) / 2(3)

x = (-1 ± sqrt(1 + 960)) / 6

x = (-1 ± sqrt(961)) / 6

We can simplify the square root to:

x = (-1 ± 31) / 6

Taking the positive value, we get:

x = 5

So the width of the rectangle is 5 meters. Using the expression we found for the length in terms of the width, the length is:

3x + 1 = 3(5) + 1 = 16

So the dimensions of the rectangle are 5 meters by 16 meters.

Explanation:

User Shubhanshu Mishra
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