Question

The area of a rectangle is 27 square meters. If the length is 6 meters less than 3 times the width, then find the dimensions of the rectangle. Round off your answers to the nearest hundredth

Answer 1

width = 4.16 meter

length = 6.49 meter

Explanation:

Area of the rectangle =27 m²

Let the width of the rectangle be x meter

So, Length = 3 * width - 6

= 3*x - 6

= 3x-6 meter

Area of the rectangle = length * width

27 = (3x-6)*x

Flipping the sides of the equation, we have

(3x-6)*x =27

Distributing the left side, we get

(3x)*(x) - (6)*(x) = 27

=> 3x² - 6x = 27

Subtract 27 from both sides,

3x² - 6x -27 = 27 - 27

=> 3x² - 6x -27 = 0

Factoring out 3 from all the terms on the left side, we have

3(x² - 2x -9) = 0

Dividing both sides by 3, we have

`(3(x^(2)-2x-9) )/(3)` =
`(0)/(3)`

Cancelling out the 3's on the left, we get

x² - 2x -9 = 0

We'll use the quadratic formula to solve for the x,

x =
`\frac{-b\pm\sqrt{b^(2)-4ac } }{2a}`

Comparing the quadratic equation x² - 2x -9 = 0 with ax² + bx + c = 0, we get

a = 1 (as x² has no coefficient)

b = -2

c = -9

Plugging in the values of a, b, and c into the quadratic formula, we get

x =
`\frac{-(-2)\pm\sqrt{(-2)^(2)-4(1)(-9) } }{2(1)}`

=> x =
`(2\pm√(4+36 ) )/(2)`

=> x =
`(2\pm√(40))/(2)`

=> x =
`(2\pm2√(10))/(2)`

Factoring out 2 from the top, we get

x =
`(2(1\pm√(10)))/(2)`

Canceling out the 2's from the top and bottom, we have

x =
`1\pm√(10)`

Either x =
`1+\sqrt10` or x=
`1-\sqrt10`

=> x = 1 + 3.162 or x = 1 - 3.162

=> x = 4.162 (possible) or x = -2.162 (not possible as width can't be negative)

So, width = 4.16 meter (rounded off to the nearest hundredth)

Now,

Area of the rectangle = length * width

27 = length * 4.16

Flipping the sides of the equation,

length * 4.16 = 27

Dividing both sides by 4.16, we get

`(length * 4.16)/(4.16) = (27)/(4.16)`

Cancelling out 4.16 from the top and bottom of the left side, we get

length = 6.490

=> length = 6.49 meter (rounded off to the nearest hundredth)