Respuesta :
Answer:
width = 4.16 meter
length = 6.49 meter
Step-by-step 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
[tex]\frac{3(x^{2}-2x-9) }{3}[/tex] = [tex]\frac{0}{3}[/tex]
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 = [tex]\frac{-b\pm\sqrt{b^{2}-4ac } }{2a}[/tex]
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 = [tex]\frac{-(-2)\pm\sqrt{(-2)^{2}-4(1)(-9) } }{2(1)}[/tex]
=> x = [tex]\frac{2\pm\sqrt{4+36 } }{2}[/tex]
=> x = [tex]\frac{2\pm\sqrt{40}}{2}[/tex]
=> x = [tex]\frac{2\pm2\sqrt{10}}{2}[/tex]
Factoring out 2 from the top, we get
x = [tex]\frac{2(1\pm\sqrt{10})}{2}[/tex]
Canceling out the 2's from the top and bottom, we have
x = [tex]1\pm\sqrt{10}[/tex]
Either x = [tex]1+\sqrt10[/tex] or x= [tex]1-\sqrt10[/tex]
=> 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
[tex]\frac{length * 4.16}{4.16} = \frac{27}{4.16}[/tex]
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)
