A rectangle is drawn so that the width is 1 feet shorter than the length. The area of the rectangle is 20 square feet. Find the length of the rectangle.

Please show your work So i can learn it

Respuesta :

Answer:

Length = 5 feet

Step-by-step explanation:

The formula for finding the area of a rectangle is:

Area = Length x Width

The problem said that the width is 1 foot shorter than the length. This can be shown like this:

Width = (Length - 1)

Going back to our formula for finding area, we can replace Width with (Length - 1). We can also put 20 as the area since the problem already told us that the area of the rectangle is 20 square feet. By now, your equation should look similar to this:

Area = Length x Width

Area = Length x (Length - 1)

20 = Length x (Length - 1)

Now, we can go on to figuring out what the length is. For now, the length is going to be referred to as ' x '.

20 = Length x (Length - 1)

20 = (x)(x - 1)

*Distribute x to (x - 1)*

20 = x² - x

*This is beginning to look like a quadratic equation. Let's move all of the terms over to one side.*

20 = x² - x

0 = x² - x - 20

*It would appear that we need to factor this equation to get our answer.*

x² - x - 20

This would be the factored version of our equation because the factors -4 and 5 give us a product of -1 but also give us a sum of -1.

(x - 5) (x + 4)

From here, our options are narrowed down to 5 and -4. As we know, lengths cannot be real which means...

Length = 5 feet

~Hope this helps!)

Lanuel

Since the area of a rectangle is 20 square feet, the length of this rectangle is equal to 4 feet.

  • Let the length of the rectangle be L.
  • Let the width of the rectangle be W.

Given the following data:

  • Area of rectangle = 20 square feet.

To find the length of the rectangle:

Translating the word problem into an algebraic equation, we have;

[tex]L=W-1[/tex]   ....equation 1.

[tex]W=L+1[/tex]    ....equation 2.

Mathematically, the area of a rectangle is given by the formula;

[tex]A = LW[/tex]    ...equation 3.

Substituting the values into the eqn. 2, we have;

[tex]20 = L(L+1)\\\\20=L^2 +L\\\\L^2 +L-20\\\\L^2 +5L-4L-20\\\\L(L+5)-4(L+5)\\\\(L+5)(L-4)[/tex]

L = -5 or L = 4

Length, L = 4 feet.

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