Rectangle: The area of a rectangle is 84 square inches and i

[Guest]

The area of a rectangle is 84 square inches and its perimeter is 38 inches. Set up an equation and find the length and width of the rectangle.

 

[skeeter]

LW = 84

2(L+W) = 38

using either one of the equations above, solve for L in terms of W. substitute your result in for L in the second equation ... you'll get an equation where the only unknown is W. solve for W, then you can find L.

 

[Guest]

ok

Here is my work: Am I right?

Here it is:

The area of a rectangle is 84 square inches and its
perimeter is 38 inches. Set up an equation and find the
length and width of the rectangle.

Note: we need to use two formulas here. We need the perimeter of a rectangle formula and the area of a rectangle formula.

Perimeter Formula: P = 2L + 2w

Area Formula: A = Lw

NOTE: L = length and w = width

A = Lw

84 = Lw

84/w = L

Now, that I know L, I will plug L into P = 2L + 2w and simplify.

After doing the algebra, I get this: w^2 - 19w + 168 = 0.

I will now need to factor.

I found the length to be 12 inches and the width to be 7 inches.

Let us check.

Area = 84, right?

A = L w

84 = 12(7)

84 = 84...IT CHECKS.

How about checking using the perimeter of a rectangle formula?

P = 38 inches, right?

P = 2L + 2w

38 = 2(12) + 2(7)

38 = 24 + 14

38 = 38...IT ALSO CHECKS.

So, the length = 12 inches and the width = 7 inches

 

[skeeter]

Here is my work: Am I right?

you checked it ... what do you think?

 

[Denis]

Re: ok

After doing the algebra, I get this: w^2 - 19w + 168 = 0.

That's not correct.

 

[Guest]

ok

Here it is:

The area of a rectangle is 84 square inches and its
perimeter is 38 inches. Set up an equation and find the
length and width of the rectangle.

Note: we need to use two formulas here. We need the perimeter of a rectangle formula and the area of a rectangle formula.

Perimeter Formula: P = 2L + 2w

Area Formula: A = Lw

NOTE: L = length and w = width

A = Lw

84 = Lw

84/w = L

Now, that I know L, I will plug L into P = 2L + 2w and simplify.

After doing the algebra, I get this: w^2 - 19w + 84 = 0.

I will now need to factor.

I found the length to be 12 inches and the width to be 7 inches.

Let us check.

Area = 84, right?

A = L w

84 = 12(7)

84 = 84...IT CHECKS.

How about checking using the perimeter of a rectangle formula?

P = 38 inches, right?

P = 2L + 2w

38 = 2(12) + 2(7)

38 = 24 + 14

38 = 38...IT ALSO CHECKS.

So, the length = 12 inches and the width = 7 inches

Is my work correct?