Algebra 1 math question

Question

The length of a rectangle is 1 cm greater than its width. the area of the rectangle is 120 cm. find the length and width

Answer 1

It says the area of the rectangle 120 cm. I am assuming the unit should be 120 sq cm.

The text says the length is 1 cm greater than the width.

thus

let w be the width this will make w+1 the legnth

area is length times width that is: w(w + 1) = 120

w^2 + w = 120

w^2 + w -120 = 0

this does not factor so we use the quadratic formula.

(-b +/- sqrt(b^2 - 4ac))/ 2a with a=1 b= 1 c= -120

(-1 +/- sqrt(1^2 - 4(1)(-120)) )/ 2(1)

(-1 +/- sqrt(1 - (-480)))/2

(-1 +/- sqrt(481))/2

(-1 +/- 21.93)/2

20.93/2 22.93/2

10.46 11.45 remember that l is one longer than w.

if 10.46 is w then 11,46 is l area product is 119,87 well within rounding errors

if 11.45 is w then 12.45 is l area product is 142.55 not the correct amount - disregard

If you realize that if this rectangle was a complete square each side would be sqrt(120) or 10.95

The rectangle is slightly off square.with one side slightly shorter and the other slightly longer.

Answer 2

It says the area of the rectangle 120 cm. I am assuming the unit should be 120 sq cm.

The text says the length is 1 cm greater than the width.

thus

let w be the width this will make w+1 the legnth

area is length times width that is: w(w + 1) = 120

w^2 + w = 120

w^2 + w -120 = 0

this does not factor so we use the quadratic formula.

(-b +/- sqrt(b^2 - 4ac))/ 2a with a=1 b= 1 c= -120

(-1 +/- sqrt(1^2 - 4(1)(-120)) )/ 2(1)

(-1 +/- sqrt(1 - (-480)))/2

(-1 +/- sqrt(481))/2

(-1 +/- 21.93)/2

20.93/2 22.93/2

10.46 11.45 remember that l is one longer than w.

if 10.46 is w then 11,46 is l area product is 119,87 well within rounding errors

if 11.45 is w then 12.45 is l area product is 142.55 not the correct amount - disregard

If you realize that if this rectangle was a complete square each side would be sqrt(120) or 10.95

The rectangle is slightly off square.with one side slightly shorter and the other slightly longer.

Answer 3

First, we must figure out how we are going to translate the words into an actual equation.

Since this problem states the area of a rectangle, we must use the formula for the area of a rectangle, which is A = L·W

Now, the length of the rectangle is 1cm greater than its width, so we can say L = W + 1

Let.s substitute this into the area formula.

A = (W + 1) · W

When you want to solve a single equation, you ONLY want one variable.

Currently, we have 2, which are A and W.

However, the area was given in the problem as 120cm²

So, we have

120 = (W + 1) · W

Now, we have just one variable, W, so let's get to solving.

Let's start off by expanding.

120 = W² + W

It is clear this equation is quadratic.

We can choose to solve by factoring, completing the square, or the quadratic formula.

Let's choose the quadratic formula.

But, before we can use this, we must move everything to one side so that the whole equation equals to 0.

So,

W² + W - 120 = 0

Hence, the coefficients are the following: a = 1, b = 1, c = -120

W = (-1 ± √[1² - 4 · 1 · (-120)]) / (2 · 1)

W = (-1 ± √481) / 2

By using a calculator, we get

W ≈ -11.47 and W ≈ 10.47

Since a width can only be a positive value, W must by 10.47.

Lastly, since the length is 1cm bigger than its width, L mus be 11.47.

Therefore, the width is 10.47cm and length is 11.47cm

*Please feel free to ask me any question for clarification.

Answer 4

Morgan, the hard part about word problems is the first step: translating from words to math.

First, identify quantities that you don’t know, and name them.

Let L be the length

Let W be the width

Next remember relevant formulas.

Area = Length * Width

Then translate any words using these names and formulas.

Area = L * W

120 = L * W

L = W + 1

Now solve these two equations for W, by substituting (W + 1) for L in the area equation.

120 = (W + 1) * W

At this stage in school year, I presume that you know how to solve this for W. If not, add a comment and I’ll be glad to show you how.