Question

The length of a rectangle is 5 mm longer than the width. If the area of the rectangle is 24 mm^2, what is the width of the rectangle?

Answer 1

Answer:The width is 3mm

Explanation:

Length= 5mm + width

Let the width be represented by W

Area=24mm^2

Since area= length x width

24=(5+ W) x W

24=5W +
`W^(2)`

`W^(2)` +5W-24=0

The factors are 8 and -3

`W^(2)`+8w-3w-24=0

Factorise

W(w+8)-3(w+8)=0

(w-3)=0 or (w+8)=0

W=3 0r w= -8

The width cannot be negative.

Hence,W=3mm

Answer 2

8

Explanation:

This question is asking you to solve for the following system of equations, let x be width and y be height of the rectangle.

`\left \{ {{x=y+5} \atop {x*y=24}} \right.`

Since the former is already equal to x lets set the second equation to x

`x*y=24`

`x=24/y`

Now we have the following system of equations:

`\left \{ {{x=y+5} \atop {x=24/y}} \right.`

Now that the two equations are equal, we can solve for y:

`y+5=24/y`

Divide both sides by y

`y^2+5y=24`

Subtract 24 from the right side setting it to 0:

`y^2+5y-24=0`

Solve for y using quadratic formula:

`y=(-5\pm √(5^2-4\cdot \:1\cdot \left(-24\right)))/(2\cdot \:1)`

`y=3,-8`

Given the y-values lets plug them into our original equations to find the intersections.

`x=3+5\\x=8`

Giving intersection vector (8,3)

`x=-8+5\\x=-3`

Giving intersection vector (-3,-8)

Since both the width and length of the rectangle must be positive lets use the first vector (8,3) as our solution.

The x value being 8 means the width must be 8, y value being 3 the height must be 3, these variable correlate to our original definitions of making the x value equal to the width of the carpet and y value the height of the carpet.