Question

The area of a rectangle is 112 and perimeter is 64 what are the dimension

Answer 1

w = 28, 4 and l = 4, 28.

Explanation:

The formula for the area of a rectangle is
`lw` where l is the length and w is the width. Since the area of the rectangle is 112,
`lw=112`. The formula for the perimeter of a rectangle is
`2(l+w)` (simplified from
`l+w+l+w`). The perimeter is 64, therefore
`2(l+w) = 64`. You can simplify the equation by dividing both sides by 2, giving you
`(2(l+w))/(2) =(64)/(2)` =
`l+w=32`.

Now you have a system of equations:
`\left \{ {{lw=112} \atop {l+w=32}} \right.` . The second equation gives you
`l=32-w` by subtracting w from both sides.

You can then plug in the expression for l into the first equation, getting
`(32-2)w=112`. Using the distributive property, you can simply the equation to get
`32w-w^(2) =112`. By subtracting 112 on both sides, you get
`32w-w^2-112=0`, or
`-w^2+32w-112` by switching the places of
`-w^2` and 32w.

Then we can factor the equation. The only possible combination you can use to get
`-w^2` is -w multiplied by w. So, we have part of the new equation: (-w )(w ) = 0. Then, we need a pair of numbers that has a sum of 32 and a product of -112. You can list out all the factors of 112: 1, 2, 4, 7, 8, 14, 16, 28, 56, and 112. To get -112, you can just pick a pair that has a product of 112, but make one of the numbers negative.

After some trial and error, we find that the pair we are looking for is 28 and -4. We are done factoring the equation to get
`(-w+28)(w-4)=0`. We don't put -4 in the first set of parentheses because the product of a negative number and another negative number is a positive number. Knowing that 28 is smaller than 32, we can multiply the -w with the -4 to get a positive 4w. Adding that to the 28w we have from the product of 28 and w, we get the 32w we're looking for.

Now we know that
`(-w+28)=0` and
`(w-4)=0` For the first equation, we subtract 28 on both sides to give us
`-w=-28`. Dividing both sides by -1 gives us
`w=28`. However, the second equation gives us
`w=4` by adding 4 to both sides.

Therefore, w has two solutions: 28 and 4. Using
`lw=112`, we can plug in w to get
`l(28)=112` or
`l(4)=112`. This also gives us two solutions for l: l = 4 or l = 28.

We now have our final answers: w = 28, 4 and l = 4, 28.