Question

A rectangle has an area of 24 sq. inches and a perimeter of 50 in. what are the dimensions of the rectangle​

Answer 1

Length of rectangle = 24 inches

Width of rectangle = 1 inches

Explanation:

Area of rectangle = 24 square inches

Perimeter of rectangle = 50 inches

Formulas:

Area of rectangle =
`l* w`

Perimeter of rectangle =
`2(l+w)`

where
`l` represents length of rectangle and
`w` represents width of rectangle.

So, we can get two equations for
`l` and
`w`

A)
`l* w=24` [Area]

B)
`2(l+w)=50` [ Perimeter ]

Simplifying equation B.

Dividing both sides by 2.

`(2(l+w))/(2)=(50)/(2)`

`l+w=25`

Solving for
`l` in terms of width.

Subtracting
`w` both sides.

`l+w-w=25-w`

∴
`l=25-w`

Substituting value of
`l` in terms of
`w` in equation A.

`(25-w)w=24`

Using distribution.

`25w-w^2=24`

Adding
`w^2` both sides.

`25w-w^2+w^2=24+w^2`

`25w=24+w^2`

subtracting
`25 w` both sides

`25w-25w=24+w^2-25w`

`0=w^2-25w+24`

We have a quadratic equation.
`w^2-25w+24=0`

We solve for
`w` by factor method.

Splitting middle term into two terms such that the sum of the two =
`-25w` and product of two =
`24w^2`

`w^2-24w-w+24=0`

Factoring in pairs.

`w(w-24)-1(w-24)=0`

Factoring as a whole

`(w-24)(w-1)=0`

So, we have

`w-24=0` and
`w-1=0`

By adding 24 to one equation and adding 1 to other we get

`w-24+24=0+24` and
`w-1+1=0+1`

`w=24` and
`w=1`

So length can be found out by plugging value of
`w` in the length equation.

`l=25-24=1` and
`l=25-1=24`

Since length is always the longer side of rectangle so, the dimensions of rectangle can be given by.

length = 24 inches (answer)

width = 1 inches (answer)