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

Answer:

Length of rectangle = 24 inches

Width of rectangle = 1 inches

Step-by-step explanation:

Area of rectangle = 24 square inches

Perimeter of rectangle = 50 inches

Formulas:

Area of rectangle = $l\times w$

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) $l\times w=24$    [Area]

B) $2(l+w)=50$     [ Perimeter ]

Simplifying equation B.

Dividing both sides by 2.

$\frac{2(l+w)}{2}=\frac{50}{2}$  

$l+w=25$  

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)