Question

Hi I need help with this math problem. I need help finding the solution and properly formatting the problems. The question is in the picture. thank you! I can provide more examples of how to format the work, but I can’t seem to add another picture. Let me know. Thank you so much.

Answer 1

Width = 4.5 yards

Length : 12 yards

Explanation

the area of a rectangle is given by:

$\text{Area}=\text{length}\cdot\text{ width}$

so

Step 1

a)

Let l represents the length

let w represents the width

so

I)the area of the rectangle is 54

replace

$\begin{gathered} \text{Area}=\text{length}\cdot\text{ width} \\ 54=l\cdot w\rightarrow equation(1) \end{gathered}$

ii)the length of the triangle is 3 yd more than twice the width ( in other words you have to add 3 to twice the width to obtain the length)

so

$l=2w+3\rightarrow\text{equation}(2)$

Step 2

Solve the equations

$\begin{gathered} 54=l\cdot w\rightarrow equation(1) \\ l=2w+3\rightarrow\text{equation}(2) \end{gathered}$

a) replace the l value form equation (2) in equaiton(1)

$\begin{gathered} 54=l\cdot w\rightarrow equation(1) \\ 54=(2w+3)\cdot w \\ 54=2w^2+3w \\ \text{subtract 54 in both sides} \\ 54-54=2w^2+3w-54 \\ 0=2w^2+3w-54 \end{gathered}$

now, we have a quadratic equation, we can use the quadratic formula

$\begin{gathered} 0=2w^2+3w-54\rightarrow0=ax^2+bx+c \\ so \\ a=2 \\ b=3 \\ c=-54 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \text{replace} \\ w=\frac{-3\pm\sqrt[]{3^2^{}-4(2)(-54)}}{2(2)} \\ w=\frac{-3\pm\sqrt[]{9+432}}{4} \\ w=\frac{-3\pm\sqrt[]{441}}{4} \\ w_1=\frac{-3+21}{4}=\frac{18}{4}=\frac{9}{2}=4.5 \\ w_1=\frac{-3-21}{4}=\frac{-24}{4}=-12=-12 \\ the\text{ valid option is the positive, so} \\ w=4.5 \end{gathered}$

Width = 4.5 yards

b) now, replace in equation(2)

$\begin{gathered} l=2w+3\rightarrow\text{equation}(2) \\ l=2(4.5)+3 \\ l=9+3 \\ l=12 \end{gathered}$

therefore, the answer is

length : 12 yards

I hope this helps you