Q34. - Algebra 1

Question

The perimeter of a square is given by the function $\mathit{P}\mathbf{=}\mathbf{4}\sqrt{\mathbf{A}}$ where A is the area of the square. 

a. Graph the function. 

b. Determine the perimeter of a square with an area of $\mathbf{225}{\mathit{m}}^{\mathbf{2}}$. 

c. When will the perimeter and the area be the same value?

Step-by-Step Solution

Answer

a. The graph of the function $P=4\sqrt{A}$ is,

b. The perimeter of the square with an area of $225m^2$ is $\mathbf{4}\mathit{m}\mathit{e}\mathit{t}\mathit{e}\mathit{r}\mathit{s}$ $\mathbf{60}\mathit{m}\mathit{e}\mathit{t}\mathit{e}\mathit{r}\mathit{s}$.

c. The square having a side of will have the same perimeter and area.

Step 1. State the concept of the perimeter of a square.

The perimeter of a square is the sum of the length of all the sides of a square. 

Suppose ‘$I$’ be the length of the side of a square.

 $\begin{array}{c}\text{Then the perimeter of the square}\left(P\right)=l+l+l+l\\ =4l\dots \left(1\right)\end{array}$

Step 2. State the concept of the Area of the square.

The area of a square is the square of its length. Suppose the length of a square is ‘$I$’, then the area of the square(A) is given as:

$\text{Area}\left(A\right)=l^2\dots \left(2\right)$

Step 3. Calculation

a. The given function is: $P=4\sqrt{A}$

To graph a function, find a few coordinates by substituting values of ‘A’ and by finding the respective values of ‘P’. 

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}A=0,\\ P=4\sqrt{0}\\ =4\left(0\right)\\ =0\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}A=1,\\ P=4\sqrt{1}\\ =4\left(1\right)\\ =4\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}A=4,\\ P=4\sqrt{4}\\ =4\left(2\right)\\ =8\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}A=9,\\ P=4\sqrt{9}\\ =4\left(3\right)\\ =12\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}A=16,\\ P=4\sqrt{16}\\ =4\left(4\right)\\ =16\end{array}$

Values of ‘A’=x-coordinate	Values of ‘P’=y-coordinate	$\left(A,P\right)=\left(x,y\right)$
0	0	(0,0)
1	4	(1,4)
4	8	(4,8)
9	12	(9,12)
16	16	(16,16)

Plot the values of area(A) on the X-axis and the values of the perimeter(P) on the Y-axis, on a coordinate plane, and join those points to get the required graph.

Hence, the required graph. 

b. 

The area of a square is the square of its length. Suppose the length of a square is ‘l’, then the area of the square(A) is given as:

$\text{Area}\left(A\right)=l^2\dots \left(2\right)$

The perimeter of a square is given by the function $P=4\sqrt{A}$ where A is the area of the square. 

Substitute $A=225$ in $P=4\sqrt{A}$ to get the value of perimeter.

$\begin{array}{c}P=4\sqrt{A}\\ =4\left(\sqrt{225}\right)\\ =4\left(15\right)\\ =60\end{array}$

Therefore, the perimeter of the square with an area of $225m^2$ is 60 meters.

c. 

The area of a square is the square of its length. Suppose the length of a square is ‘l’, then the area of the square(A) is given as:

$\text{Area}\left(A\right)=l^2\dots \left(2\right)$

Consider a square of length ‘l’. 

Suppose the perimeter and area of this square are the same. That is,

$\begin{array}{c}P=A\\ 4l=l^2\left[\text{From\hspace{0.17em}\hspace{0.17em}}\left(1\right)\&\left(2\right)\right]\\ \frac{4l}{l}=\frac{l^2}{l}\left[\text{Dividing\hspace{0.17em}\hspace{0.17em}throughout\hspace{0.17em}\hspace{0.17em}by\hspace{0.17em}\hspace{0.17em}}\text{'}l\text{'}\right]\\ 4=l\\ l=4\end{array}$

Therefore, a square whose length of the side is 4 meters, will have the same perimeter and area.