Chapter 8, Problem 18TP

Solution

(a)

To find: Explain how to use the midpoint formula to find the coordinates of the vertices for each new square in the pattern

The explanation is given.

Given information:

The given figure is below:

  

The points one quarter of the way along the sides of the large square are connected to form an inscribed square. The vertices of the largest square is given as $\left(32,48\right),\left(48,16\right),\left(16,0\right),\left(0,32\right)$

Calculation:

The first inscribe square’s vertices are one quarter of the way of the largest square therefore the innermost inscribed square’s vertices are half of the way of largest square.

By using mid-point formula, the vertices of smallest inscribed square can be found. After that the second square’s vertices can be found using mid-point formula with the largest and smallest squares vertices.

(b)

To find: The missing vertices of squares.

The four vertices of the smallest square are , $\left(30,32\right)$ , $\left(16,40\right)$ , $\left(8,16\right)$ , $\left(32,8\right)$ . The four vertices of the middle square are , $\left(39,24\right)$ , $\left(24,49\right)$ , $\left(9,24\right)$ , $\left(24,4\right)$ .

Given information:

Given that the points one quarter of the way along the sides of the large square are connected to form an inscribed square .The vertices of the largest square is given as $\left(32,48\right),\left(48,16\right),\left(16,0\right),\left(0,32\right)$

Calculation:

The vertices of the innermost square are the mid-point of the outermost square’s vertices.

The mid-point formula for the two points is given as:

  $x=\frac{x_1+x_2}{2},y=\frac{y_1+y_2}{2}$

Let the first two vertices of the square are $\left(32,48\right),\left(48,16\right)$ .

Substitute $x_1=32,x_2=48,y_1=48,y_2=16$ in the mid-point formula:

  $\begin{array}{l}x=\frac{32+48}{2}\\ x=\frac{60}{2}\\ x=30\\ y=\frac{y_1+y_2}{2}\\ y=\frac{48+16}{2}\\ y=\frac{64}{2}\\ y=32\end{array}$

First vertex is $\left(30,32\right)$ .

Similarly, other vertices of the smallest square can be found.

Let the second two vertices of the square are $\left(32,48\right),\left(0,32\right)$ .

Substitute $x_1=32,x_2=0,y_1=48,y_2=32$ in the mid-point formula:

  $\begin{array}{l}x=\frac{32+0}{2}\\ x=\frac{32}{2}\\ x=16\\ y=\frac{y_1+y_2}{2}\\ y=\frac{48+32}{2}\\ y=\frac{80}{2}\\ y=40\end{array}$

Second vertex is $\left(16,40\right)$ .

Let take the two vertices of the square are $\left(32,48\right),\left(0,32\right)$ to find the third vertex f the smallest square. $\left(16,0\right),\left(0,32\right)$ .

Substitute $x_1=16,x_2=0,y_1=0,y_2=32$ in the mid-point formula:

  $\begin{array}{l}x=\frac{16+0}{2}\\ x=\frac{16}{2}\\ x=8\\ y=\frac{y_1+y_2}{2}\\ y=\frac{0+32}{2}\\ y=\frac{32}{2}\\ y=16\end{array}$

Second vertex is $\left(8,16\right)$ .

Let take the two vertices of the square are $\left(16,0\right),\left(48,16\right)$ to find the third vertex of the smallest square..

Substitute $x_1=16,x_2=48,y_1=0,y_2=16$ in the mid-point formula:

  $\begin{array}{l}x=\frac{16+48}{2}\\ x=\frac{64}{2}\\ x=32\\ y=\frac{y_1+y_2}{2}\\ y=\frac{0+16}{2}\\ y=\frac{16}{2}\\ y=8\end{array}$

Second vertex is $\left(32,8\right)$ .

The four vertices of the smallest square are , $\left(30,32\right)$ , $\left(16,40\right)$ , $\left(8,16\right)$ , $\left(32,8\right)$ .

In order to find the vertices of the middle square find the mid-point of the vertices $[\left(48,16\right),\left(30,32\right)],\left[\left(32,48\right),\left(16,40\right)\right],\left[\left(10,32\right),\left(8,16\right)\right]\left[\left(16,0\right),\left(32,8\right)\right]$

First vertices will be found using mid-point formula:

  $\begin{array}{l}x=\frac{48+30}{2}\\ x=\frac{78}{2}\\ x=39\\ y=\frac{16+32}{2}\\ y=\frac{48}{2}\\ y=24\end{array}$

The vertex of the middle square is $\left(39,24\right)$ .

Second vertex will be found using mid-point formula take the vertex $\left[\left(32,48\right),\left(16,40\right)\right]$

  $\begin{array}{l}x=\frac{32+16}{2}\\ x=\frac{48}{2}\\ x=24\\ y=\frac{48+40}{2}\\ y=\frac{98}{2}\\ y=49\end{array}$

The second vertex of the middle square is $\left(24,49\right)$ .

The third vertex will be found using mid-point formula take the vertex $\left(10,32\right),\left(8,16\right)$

  $\begin{array}{l}x=\frac{10+8}{2}\\ x=\frac{18}{2}\\ x=9\\ y=\frac{32+16}{2}\\ y=\frac{48}{2}\\ y=24\end{array}$

The third vertex of the middle square is $\left(9,24\right)$ .

The fourth vertex will be found using mid-point formula take the vertex $\left(16,0\right),\left(32,8\right)$

  $\begin{array}{l}x=\frac{16+32}{2}\\ x=\frac{48}{2}\\ x=24\\ y=\frac{0+8}{2}\\ y=\frac{8}{2}\\ y=4\end{array}$

The fourth vertex of the middle square is $\left(24,4\right)$ .

(c)

To find: The areas of squares.

The explanation is given.

Given information:

The four vertices of the smallest square are , $\left(30,32\right)$ , $\left(16,40\right)$ , $\left(8,16\right)$ , $\left(32,8\right)$ . The four vertices of the middle square are , $\left(39,24\right)$ , $\left(24,49\right)$ , $\left(9,24\right)$ , $\left(24,4\right)$ . The four vertices of the outer square are $\left(32,48\right),\left(48,16\right),\left(16,0\right),\left(0,32\right)$

Calculation:

In order to find the area of the square it is required to find the length of side of the square.

Use distance formula to find the length of the side of the square.

The distance formula is given below:

  $D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

For the outer most square let take any two points be: $\left(32,48\right),\left(48,16\right)$

Substitute $x_1=32,x_2=48,y_1=48,y_2=16$ in the distance formula:

  $\begin{array}{l}D=\sqrt{{\left(48-32\right)}^2+{\left(16-48\right)}^2}\\ D=\sqrt{{\left(16\right)}^2+{\left(-32\right)}^2}\\ D=\sqrt{256+1024}\\ D=\sqrt{1280}\\ D\approx 35.77\end{array}$

The length of the side of the outermost square is 35.77.

Area of the outermost square= $side\cdot side$ .

  $\begin{array}{l}A=\left(35.77\right)\cdot \left(35.77\right)\\ A=1280c{m}^2\end{array}$

For the second square let take any two points be: $\left(39,24\right),\left(24,49\right)$

Substitute $x_1=39,x_2=24,y_1=24,y_2=49$ in the distance formula:

  $\begin{array}{l}D=\sqrt{{\left(24-39\right)}^2+{\left(49-24\right)}^2}\\ D=\sqrt{{\left(-15\right)}^2+{\left(25\right)}^2}\\ D=\sqrt{225+625}\\ D=\sqrt{850}\\ D=29.49\end{array}$

The area of the second square will be:

  $\begin{array}{l}A=\left(29.49\right)\cdot \left(29.49\right)\\ A=850c{m}^2\end{array}$

For the inner most square let take any two points be: $\left(30,32\right),\left(16,40\right)$

Substitute $x_1=30,x_2=16,y_1=32,y_2=40$ in the distance formula:

  $\begin{array}{l}D=\sqrt{{\left(16-30\right)}^2+{\left(40-32\right)}^2}\\ D=\sqrt{{\left(-14\right)}^2+{\left(8\right)}^2}\\ D=\sqrt{196+64}\\ D=\sqrt{260}\\ D\approx 16.12\end{array}$

The area of the second square will be:

  $\begin{array}{l}A=\left(16.12\right)\cdot \left(16.12\right)\\ A\approx 260c{m}^2\end{array}$

It is observed that the area of the outer square is greater than the middle square. The area of middle square is greater than the smallest square.