Chapter 12.4, Problem 1CK

a.

To determine

To find: The area to the nearest millimetre.

Answer to Problem 1CK

Side of the polygon is 20 mm and area is 173.2 mm².

Explanation of Solution

Given information:

The polygon has a perimeter 60 mm.

Calculation:

The given polygon is an equilateral triangle.

And perimeter of the triangle is given by

  $P=3s$ where s is the measure of each sides.

Since, the perimeter is 60 mm. Therefore,

  $\begin{array}{l}60=3s\\ s=\frac{60}{3}\\ s=20\text{ mm}\end{array}$

Also, area of the equilateral triangle is given by

  $A=\frac{\sqrt{3}}{4}{s}^2$ where s is the side of triangle.

Here, $s=20\text{ mm}$ . Therefore, area is

  $\begin{array}{l}A=\frac{\sqrt{3}}{4}{\left(20\right)}^2\\ A=\frac{\sqrt{3}}{4}\times 400\\ A=100\sqrt{3}\\ A=173.2{\text{ mm}}^2\end{array}$

Hence,

Side of the polygon is 20 mm and area is 173.2 mm².

b.

To determine

To find: The area to the nearest millimetre.

Answer to Problem 1CK

Side of the polygon is 15 mm and area is 225 mm².

Explanation of Solution

Given information:

The polygon has a perimeter 60 mm.

Calculation:

The given polygon is a square.

And perimeter of the square is given by

  $P=4x$ where s is the measure of each sides.

Since, the perimeter is 60 mm. Therefore,

  $\begin{array}{l}60=4x\\ x=\frac{60}{4}\\ x=15\text{ mm}\end{array}$

Also, area of the square is given by

  $A=x^2$ where x is the side of square.

Here, $x=15\text{ mm}$ . Therefore, area is

  $\begin{array}{l}A=x^2\\ A={\left(15\right)}^2\\ A=225{\text{ mm}}^2\end{array}$

Hence,

Side of the polygon is 15 mm and area is 225 mm².

c.

To determine

To find: The area to the nearest millimetre.

Answer to Problem 1CK

Side of the polygon is 15 mm and area is 225 mm².

Explanation of Solution

Given information:

The polygon has a perimeter 60 mm.

Calculation:

The given polygon is a regular hexagon.

And perimeter of the regular hexagon is given by

  $P=6y$ where y is the measure of each sides.

Since, the perimeter is 60 mm. Therefore,

  $\begin{array}{l}60=6y\\ y=\frac{60}{6}\\ y=10\text{ mm}\end{array}$

Also, area of the regular hexagon is given by

  $A=\frac{3\sqrt{3}}{2}{y}^2$ where y is the side of hexagon.

Here, $y=10\text{ mm}$ . Therefore, area is

  $\begin{array}{l}A=\frac{3\sqrt{3}}{2}{\left(10\right)}^2\\ A=\frac{3\sqrt{3}}{2}\times 100\\ A=150\sqrt{3}\\ A=259.81{\text{ mm}}^2\end{array}$

Hence,

Side of the polygon is 10 mm and area is 259.81 mm².

d.

To determine

To find: The area to the nearest millimetre.

Answer to Problem 1CK

The radius of the circle is $\frac{30}{\pi }$ mm and area is $286.48{\text{ mm}}^2$ .

Explanation of Solution

Given information:

The circle has a circumference 60 mm.

Calculation:

The circumference of the circle is given by

  $C=2\pi r$ where r is the radius of the circle.

Since, the circumference is 60 mm. Therefore,

  $\begin{array}{l}60=2\pi r\\ r=\frac{60}{2\pi }\\ r=\frac{30}{\pi }\text{ mm}\end{array}$

Also, area of the circle is given by

  $A=\pi r^2$ where r is the radius of the circle.

Here, $r=\frac{30}{\pi }$ . Therefore, area is

  $\begin{array}{l}A=\pi {\left(\frac{30}{\pi }\right)}^2\\ A=\frac{900}{\pi }\\ A=286.48{\text{ mm}}^2\end{array}$

Hence,

The radius of the circle is $\frac{30}{\pi }$ mm and area is $286.48{\text{ mm}}^2$ .