Chapter 11, Problem 22RP

Solution

a.

To calculate: The circumference of the circle.

The circumference of the circle is $40.2$

Given:

The coordinates of a point “B” on the circle is $\left(4,5\right)$

Concept used:

Here, we are using distance formula:

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

Calculation:

  

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

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

Here, the points are $\left(0,0\right)$ and $\left(4,5\right)$

So, $x_1=0,x_2=4$

And, $y_1=0,y_2=5$

OB = $\sqrt{{\left(4-0\right)}^2+{\left(5-0\right)}^2}$

➔$\sqrt{16+25}$

➔$\sqrt{41}$

OB = Radius of the circle.

So, Radius = $\sqrt{41}$

Now, circumference of the circle = $2\pi r$

(Where, “ r ” is radius)

➔$2\times \frac{22}{7}\times \sqrt{41}$

➔$\frac{44}{7}\sqrt{41}$

➔$\frac{44}{7}\times 6.403$

➔$40.247$

Nearest tenth value = $40.2$

Conclusion: Hence, the circumference of the circle is $40.2$

b.

To calculate: The area of the circle

The area of the circle is $128.9$ unit ${}^2$

Given:

The coordinates of a point “B” on the circle is $\left(4,5\right)$

Concept used:

Here, we are using the formula

Area of the circle = $\pi r^2$ (Where, “ r ” is radius)

Calculation:

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

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

Here, the points are $\left(0,0\right)$ and $\left(4,5\right)$

So, $x_1=0,x_2=4$

And, $y_1=0,y_2=5$

OB = $\sqrt{{\left(4-0\right)}^2+{\left(5-0\right)}^2}$

➔$\sqrt{16+25}$

➔$\sqrt{41}$

OB = Radius of the circle.

So, Radius = $\sqrt{41}$

We know that,

Area of the circle = $\pi r^2$

(Where, “ r ” is radius)

➔$\frac{22}{7}\times \sqrt{41}\times \sqrt{41}$

➔$\frac{22}{7}\times 41$

➔$\frac{902}{7}=128.85$

Nearest tenth value = $128.9$

Conclusion: Hence, the area of the circle is $128.9$ unit ${}^2$