Chapter 12.6, Problem 15PSC

To determine

To find: What percentage of the area of the circle is equal to the area of $ABGH.$

Answer to Problem 15PSC

11% of the area of a circle is the area of a rectangle.

Explanation of Solution

Given Information:

  $\overline{AB}=\overline{BC}=\overline{CD}$

  $H\left(12,-2\right)\text{and}E\left(15,-2\right)$

Formula used:

Area of a circle $\left(A\right)=\pi r^2$

  $"r"$ is the radius of a circle

Area of a rectangle $\left(A\text{'}\right)=lb$

  $l\text{and}b$ are the length and breadth of a rectangle

Distance between two points $\left(x_1,y_1\right)\text{and}\left(x_2,y_2\right)$ is $\left(D\right)=\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}$

Calculation:

Let’s draw a line segment $PQ$ passing through the centre $O$ and perpendicular to x-axis

  

Here, $OP$ and $OQ$ becomes the radii of the circle.

Now, from the points $H$ and $R,$ we get the coordinates of point $A$ as (12, 10)

Distance between $A$ and $R$ is same as the length of the diameter of a circle.

Distance between two points $\left(x_1,y_1\right)\text{and}\left(x_2,y_2\right)$ is $\left(D\right)=\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}$

  $\begin{array}{l}AR=\sqrt{{\left(12-0\right)}^2+{\left(10-10\right)}^2}=\sqrt{{12}^2}\\ AR=12\\ \therefore PQ=AR=12\\ OP=OQ=\frac{PQ}{2}=6\end{array}$

Therefore, radius of the circle $\left(r\right)=6$

We know that, area of a circle $\left(A\right)=\pi r^2$

  $\begin{array}{l}\therefore A=\frac{22}{7}\times 6^2\\ \therefore A=113.14\\ \therefore A\approx 113\end{array}$

As $\overline{AB}=\overline{BC}=\overline{CD}$

  $\begin{array}{l}\therefore \overline{HG}=\overline{GF}=\overline{EF}\\ \text{But,}\overline{HE}=\overline{HG}+\overline{GF}+\overline{EF}\\ \therefore \sqrt{{\left(12-15\right)}^2+{\left(-2+2\right)}^2}=3\overline{HG}\\ \therefore \overline{HG}=\frac{\sqrt{{\left(12-15\right)}^2+{\left(-2+2\right)}^2}}{3}\\ \therefore \overline{HG}=\frac{\sqrt{{\left(-3\right)}^2}}{3}=\frac{3}{3}\\ \therefore \overline{HG}=1\\ \therefore \overline{HG}=\overline{GF}=\overline{EF}=1\\ \therefore \overline{AB}=\overline{BC}=\overline{CD}=1\end{array}$

Also, $\overline{AH}=\overline{BG}=PQ=12$

  $\therefore \overline{AH}=\overline{BG}=12$

We know that, area of a rectangle $\left(A\text{'}\right)=lb$

  $l\text{and}b$ are the length and breadth of a rectangle

  $\therefore$ Area of rectangle $ABGH\text{is}\left(A\text{'}\right)=BG\times GH$

  $\begin{array}{l}\therefore A\text{'}=12\times 1\\ \therefore A\text{'}=12\end{array}$

Now, $\left(\frac{A\text{'}}{A}\times 100\right)\%=\left(\frac{12}{113}\times 100\right)\%=10.61\%\approx 11\%$

Hence, 11% of the area of a circle is the area of a rectangle.