Exercise 16.3: Measurement Regarding Circles
1. An arc of a circle subtends an angle of $30^\circ$ at the center. If the diameter of the circle is $126\text{ cm}$, find the length of the arc.
Solution: Given,
Central angle, $\theta = 30^\circ$
Diameter of the circle, $2r = 126\text{ cm}$
$\implies \text{Radius}, r = \frac{126}{2} = 63\text{ cm}$
Length of the arc, $s = \frac{\pi r \theta}{180^\circ}$
$\implies s = \frac{3.1416 \times 63 \times 30}{180}$
$\implies s = \frac{5937.624}{180} = 32.9868 \approx 32.99\text{ cm}$
Answer: The length of the arc is approximately $32.99\text{ cm}$.
2. A horse ran around a field in $1\frac{1}{2}$ minutes at a speed of $66\text{ meters}$ per minute. Find the diameter of the field.
Solution:
In $1\text{ minute}$, the horse covers $66\text{ meters}$.
$\therefore \text{In } 1\frac{1}{2} \text{ or } \frac{3}{2} \text{ minutes, it covers } = 66 \times \frac{3}{2} = 99\text{ meters}$.
Since the horse completes one full round of the field, the circumference of the field $= 99\text{ meters}$.
Let the diameter of the field be $d\text{ meters}$.
We know that, $\text{Circumference} = \pi d$
According to the question,
$\pi d = 99$
$\implies d = \frac{99}{\pi} = \frac{99}{3.1416} = 31.512605 \approx 31.51\text{ meters}$
Answer: The diameter of the field is approximately $31.51\text{ meters}$.
3. The area of a sector of a circle is $77\text{ sq. meters}$ and the radius of the circle is $21\text{ meters}$. Find the angle subtended by the arc at the center.
Solution:
Given, Area of the sector $= 77\text{ sq. meters}$
Radius of the circle, $r = 21\text{ meters}$
Let the subtended angle be $=\theta$
We know that, $\text{Area of a sector} = \frac{\theta}{360^\circ} \times \pi r^2$
According to the question,
$\frac{\theta}{360} \times 3.1416 \times (21)^2 = 77$
$\implies \frac{\theta \times 3.1416 \times 441}{360} = 77$
$\implies \theta \times 1385.4456 = 77 \times 360$
$\implies \theta \times 1385.4456 = 27720$
$\implies \theta = \frac{27720}{1385.4456} = 20.0080032 \approx 20^\circ$
Answer: The subtended angle is approximately $20^\circ$ (or $20.008^\circ$).
4. The radius of a circle is $14\text{ cm}$ and an arc subtends an angle of $75^\circ$ at the center. Find the area of the sector.
Solution: Given, Radius, $r = 14\text{ cm}$
Central angle, $\theta = 75^\circ$
$$\text{Area of the sector} = \frac{\theta}{360^\circ} \times \pi r^2$$
$$= \frac{75}{360} \times 3.1416 \times (14)^2$$
$$= \frac{75}{360} \times 3.1416 \times 196$$
$$= \frac{46181.52}{360} = 128.282 \approx 128.28\text{ sq. cm}$$
Answer: Area is approximately $128.28\text{ sq. cm}$.
5. A circular field is surrounded by a road. The outer circumference of the road is $44\text{ meters}$ greater than the inner circumference. Find the width of the road.
Solution:
Let the outer radius of the circular field be $R\text{ meters}$ and the inner radius be $r\text{ meters}$.
$\therefore \text{Width of the road} = (R – r)\text{ meters}$.
Outer circumference $= 2\pi R$ and inner circumference $= 2\pi r$
According to the question,
$2\pi R – 2\pi r = 44$
$\implies 2\pi (R – r) = 44$
$\implies R – r = \frac{44}{2\pi}$
$\implies R – r = \frac{44}{2 \times 3.1416}$
$\implies R – r = \frac{44}{6.2832} = 7.00280112 \approx 7.003\text{ meters}$
Answer: The width of the road is approximately $7.003\text{ meters}$.
6. The diameter of a circular park is $26\text{ meters}$. Surrounding the park on the outside, there is a pathway of width $2\text{ meters}$. Find the area of the path.
Solution:
Diameter of the park $= 26\text{ meters}$
$\implies \text{Inner radius}, r = \frac{26}{2} = 13\text{ meters}$.
Width of the path $= 2\text{ meters}$.
$\therefore \text{Outer radius including the path}, R = 13 + 2 = 15\text{ meters}$.
Area of the park $= \pi r^2 = \pi \times 13^2 = 169\pi\text{ sq. meters}$.
Area of the park including the path $= \pi R^2 = \pi \times 15^2 = 225\pi\text{ sq. meters}$.
$\therefore \text{Area of the path} = 225\pi – 169\pi = 56\pi\text{ sq. meters}$
$$= 56 \times 3.1416 = 175.9296 \approx 175.93\text{ sq. meters}$$
Answer: The area of the path is approximately $175.93\text{ sq. meters}$.
7. The diameter of the front wheel of a car is $28\text{ cm}$ and that of the rear wheel is $35\text{ cm}$. In traveling a distance of $88\text{ meters}$, how many complete revolutions will the front wheel make more than the rear wheel?
Solution:
Total distance $= 88\text{ meters} = 88 \times 100 = 8800\text{ cm}$.
For the front wheel:
$\text{Diameter} = 28\text{ cm} \implies \text{Circumference} = \pi d = 3.1416 \times 28 \approx 87.9648\text{ cm}$
$$\text{Number of revolutions} = \frac{8800}{87.9648} \approx 100.04 \approx 100\text{ times.}$$
For the rear wheel:
$\text{Diameter} = 35\text{ cm} \implies \text{Circumference} = \pi d = 3.1416 \times 35 \approx 109.956\text{ cm}$
$$\text{Number of revolutions} = \frac{8800}{109.956} \approx 80.03 \approx 80\text{ times.}$$
$\therefore \text{The front wheel rotates more by} = 100 – 80 = 20\text{ times.}$
Answer: It will rotate $20$ times more.
8. The circumference of a circle is $220\text{ meters}$. Find the length of a side of the square inscribed in the circle.
Solution:
Let the radius of the circle be $r\text{ meters}$.
$\therefore \text{Circumference of the circle} = 2\pi r = 220$
$\implies 2r = \frac{220}{\pi} = \frac{220}{3.1416} \approx 70.028\text{ meters}$.
Here, $2r$ is the diameter of the circle.
The diagonal of the square inscribed in the circle = Diameter of the circle ($2r$).
Let the length of one side of the square be $a\text{ meters}$.
$\therefore \text{Diagonal of the square} = a\sqrt{2}$
According to the question,
$a\sqrt{2} = 70.028$
$\implies a = \frac{70.028}{\sqrt{2}} = \frac{70.028}{1.4142} = 49.5177486 \approx 49.52\text{ meters}$.
Answer: The length of a side of the square is approximately $49.52\text{ meters}$.
9. The circumference of a circle is equal to the perimeter of an equilateral triangle. Find the ratio of their areas.
Solution:
Let the radius of the circle $= r \implies \text{Circumference} = 2\pi r$
Let the side length of the equilateral triangle $= a \implies \text{Perimeter} = 3a$
According to the question,
$2\pi r = 3a$
$\implies a = \frac{2\pi r}{3}$
$$\text{Area of the circle} = \pi r^2$$
$$\text{Area of the equilateral triangle} = \frac{\sqrt{3}}{4} a^2$$
$$= \frac{\sqrt{3}}{4} \left(\frac{2\pi r}{3}\right)^2$$
$$= \frac{\sqrt{3}}{4} \times \frac{4\pi^2 r^2}{9}$$
$$= \frac{\sqrt{3}\pi^2 r^2}{9} = \frac{\pi^2 r^2}{3\sqrt{3}}$$
$\therefore \text{Ratio of their areas} = \pi r^2 : \frac{\pi^2 r^2}{3\sqrt{3}}$
$= 1 : \frac{\pi}{3\sqrt{3}}$ [Dividing both sides by $\pi r^2$]
$= 3\sqrt{3} : \pi$
Answer: The ratio of the area of the circle to that of the equilateral triangle is $3\sqrt{3} : \pi$.
10. Based on the information in the figures below, find the area of the shaded regions.
First Figure
The figure consists of a right-angled triangle and a semicircle.
- Area of the right-angled triangle:
According to the figure, Perpendicular $= 8\text{ cm}$ and Hypotenuse $= 10\text{ cm}$
By Pythagoras’ theorem,
$$\text{Base} = \sqrt{10^2 – 8^2} = \sqrt{100 – 64} = \sqrt{36} = 6\text{ cm}$$$$\text{Area of triangle} = \frac{1}{2} \times \text{Base} \times \text{Perpendicular} = \frac{1}{2} \times 6 \times 8 = 24\text{ sq. cm}$$ - Area of the semicircle:
Diameter of the semicircle $= 9\text{ cm}$
$\therefore \text{Radius } r = \frac{9}{2} = 4.5\text{ cm}$
$$\text{Area of semicircle} = \frac{1}{2} \pi r^2 = \frac{1}{2} \times 3.1416 \times (4.5)^2 = 31.8087 \approx 31.81\text{ sq. cm}$$
Total shaded area:
$$= 24 + 31.81 = 55.81\text{ sq. cm}$$
Answer:$55.81\text{ sq. cm}$ (approx.)
Second Figure
The figure consists of a square with a semicircle on top. The sides of the square are marked as equal.
- Area of the square:
Side length of the square, $a = 4\text{ cm}$
$$\text{Area of square} = a^2 = 4^2 = 16\text{ sq. cm}$$ - Area of the semicircle:
Diameter of the semicircle $= 4\text{ cm}$
$\therefore \text{Radius } r = \frac{4}{2} = 2\text{ cm}$
$$\text{Area of semicircle} = \frac{1}{2} \pi r^2 = \frac{1}{2} \times 3.1416 \times (2)^2 \approx 6.28\text{ sq. cm}$$
Total shaded area:
$$= 16 + 6.28 = 22.28\text{ sq. cm}$$
Answer:$22.28\text{ sq. cm}$ (approx.)
Third Figure
In this figure, a semicircle is removed from a rectangle.
- Area of the rectangle:
Length $= 12\text{ cm}$ and Width $= 10\text{ cm}$
$$\text{Area of rectangle} = 12 \times 10 = 120\text{ sq. cm}$$ - Area of the semicircle:
Diameter of the semicircle $= 12\text{ cm}$
$\therefore \text{Radius } r = \frac{12}{2} = 6\text{ cm}$
$$\text{Area of semicircle} = \frac{1}{2} \pi r^2 = \frac{1}{2} \times 3.1416 \times (6)^2 = 56.5488 \approx 56.55\text{ sq. cm}$$
Shaded area:
$$= 120 – 56.55 = 63.45\text{ sq. cm}$$
Answer:$63.45\text{ sq. cm}$ (approx.)
Fourth Figure
In this figure, an inscribed full circle is removed from a square.
- Area of the square:
Side length, $a = 12\text{ cm}$
$$\text{Area of square} = 12^2 = 144\text{ sq. cm}$$ - Area of the inscribed circle:
Diameter of the circle = Side length of the square $= 12\text{ cm}$
$\therefore \text{Radius } r = \frac{12}{2} = 6\text{ cm}$
$$\text{Area of circle} = \pi r^2 = 3.1416 \times (6)^2 = 113.0976 \approx 113.10\text{ sq. cm}$$
Shaded area:
$$= 144 – 113.10 = 30.90\text{ sq. cm}$$
Answer:$30.90\text{ sq. cm}$ (approx.)