Exercise 6.3 Practice
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Overview
This page provides comprehensive Ch 6: Measuring Space: Perimeter and Area - Exercise Set 6.3 Practice. Solve questions on circular sectors, segments, clock sweep areas, car wiper coverage, and inscribed regular polygon area ratios with step-by-step solutions.
Circular Sector Areas, Segments & Inscribed Regular Polygons (Use $\pi \approx 22/7$ unless stated)
Q1: Sector Area basic
Find the area of a sector of a circle with radius $7\text{ cm}$ if the angle of the sector is $60^\circ$.
The area of a sector is given by $A = \frac{\theta}{360^\circ} \times \pi r^2$.
$$A = \frac{60^\circ}{360^\circ} \times \frac{22}{7} \times 7^2 = \frac{1}{6} \times 22 \times 7 = \frac{77}{3} \approx \mathbf{25.67\text{ sq. cm}}$$
$$A = \frac{60^\circ}{360^\circ} \times \frac{22}{7} \times 7^2 = \frac{1}{6} \times 22 \times 7 = \frac{77}{3} \approx \mathbf{25.67\text{ sq. cm}}$$
Area = 25.67 sq. cm
Q2: Quadrant Area from Circumference
Find the area of a quadrant of a circle whose circumference is $44\text{ cm}$.
First find the radius $r$ using the circumference:
$$C = 2\pi r = 44 \implies 2 \times \frac{22}{7} \times r = 44 \implies r = 7\text{ cm}$$
$$C = 2\pi r = 44 \implies 2 \times \frac{22}{7} \times r = 44 \implies r = 7\text{ cm}$$
The area of a quadrant is $A = \frac{1}{4}\pi r^2$:
$$A = \frac{1}{4} \times \frac{22}{7} \times 7^2 = \frac{77}{2} = \mathbf{38.5\text{ sq. cm}}$$
$$A = \frac{1}{4} \times \frac{22}{7} \times 7^2 = \frac{77}{2} = \mathbf{38.5\text{ sq. cm}}$$
Area = 38.5 sq. cm
Q3: Clock minute hand sweep
The length of the minute hand of a clock is $7\text{ cm}$. Find the area swept by the minute hand in $10\text{ minutes}$.
The minute hand sweeps $360^\circ$ in 60 minutes.
Angle swept in 10 minutes:
$$\theta = \frac{10}{60} \times 360^\circ = 60^\circ$$
Angle swept in 10 minutes:
$$\theta = \frac{10}{60} \times 360^\circ = 60^\circ$$
Area swept $= \text{Area of sector of radius } 7\text{ cm and angle } 60^\circ$:
$$A = \frac{60^\circ}{360^\circ} \times \frac{22}{7} \times 7^2 = \frac{77}{3} \approx \mathbf{25.67\text{ sq. cm}}$$
$$A = \frac{60^\circ}{360^\circ} \times \frac{22}{7} \times 7^2 = \frac{77}{3} \approx \mathbf{25.67\text{ sq. cm}}$$
Area swept = 25.67 sq. cm
Q4: Chord subtending 90°
A chord of a circle of radius $10\text{ cm}$ subtends $90^\circ$ at the centre. Find the area of the corresponding:
(i) minor sector (subtends $90^\circ$).
(ii) major sector (subtends $270^\circ$).
(Use $\pi \approx 3.14$)
(i) minor sector (subtends $90^\circ$).
(ii) major sector (subtends $270^\circ$).
(Use $\pi \approx 3.14$)
(i) Minor Sector Area ($\theta = 90^\circ$):
$$A_{minor} = \frac{90^\circ}{360^\circ} \times 3.14 \times 10^2 = 0.25 \times 314 = \mathbf{78.5\text{ sq. cm}}$$
$$A_{minor} = \frac{90^\circ}{360^\circ} \times 3.14 \times 10^2 = 0.25 \times 314 = \mathbf{78.5\text{ sq. cm}}$$
(ii) Major Sector Area ($\theta = 270^\circ$):
$$A_{major} = \frac{270^\circ}{360^\circ} \times 3.14 \times 10^2 = 0.75 \times 314 = \mathbf{235.5\text{ sq. cm}}$$
(Alternatively, $A_{major} = \text{Area of circle} - A_{minor} = 314 - 78.5 = 235.5\text{ sq. cm}$).
$$A_{major} = \frac{270^\circ}{360^\circ} \times 3.14 \times 10^2 = 0.75 \times 314 = \mathbf{235.5\text{ sq. cm}}$$
(Alternatively, $A_{major} = \text{Area of circle} - A_{minor} = 314 - 78.5 = 235.5\text{ sq. cm}$).
(i) 78.5 sq. cm (ii) 235.5 sq. cm
Q5: Minor & Major segments
A chord of a circle of radius $15\text{ cm}$ subtends an angle of $60^\circ$ at the centre of the circle. Find the areas of the corresponding minor and major segments of the circle. (Use $\pi \approx 3.14$ and $\sqrt{3} \approx 1.73$)
Step 1: Area of the corresponding sector:
$$A_{sector} = \frac{60^\circ}{360^\circ} \times \pi r^2 = \frac{1}{6} \times 3.14 \times 15^2 = \frac{706.5}{6} = 117.75\text{ sq. cm}$$
$$A_{sector} = \frac{60^\circ}{360^\circ} \times \pi r^2 = \frac{1}{6} \times 3.14 \times 15^2 = \frac{706.5}{6} = 117.75\text{ sq. cm}$$
Step 2: Area of the central triangle:
Since the angle at the center is $60^\circ$ and two sides are radii ($15\text{ cm}$), the triangle is equilateral.
$$\text{Area of Triangle} = \frac{\sqrt{3}}{4} r^2 = \frac{1.73}{4} \times 15^2 = \frac{1.73 \times 225}{4} = 97.3125\text{ sq. cm}$$
Since the angle at the center is $60^\circ$ and two sides are radii ($15\text{ cm}$), the triangle is equilateral.
$$\text{Area of Triangle} = \frac{\sqrt{3}}{4} r^2 = \frac{1.73}{4} \times 15^2 = \frac{1.73 \times 225}{4} = 97.3125\text{ sq. cm}$$
Step 3: Area of Minor Segment:
$$\text{Area of Minor Segment} = A_{sector} - \text{Area of Triangle}$$
$$\text{Area of Minor Segment} = 117.75 - 97.3125 = \mathbf{20.44\text{ sq. cm}}$$
$$\text{Area of Minor Segment} = A_{sector} - \text{Area of Triangle}$$
$$\text{Area of Minor Segment} = 117.75 - 97.3125 = \mathbf{20.44\text{ sq. cm}}$$
Step 4: Area of Major Segment:
$$\text{Area of Circle} = 3.14 \times 15^2 = 706.5\text{ sq. cm}$$
$$\text{Area of Major Segment} = \text{Area of Circle} - \text{Area of Minor Segment}$$
$$\text{Area of Major Segment} = 706.5 - 20.44 = \mathbf{686.06\text{ sq. cm}}$$
$$\text{Area of Circle} = 3.14 \times 15^2 = 706.5\text{ sq. cm}$$
$$\text{Area of Major Segment} = \text{Area of Circle} - \text{Area of Minor Segment}$$
$$\text{Area of Major Segment} = 706.5 - 20.44 = \mathbf{686.06\text{ sq. cm}}$$
Minor segment = 20.44 sq. cm Major segment = 686.06 sq. cm
Q6: Car wiper blades area
A car has two wipers which do not overlap. Each wiper has a blade of length $28\text{ cm}$ and sweeps through an angle of $120^\circ$. Find the total area cleaned at each sweep of the blades.
Each wiper sweeps a sector of radius $r = 28\text{ cm}$ and angle $\theta = 120^\circ$.
$$\text{Area of 1 sector} = \frac{120^\circ}{360^\circ} \times \pi r^2 = \frac{1}{3} \times \frac{22}{7} \times 28^2 = \frac{1}{3} \times 22 \times 4 \times 28 = \frac{2464}{3}\text{ sq. cm}$$
$$\text{Area of 1 sector} = \frac{120^\circ}{360^\circ} \times \pi r^2 = \frac{1}{3} \times \frac{22}{7} \times 28^2 = \frac{1}{3} \times 22 \times 4 \times 28 = \frac{2464}{3}\text{ sq. cm}$$
Since there are two wipers:
$$\text{Total Area} = 2 \times \frac{2464}{3} = \frac{4928}{3} \approx \mathbf{1642.67\text{ sq. cm}}$$
$$\text{Total Area} = 2 \times \frac{2464}{3} = \frac{4928}{3} \approx \mathbf{1642.67\text{ sq. cm}}$$
Total Area Cleaned = 1642.67 sq. cm
Q7: Minor segment general proof
A chord of a circle of radius $r$ subtends an angle of $60^\circ$ at the centre of the circle. Show that the area of the corresponding minor segment of the circle is equal to:
$$A = r^2\left(\frac{\pi}{6} - \frac{\sqrt{3}}{4}\right)$$
$$A = r^2\left(\frac{\pi}{6} - \frac{\sqrt{3}}{4}\right)$$
• **Area of Sector:**
$$\text{Area of Sector} = \frac{60^\circ}{360^\circ} \times \pi r^2 = \frac{1}{6}\pi r^2$$
$$\text{Area of Sector} = \frac{60^\circ}{360^\circ} \times \pi r^2 = \frac{1}{6}\pi r^2$$
• **Area of Equilateral Triangle:**
Since the central angle is $60^\circ$ and two sides are radii ($r$), the triangle is equilateral.
$$\text{Area of Triangle} = \frac{\sqrt{3}}{4} r^2$$
Since the central angle is $60^\circ$ and two sides are radii ($r$), the triangle is equilateral.
$$\text{Area of Triangle} = \frac{\sqrt{3}}{4} r^2$$
• **Area of Segment:**
$$\text{Area of Minor Segment} = \text{Area of Sector} - \text{Area of Triangle}$$
$$\text{Area of Minor Segment} = \frac{1}{6}\pi r^2 - \frac{\sqrt{3}}{4} r^2 = \mathbf{r^2\left(\frac{\pi}{6} - \frac{\sqrt{3}}{4}\right)}$$
$$\text{Area of Minor Segment} = \text{Area of Sector} - \text{Area of Triangle}$$
$$\text{Area of Minor Segment} = \frac{1}{6}\pi r^2 - \frac{\sqrt{3}}{4} r^2 = \mathbf{r^2\left(\frac{\pi}{6} - \frac{\sqrt{3}}{4}\right)}$$
Proved: A = r²(π/6 - √3/4)
Q8: Inscribed Equilateral Triangle ratio
An equilateral triangle is inscribed in a circle of radius $r$. Show that the ratio of the area of the triangle to the area of the circle is equal to $\frac{3\sqrt{3}}{4\pi} \approx 0.413$.
For an equilateral triangle inscribed in a circle of radius $r$, the side of the triangle $a$ is related to the radius by:
$$a = r\sqrt{3}$$
$$a = r\sqrt{3}$$
$$\text{Area of Equilateral Triangle} = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} (r\sqrt{3})^2 = \frac{3\sqrt{3}}{4} r^2$$
$$\text{Area of Circle} = \pi r^2$$
Ratio of areas:
$$\text{Ratio} = \frac{\frac{3\sqrt{3}}{4}r^2}{\pi r^2} = \mathbf{\frac{3\sqrt{3}}{4\pi} \approx 0.413}$$
$$\text{Ratio} = \frac{\frac{3\sqrt{3}}{4}r^2}{\pi r^2} = \mathbf{\frac{3\sqrt{3}}{4\pi} \approx 0.413}$$
Proved: Ratio = 3√3 / 4π (≈ 0.413)
Q9: Inscribed Square ratio
A square is inscribed in a circle of radius $r$. Show that the ratio of the area of the square to the area of the circle is equal to $\frac{2}{\pi} \approx 0.637$.
For a square inscribed in a circle of radius $r$, the diagonal of the square $d$ is equal to the diameter of the circle:
$$d = 2r$$
$$d = 2r$$
$$\text{Area of Square} = \frac{1}{2} d^2 = \frac{1}{2} (2r)^2 = 2r^2$$
$$\text{Area of Circle} = \pi r^2$$
Ratio of areas:
$$\text{Ratio} = \frac{2r^2}{\pi r^2} = \mathbf{\frac{2}{\pi} \approx 0.637}$$
$$\text{Ratio} = \frac{2r^2}{\pi r^2} = \mathbf{\frac{2}{\pi} \approx 0.637}$$
Proved: Ratio = 2 / π (≈ 0.637)
Q10: Inscribed Regular Hexagon ratio
A hexagon is inscribed in a circle of radius $r$. Show that the ratio of the area of the hexagon to the area of the circle is equal to $\frac{3\sqrt{3}}{2\pi} \approx 0.827$. Can you see why the answer is exactly twice the answer to Question 8?
A regular hexagon inscribed in a circle of radius $r$ can be divided into 6 congruent equilateral triangles, each with side length equal to the radius $r$.
$$\text{Area of Hexagon} = 6 \times \left(\frac{\sqrt{3}}{4} r^2\right) = \frac{3\sqrt{3}}{2} r^2$$
$$\text{Area of Hexagon} = 6 \times \left(\frac{\sqrt{3}}{4} r^2\right) = \frac{3\sqrt{3}}{2} r^2$$
$$\text{Area of Circle} = \pi r^2$$
Ratio of areas:
$$\text{Ratio} = \frac{\frac{3\sqrt{3}}{2}r^2}{\pi r^2} = \mathbf{\frac{3\sqrt{3}}{2\pi} \approx 0.827}$$
$$\text{Ratio} = \frac{\frac{3\sqrt{3}}{2}r^2}{\pi r^2} = \mathbf{\frac{3\sqrt{3}}{2\pi} \approx 0.827}$$
Relation to Question 8:
This is exactly twice the ratio of the inscribed equilateral triangle ($\frac{3\sqrt{3}}{4\pi}$). An inscribed equilateral triangle is formed by connecting every second vertex of the inscribed regular hexagon, occupying exactly half of the hexagon's area.
This is exactly twice the ratio of the inscribed equilateral triangle ($\frac{3\sqrt{3}}{4\pi}$). An inscribed equilateral triangle is formed by connecting every second vertex of the inscribed regular hexagon, occupying exactly half of the hexagon's area.
Proved: Ratio = 3√3 / 2π (≈ 0.827); exactly twice Q8's ratio.