Exercise 6.1 Practice
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Overview
This page provides comprehensive Ch 6: Measuring Space: Perimeter and Area - Exercise Set 6.1 Practice. Solve problems on circle perimeters, sector arc lengths, compound shapes, tyre rotation mechanics, and petal flower boundaries with step-by-step solutions.
Circle Circumference, Sector Arc Length & Petal Flower Perimeters (Use $\pi \approx 22/7$)
Q1: Radius from Perimeter
The perimeter of a circle is $44\text{ cm}$. What is its radius?
The perimeter (circumference) of a circle is $C = 2\pi r$.
$$2\pi r = 44$$
$$2 \times \frac{22}{7} \times r = 44$$
$$\frac{44}{7} \times r = 44 \implies r = \mathbf{7\text{ cm}}$$
$$2\pi r = 44$$
$$2 \times \frac{22}{7} \times r = 44$$
$$\frac{44}{7} \times r = 44 \implies r = \mathbf{7\text{ cm}}$$
Radius = 7 cm
Q2: Circumference Calculations
Calculate, correct to 3 significant figures, the circumference of a circle with:
(i) radius $7\text{ cm}$
(ii) radius $10\text{ cm}$
(iii) radius $12\text{ cm}$.
(i) radius $7\text{ cm}$
(ii) radius $10\text{ cm}$
(iii) radius $12\text{ cm}$.
(i) Radius $7\text{ cm}$:
$$C = 2\pi r = 2 \times \frac{22}{7} \times 7 = \mathbf{44.0\text{ cm}}$$
$$C = 2\pi r = 2 \times \frac{22}{7} \times 7 = \mathbf{44.0\text{ cm}}$$
(ii) Radius $10\text{ cm}$:
$$C = 2\pi r = 2 \times 3.1416 \times 10 = \mathbf{62.8\text{ cm}}$$
$$C = 2\pi r = 2 \times 3.1416 \times 10 = \mathbf{62.8\text{ cm}}$$
(iii) Radius $12\text{ cm}$:
$$C = 2\pi r = 2 \times 3.1416 \times 12 = \mathbf{75.4\text{ cm}}$$
$$C = 2\pi r = 2 \times 3.1416 \times 12 = \mathbf{75.4\text{ cm}}$$
(i) 44.0 cm (ii) 62.8 cm (iii) 75.4 cm
Q3: Sector Arc Lengths
Calculate the length of the arc of a circle if:
(i) the radius is $3.5\text{ cm}$ and the angle at the centre is $60^\circ$.
(ii) the radius is $6.3\text{ m}$ and the angle at the centre is $120^\circ$.
(i) the radius is $3.5\text{ cm}$ and the angle at the centre is $60^\circ$.
(ii) the radius is $6.3\text{ m}$ and the angle at the centre is $120^\circ$.
The length of an arc subtending angle $\theta$ is $L = \frac{\theta}{360^\circ} \times 2\pi r$.
(i) $r = 3.5\text{ cm}, \theta = 60^\circ$:
$$L = \frac{60}{360} \times 2 \times \frac{22}{7} \times 3.5 = \frac{1}{6} \times 22 = \mathbf{3.67\text{ cm}} \quad \text{(to 3 sig figs)}$$
$$L = \frac{60}{360} \times 2 \times \frac{22}{7} \times 3.5 = \frac{1}{6} \times 22 = \mathbf{3.67\text{ cm}} \quad \text{(to 3 sig figs)}$$
(ii) $r = 6.3\text{ m}, \theta = 120^\circ$:
$$L = \frac{120}{360} \times 2 \times \frac{22}{7} \times 6.3 = \frac{1}{3} \times 44 \times 0.9 = 44 \times 0.3 = \mathbf{13.2\text{ m}}$$
$$L = \frac{120}{360} \times 2 \times \frac{22}{7} \times 6.3 = \frac{1}{3} \times 44 \times 0.9 = 44 \times 0.3 = \mathbf{13.2\text{ m}}$$
(i) 3.67 cm (ii) 13.2 m
Q4: Sector Perimeter
Find the perimeter of a sector of a circle of radius $14\text{ cm}$ and sector angle $75^\circ$.
The perimeter of a sector consists of the arc length plus two radii: $P = L + 2r$.
$$\text{Arc Length } L = \frac{75^\circ}{360^\circ} \times 2\pi r = \frac{5}{24} \times 2 \times \frac{22}{7} \times 14 = \frac{5}{24} \times 88 = \frac{55}{3} \approx 18.33\text{ cm}$$
$$\text{Arc Length } L = \frac{75^\circ}{360^\circ} \times 2\pi r = \frac{5}{24} \times 2 \times \frac{22}{7} \times 14 = \frac{5}{24} \times 88 = \frac{55}{3} \approx 18.33\text{ cm}$$
$$\text{Perimeter } P = 18.33 + 2(14) = 18.33 + 28 = \mathbf{46.33\text{ cm}}$$
Perimeter = 46.33 cm (or 46 1/3 cm)
Q5: Compound Shapes Perimeters
Find the perimeters of the following shapes in Fig. 6.14 (taking arcs to be quarters, halves, or three-quarters of circles):
• (i) Stadium capsule: Rectangle $80\text{ m} \times 60\text{ m}$ with semicircle ends.
• (ii) Concentric semicircle tunnel: Outer diameter $12\text{ cm}$, inner diameter $8\text{ cm}$.
• (iii) Clover: Square of side $10\text{ cm}$ with 4 outer semicircles.
• (iv) 3-Lobed flower: Equilateral triangle of side $12\text{ cm}$ with 3 outer semicircles.
• (v) Clover-14: Square of side $14\text{ cm}$ with 4 outer semicircles.
• (vi) Wavy outline: Large semicircle diameter $28\text{ cm}$, and two small semicircles of diameter $14\text{ cm}$ along the base.
• (vii) Right triangle boundary: Semicircles on sides $6\text{ cm}$ and $8\text{ cm}$, and hypotenuse.
• (viii) Arch bridge: Semicircle diameter $12\text{ cm}$ on top, three small semicircles diameter $4\text{ cm}$ under.
• (ix) Yin-Yang loop: Large semicircle diameter $20\text{ cm}$, and two small semicircles diameter $10\text{ cm}$ inside.
(i) Stadium Capsule:
• 2 straight edges $= 80 + 80 = 160\text{ m}$.
• 2 semicircle ends of diameter 60m (radius 30m) form 1 complete circle: $2\pi r = 2 \times \frac{22}{7} \times 30 = 188.57\text{ m}$.
• Total Perimeter $= 160 + 188.57 = \mathbf{348.57\text{ m}}$.
• 2 straight edges $= 80 + 80 = 160\text{ m}$.
• 2 semicircle ends of diameter 60m (radius 30m) form 1 complete circle: $2\pi r = 2 \times \frac{22}{7} \times 30 = 188.57\text{ m}$.
• Total Perimeter $= 160 + 188.57 = \mathbf{348.57\text{ m}}$.
(ii) Concentric Tunnel:
• Outer semicircle arc (radius 6): $\pi \times 6 = 18.86\text{ cm}$.
• Inner semicircle arc (radius 4): $\pi \times 4 = 12.57\text{ cm}$.
• 2 flat bottom connection steps: $2 \times \frac{12-8}{2} = 4\text{ cm}$.
• Total Perimeter $= 18.86 + 12.57 + 4 = \mathbf{35.43\text{ cm}}$.
• Outer semicircle arc (radius 6): $\pi \times 6 = 18.86\text{ cm}$.
• Inner semicircle arc (radius 4): $\pi \times 4 = 12.57\text{ cm}$.
• 2 flat bottom connection steps: $2 \times \frac{12-8}{2} = 4\text{ cm}$.
• Total Perimeter $= 18.86 + 12.57 + 4 = \mathbf{35.43\text{ cm}}$.
(iii) Clover (side 10):
• 4 semicircles of diameter 10cm form 2 complete circles of radius 5cm:
• Total Perimeter $= 2 \times (2\pi \times 5) = 20\pi \approx 20 \times \frac{22}{7} = \mathbf{62.86\text{ cm}}$.
• 4 semicircles of diameter 10cm form 2 complete circles of radius 5cm:
• Total Perimeter $= 2 \times (2\pi \times 5) = 20\pi \approx 20 \times \frac{22}{7} = \mathbf{62.86\text{ cm}}$.
(iv) 3-Lobed flower (side 12):
• 3 semicircles of diameter 12cm form 1.5 complete circles of radius 6cm:
• Total Perimeter $= 1.5 \times (2\pi \times 6) = 18\pi \approx 18 \times \frac{22}{7} = \mathbf{56.57\text{ cm}}$.
• 3 semicircles of diameter 12cm form 1.5 complete circles of radius 6cm:
• Total Perimeter $= 1.5 \times (2\pi \times 6) = 18\pi \approx 18 \times \frac{22}{7} = \mathbf{56.57\text{ cm}}$.
(v) Clover-14 (side 14):
• 4 semicircles of diameter 14cm form 2 complete circles of radius 7cm:
• Total Perimeter $= 2 \times (2\pi \times 7) = 28\pi \approx 28 \times \frac{22}{7} = \mathbf{88.0\text{ cm}}$.
• 4 semicircles of diameter 14cm form 2 complete circles of radius 7cm:
• Total Perimeter $= 2 \times (2\pi \times 7) = 28\pi \approx 28 \times \frac{22}{7} = \mathbf{88.0\text{ cm}}$.
(vi) Wavy Outline:
• Large semicircle arc (diameter 28): $\frac{1}{2} \pi d = 14\pi$.
• 2 small semicircle arcs (diameter 14): $2 \times (\frac{1}{2} \pi \times 14) = 14\pi$.
• Total Perimeter $= 14\pi + 14\pi = 28\pi \approx 28 \times \frac{22}{7} = \mathbf{88.0\text{ cm}}$.
• Large semicircle arc (diameter 28): $\frac{1}{2} \pi d = 14\pi$.
• 2 small semicircle arcs (diameter 14): $2 \times (\frac{1}{2} \pi \times 14) = 14\pi$.
• Total Perimeter $= 14\pi + 14\pi = 28\pi \approx 28 \times \frac{22}{7} = \mathbf{88.0\text{ cm}}$.
(vii) Right Triangle boundary:
• Semicircles on sides 6cm, 8cm and hypotenuse 10cm:
• Total Perimeter $= \frac{1}{2}\pi(6) + \frac{1}{2}\pi(8) + \frac{1}{2}\pi(10) = 12\pi \approx 12 \times \frac{22}{7} = \mathbf{37.71\text{ cm}}$.
• Semicircles on sides 6cm, 8cm and hypotenuse 10cm:
• Total Perimeter $= \frac{1}{2}\pi(6) + \frac{1}{2}\pi(8) + \frac{1}{2}\pi(10) = 12\pi \approx 12 \times \frac{22}{7} = \mathbf{37.71\text{ cm}}$.
(viii) Arch Bridge:
• Top large semicircle (diameter 12): $\frac{1}{2}\pi(12) = 6\pi$.
• Bottom three semicircles (diameter 4): $3 \times \frac{1}{2}\pi(4) = 6\pi$.
• Total Perimeter $= 6\pi + 6\pi = 12\pi \approx \mathbf{37.71\text{ cm}}$.
• Top large semicircle (diameter 12): $\frac{1}{2}\pi(12) = 6\pi$.
• Bottom three semicircles (diameter 4): $3 \times \frac{1}{2}\pi(4) = 6\pi$.
• Total Perimeter $= 6\pi + 6\pi = 12\pi \approx \mathbf{37.71\text{ cm}}$.
(ix) Yin-Yang Loop:
• Outer large semicircle (diameter 20): $\frac{1}{2}\pi(20) = 10\pi$.
• Inner two semicircles (diameter 10): $2 \times \frac{1}{2}\pi(10) = 10\pi$.
• Total Perimeter $= 10\pi + 10\pi = 20\pi \approx \mathbf{62.86\text{ cm}}$.
• Outer large semicircle (diameter 20): $\frac{1}{2}\pi(20) = 10\pi$.
• Inner two semicircles (diameter 10): $2 \times \frac{1}{2}\pi(10) = 10\pi$.
• Total Perimeter $= 10\pi + 10\pi = 20\pi \approx \mathbf{62.86\text{ cm}}$.
(i) 348.57 m (ii) 35.43 cm (iii) 62.86 cm (iv) 56.57 cm (v) 88.0 cm (vi) 88.0 cm (vii) 37.71 cm (viii) 37.71 cm (ix) 62.86 cm
Q6: Car Tyre Rotations
If the diameter of a car tyre is $56\text{ cm}$, then:
(i) How far does the car need to travel for the tyre to complete one revolution?
(ii) How many revolutions does the tyre make if the car travels $10\text{ km}$?
(i) How far does the car need to travel for the tyre to complete one revolution?
(ii) How many revolutions does the tyre make if the car travels $10\text{ km}$?
(i) Distance per revolution:
Distance $= \text{Circumference} = \pi d = \frac{22}{7} \times 56 = \mathbf{176\text{ cm} = 1.76\text{ m}}$.
Distance $= \text{Circumference} = \pi d = \frac{22}{7} \times 56 = \mathbf{176\text{ cm} = 1.76\text{ m}}$.
(ii) Revolutions in 10 km:
$$10\text{ km} = 10 \times 1000\text{ m} = 10,000\text{ m}$$
$$\text{Revolutions} = \frac{\text{Total Distance}}{\text{Circumference}} = \frac{10,000\text{ m}}{1.76\text{ m}} \approx \mathbf{5682\text{ revolutions}} \quad \text{(or 5681.82)}$$
$$10\text{ km} = 10 \times 1000\text{ m} = 10,000\text{ m}$$
$$\text{Revolutions} = \frac{\text{Total Distance}}{\text{Circumference}} = \frac{10,000\text{ m}}{1.76\text{ m}} \approx \mathbf{5682\text{ revolutions}} \quad \text{(or 5681.82)}$$
(i) 176 cm (1.76 m) (ii) 5682 revolutions
Q7: Petal Flowers
Find the total perimeter of all the petals in each of the given flowers:
• **Fig. 6.15A:** Square side $14\text{ cm}$, centers of arcs are midpoints of the sides.
• **Fig. 6.15B:** Hexagon side $42\text{ cm}$, centers of arcs are vertices of the hexagon.
• **Fig. 6.15A:** Square side $14\text{ cm}$, centers of arcs are midpoints of the sides.
• **Fig. 6.15B:** Hexagon side $42\text{ cm}$, centers of arcs are vertices of the hexagon.
Fig. 6.15A (4-Petalled square flower):
The square has side $14\text{ cm}$. The arcs go from midpoints to midpoints. The radius of each arc is $7\text{ cm}$ (quarter circle).
The 4 petals are formed by 8 quarter-circle arcs:
$$\text{Total Perimeter} = 8 \times \left(\frac{1}{4} \times 2\pi r\right) = 4\pi r = 4 \times \frac{22}{7} \times 7 = \mathbf{88\text{ cm}}$$
The square has side $14\text{ cm}$. The arcs go from midpoints to midpoints. The radius of each arc is $7\text{ cm}$ (quarter circle).
The 4 petals are formed by 8 quarter-circle arcs:
$$\text{Total Perimeter} = 8 \times \left(\frac{1}{4} \times 2\pi r\right) = 4\pi r = 4 \times \frac{22}{7} \times 7 = \mathbf{88\text{ cm}}$$
Fig. 6.15B (6-Petalled hexagonal flower):
The hexagon has side $R = 42\text{ cm}$. The 6 petals are formed by 12 arcs meeting at the center. Each arc is a $60^\circ$ sector arc (since regular hexagon vertex angle is partitioned or center angle is $60^\circ$).
Radius $r = 42\text{ cm}$.
$$\text{Total Perimeter} = 12 \times \left(\frac{60^\circ}{360^\circ} \times 2\pi r\right) = 12 \times \frac{1}{6} \times 2\pi r = 4\pi r = 4 \times \frac{22}{7} \times 42 = \mathbf{528\text{ cm}}$$
The hexagon has side $R = 42\text{ cm}$. The 6 petals are formed by 12 arcs meeting at the center. Each arc is a $60^\circ$ sector arc (since regular hexagon vertex angle is partitioned or center angle is $60^\circ$).
Radius $r = 42\text{ cm}$.
$$\text{Total Perimeter} = 12 \times \left(\frac{60^\circ}{360^\circ} \times 2\pi r\right) = 12 \times \frac{1}{6} \times 2\pi r = 4\pi r = 4 \times \frac{22}{7} \times 42 = \mathbf{528\text{ cm}}$$
Fig. 6.15A = 88 cm Fig. 6.15B = 528 cm
Q8: Radii Ratio from Perimeters
The ratio of the perimeters of two circles is $5:4$. What is the ratio of their radii?
Let perimeters be $C_1, C_2$ and radii be $r_1, r_2$.
$$\frac{C_1}{C_2} = \frac{2\pi r_1}{2\pi r_2} = \frac{r_1}{r_2}$$
Since $\frac{C_1}{C_2} = \frac{5}{4}$, we have:
$$\frac{r_1}{r_2} = \frac{5}{4}$$
$$\frac{C_1}{C_2} = \frac{2\pi r_1}{2\pi r_2} = \frac{r_1}{r_2}$$
Since $\frac{C_1}{C_2} = \frac{5}{4}$, we have:
$$\frac{r_1}{r_2} = \frac{5}{4}$$
Ratio of radii = 5 : 4