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📐 Chapter 6: Measuring Space: Perimeter and Area • Exercise Set 6.1

Verified step-by-step solutions for your class notes.

Exercise Set 6.1 4Exercise Set 6.2 4Exercise Set 6.3 3End-of-Chapter Exercises 5All Exercises 16
Q1

The perimeter of a circle is 44 cm. What is its radius?

Answer: Let the radius be $r$.
Perimeter (Circumference) $= 2\pi r = 44$.
Using $\pi \approx \frac{22}{7}$:
$2 \times \frac{22}{7} \times r = 44$
$\frac{44}{7} \times r = 44$
$r = 7$ cm.

Q2

Calculate, correct to 3 significant figures, the circumference of a circle with: (i) radius 7 cm (ii) radius 10 cm (iii) radius 12 cm.

Answer: Circumference $C = 2\pi r$. Using $\pi \approx \frac{22}{7}$:
(i) $C = 2 \times \frac{22}{7} \times 7 = 44$ cm $\rightarrow$ 44.0 cm.
(ii) $C = 2 \times \frac{22}{7} \times 10 = \frac{440}{7} \approx 62.857$ $\rightarrow$ 62.9 cm.
(iii) $C = 2 \times \frac{22}{7} \times 12 = \frac{528}{7} \approx 75.428$ $\rightarrow$ 75.4 cm.

Q3

Calculate the length of the arc of a circle if: (i) the radius is 3.5 cm and the angle at the centre is $60^\circ$, and (ii) the radius is 6.3 m and the angle at the centre is $120^\circ$.

Answer: Arc length $l = 2\pi r \times \frac{\theta}{360}$.
(i) $l = 2 \times \frac{22}{7} \times 3.5 \times \frac{60}{360} = 2 \times \frac{22}{7} \times \frac{7}{2} \times \frac{1}{6} = 22 \times \frac{1}{6} = \frac{11}{3} \approx$ 3.67 cm.
(ii) $l = 2 \times \frac{22}{7} \times 6.3 \times \frac{120}{360} = 2 \times \frac{22}{7} \times \frac{63}{10} \times \frac{1}{3} = 2 \times 22 \times \frac{9}{10} \times \frac{1}{3} = 2 \times 22 \times \frac{3}{10} = \frac{132}{10} =$ 13.2 m.

Q4

Find the perimeter of a sector (i.e., the curved portion as well as the two straight portions) of a circle of radius 14 cm and sector angle $75^\circ$.

Answer: Perimeter of sector = Arc length + 2(Radius).
Arc length $l = 2\pi r \times \frac{\theta}{360} = 2 \times \frac{22}{7} \times 14 \times \frac{75}{360}$.
$l = 2 \times 22 \times 2 \times \frac{5}{24} = 88 \times \frac{5}{24} = \frac{11 \times 5}{3} = \frac{55}{3} \approx 18.33$ cm.
Perimeter $= \frac{55}{3} + 14 + 14 = 18.33 + 28 = $ 46.33 cm.

75° r = 14 cm 14 cm Arc ≈ 18.33 cm
Perimeter of sector = Arc length (18.33 cm) + 2r (28 cm) = 46.33 cm