Unless stated otherwise, use the approximation for . 5. Find the perimeters of the following shapes (taking the arcs to be quarter or half or three-quarters of a circle, as appropriate) (Fig. 6.14i to 6.14ix):

Understand the Question Perimeter of any composite geometric figure is the total length around its outer boundary, obtained by adding the lengths of all its curved arcs and straight boundary edges. Useful arc length formulas (using ): Semicircular arc: Quarter-circular arc: Part (i) Find the perimeter of the shape in Fig. 6.14i. Step 1 · Calculate Perimeter of Shape (i) The shape consists of two straight sides of length each and two semicircular ends with diameter . Answer > (i) Part (ii) Find the perimeter of the shape in Fig. 6.14ii. Step 1 · Calculate Perimeter of Shape (ii) Outer radius , inner radius , and two straight connecting segments of length . Answer > (ii) Part (iii) Find the perimeter of the shape in Fig. 6.14iii. Step 1 · Calculate Perimeter of Shape (iii) The boundary has identical semicircles, each of diameter (radius ). Answer > (iii) Part (iv) Find the perimeter of the shape in Fig. 6.14iv. Step 1 · Calculate Perimeter of Shape (iv) The boundary consists of identical semicircles, each of diameter (radius ). Answer > (iv) Part (v) Find the perimeter of the shape in Fig. 6.14v. Step 1 · Calculate Perimeter of Shape (v) The shape has small semicircles of radius and large quarter circles of radius . Answer > (v) Part (vi) Find the perimeter of the shape in Fig. 6.14vi. Step 1 · Calculate Perimeter of Shape (vi) Large semicircle radius (diameter ); small lower semicircles with diameter (radius ). Answer > (vi) Part (vii) Find the perimeter of the shape in Fig. 6.14vii. Step 1 · Calculate Perimeter of Shape (vii) First, find the hypotenuse using the Pythagoras theorem The diameters are , , and , giving radii , , and . Answer > (vii) Part (viii) Find the perimeter of the shape in Fig. 6.14viii. Step 1 · Calculate Perimeter of Shape (viii) Large semicircle radius ; small semicircles each with diameter (radius ). Answer > (viii) Part (ix) Find the perimeter of the shape in Fig. 6.14ix. Step 1 · Calculate Perimeter of Shape (ix) Large semicircle radius ; inner semicircles each with diameter (radius ). Answer > (ix) Common Mistakes Diameter vs. Radius Confusion: Using diameter directly in the formula (or radius in ). The arc length of a semicircle is . Omitting Straight Edges: In shapes like (i) and (ii), forgetting to add the straight boundary segments to the curved arc lengths. Internal Boundaries: Adding inner division lines that do not form part of the outer perimeter.

Measuring Space: Perimeter and Area | Exercise 6.1

Question 5

Unless stated otherwise, use the approximation $\dfrac{22}{7}$ for $\pi$ .

Find the perimeters of the following shapes (taking the arcs to be quarter or half or three-quarters of a circle, as appropriate) (Fig. 6.14i to 6.14ix):

Solution

Understand the Question

Perimeter of any composite geometric figure is the total length around its outer boundary, obtained by adding the lengths of all its curved arcs and straight boundary edges.
Useful arc length formulas (using $\pi = \dfrac{22}{7}$ $π = \frac{22}{7}$ ):
- Semicircular arc: $\text{Length} = \pi r = \dfrac{1}{2}\pi d$
- Quarter-circular arc: $\text{Length} = \dfrac{1}{4}(2\pi r) = \dfrac{1}{2}\pi r$

(i) Find the perimeter of the shape in Fig. 6.14i.

Step 1 · Calculate Perimeter of Shape (i)

Diagram 1

The shape consists of two straight sides of length $80\text{ m}$ each and two semicircular ends with diameter $d = 60\text{ m}$ .

\begin{aligned} \text{Perimeter} &= 2 \times \text{straight length} + \pi \times \text{diameter} \\[0.6em] &= 2 \times 80 + \dfrac{22}{7} \times 60 \\[0.6em] &= 160 + \dfrac{1320}{7} \\[0.6em] &= \dfrac{1120 + 1320}{7} \\[0.6em] &= \dfrac{2440}{7} \approx 348.57\text{ m} \end{aligned}

Answer

(i) $348.57\text{ m}$

(ii) Find the perimeter of the shape in Fig. 6.14ii.

Step 1 · Calculate Perimeter of Shape (ii)

Diagram 2

Outer radius $R = \dfrac{12}{2} = 6\text{ cm}$ , inner radius $r = \dfrac{8}{2} = 4\text{ cm}$ , and two straight connecting segments of length $6 - 4 = 2\text{ cm}$ .

\begin{aligned} \text{Perimeter} &= \pi \times \text{outer radius} + \pi \times \text{inner radius} + 2 \times \text{straight part} \\[0.6em] &= \pi \times 6 + \pi \times 4 + 2 \times 2 \\[0.6em] &= 10 \pi + 4 \\[0.6em] &= 10 \times \dfrac{22}{7} + 4 \\[0.6em] &= \dfrac{220}{7} + \dfrac{28}{7} \\[0.6em] &= \dfrac{248}{7} \approx 35.43\text{ cm} \end{aligned}

Answer

(ii) $35.43\text{ cm}$

(iii) Find the perimeter of the shape in Fig. 6.14iii.

Step 1 · Calculate Perimeter of Shape (iii)

Diagram 3

The boundary has $4$ identical semicircles, each of diameter $d = 10\text{ cm}$ (radius $r = 5\text{ cm}$ ).

\begin{aligned} \text{Perimeter} &= 4 \times (\pi \times \text{radius}) \\[0.6em] &= 4 \times \pi \times 5 \\[0.6em] &= 20 \pi \\[0.6em] &= 20 \times \dfrac{22}{7} \\[0.6em] &= \dfrac{440}{7} \approx 62.86\text{ cm} \end{aligned}

Answer

(iii) $62.86\text{ cm}$

(iv) Find the perimeter of the shape in Fig. 6.14iv.

Step 1 · Calculate Perimeter of Shape (iv)

Diagram 4

The boundary consists of $3$ identical semicircles, each of diameter $d = 12\text{ cm}$ (radius $r = 6\text{ cm}$ ).

\begin{aligned} \text{Perimeter} &= 3 \times (\pi \times \text{radius}) \\[0.6em] &= 3 \times \pi \times 6 \\[0.6em] &= 18 \pi \\[0.6em] &= 18 \times \dfrac{22}{7} \\[0.6em] &= \dfrac{396}{7} \approx 56.57\text{ cm} \end{aligned}

Answer

(iv) $56.57\text{ cm}$

(v) Find the perimeter of the shape in Fig. 6.14v.

Step 1 · Calculate Perimeter of Shape (v)

Diagram 5

The shape has $4$ small semicircles of radius $r = 7\text{ cm}$ and $4$ large quarter circles of radius $R = 14\text{ cm}$ .

\begin{aligned} \text{Perimeter} &= 4 \times (\pi \times \text{small radius}) + 4 \times \left(\dfrac{1}{2} \pi \times \text{large radius}\right) \\[0.6em] &= 4 \times \pi \times 7 + 4 \times \dfrac{1}{2} \times \pi \times 14 \\[0.6em] &= 28 \pi + 28 \pi \\[0.6em] &= 56 \pi \\[0.6em] &= 56 \times \dfrac{22}{7} \\[0.6em] &= 8 \times 22 = 176\text{ cm} \end{aligned}

Answer

(v) $176\text{ cm}$

(vi) Find the perimeter of the shape in Fig. 6.14vi.

Step 1 · Calculate Perimeter of Shape (vi)

Diagram 6

Large semicircle radius $R = 14\text{ cm}$ (diameter $28\text{ cm}$ ); $4$ small lower semicircles with diameter $d = \dfrac{28}{4} = 7\text{ cm}$ (radius $r = 3.5\text{ cm}$ ).

\begin{aligned} \text{Perimeter} &= \pi \times \text{large radius} + 4 \times (\pi \times \text{small radius}) \\[0.6em] &= \pi \times 14 + 4 \times \pi \times 3.5 \\[0.6em] &= 14 \pi + 14 \pi \\[0.6em] &= 28 \pi \\[0.6em] &= 28 \times \dfrac{22}{7} \\[0.6em] &= 4 \times 22 = 88\text{ cm} \end{aligned}

Answer

(vi) $88\text{ cm}$

(vii) Find the perimeter of the shape in Fig. 6.14vii.

Step 1 · Calculate Perimeter of Shape (vii)

Diagram 7

First, find the hypotenuse using the Pythagoras theorem

\begin{aligned} \text{Hypotenuse}^2 &= 6^2 + 8^2 \\[0.6em] &= 36 + 64 \\[0.6em] &= 100 \\[0.6em] \text{Hypotenuse} &= \sqrt{100} = 10\text{ cm} \end{aligned}

The diameters are $6\text{ cm}$ , $8\text{ cm}$ , and $10\text{ cm}$ , giving radii $3\text{ cm}$ , $4\text{ cm}$ , and $5\text{ cm}$ .

\begin{aligned} \text{Perimeter} &= \pi \times 3 + \pi \times 4 + \pi \times 5 \\[0.6em] &= \pi \times (3 + 4 + 5) \\[0.6em] &= 12 \pi \\[0.6em] &= 12 \times \dfrac{22}{7} \\[0.6em] &= \dfrac{264}{7} \approx 37.71\text{ cm} \end{aligned}

Answer

(vii) $37.71\text{ cm}$

(viii) Find the perimeter of the shape in Fig. 6.14viii.

Step 1 · Calculate Perimeter of Shape (viii)

Diagram 8

Large semicircle radius $R = \dfrac{4 + 4 + 4}{2} = 6\text{ cm}$ ; $3$ small semicircles each with diameter $4\text{ cm}$ (radius $r = 2\text{ cm}$ ).

\begin{aligned} \text{Perimeter} &= \pi \times \text{large radius} + 3 \times (\pi \times \text{small radius}) \\[0.6em] &= \pi \times 6 + 3 \times \pi \times 2 \\[0.6em] &= 6 \pi + 6 \pi \\[0.6em] &= 12 \pi \\[0.6em] &= 12 \times \dfrac{22}{7} \\[0.6em] &= \dfrac{264}{7} \approx 37.71\text{ cm} \end{aligned}

Answer

(viii) $37.71\text{ cm}$

(ix) Find the perimeter of the shape in Fig. 6.14ix.

Step 1 · Calculate Perimeter of Shape (ix)

Diagram 9

Large semicircle radius $R = \dfrac{10 + 10}{2} = 10\text{ cm}$ ; $2$ inner semicircles each with diameter $10\text{ cm}$ (radius $r = 5\text{ cm}$ ).

\begin{aligned} \text{Perimeter} &= \pi \times \text{large radius} + 2 \times (\pi \times \text{small radius}) \\[0.6em] &= \pi \times 10 + 2 \times \pi \times 5 \\[0.6em] &= 10 \pi + 10 \pi \\[0.6em] &= 20 \pi \\[0.6em] &= 20 \times \dfrac{22}{7} \\[0.6em] &= \dfrac{440}{7} \approx 62.86\text{ cm} \end{aligned}

Answer

(ix) $62.86\text{ cm}$

Common Mistakes

Diameter vs. Radius Confusion: Using diameter $d$ directly in the formula $\pi r$ (or radius in $\pi d$ ). The arc length of a semicircle is $\pi r = \dfrac{1}{2}\pi d$ .
Omitting Straight Edges: In shapes like (i) and (ii), forgetting to add the straight boundary segments to the curved arc lengths.
Internal Boundaries: Adding inner division lines that do not form part of the outer perimeter.

Question 5

More questions in Exercise 6.1