NCERT Solutions for Class 9 Maths Chapter 6 — Exercise Set 6.1#

The perimeter (circumference) is $2\pi r$, so:

$$2\pi r = 44$$

$$\Rightarrow 2 \times \dfrac{22}{7} \times r = 44$$

$$\Rightarrow \dfrac{44}{7}r = 44$$

$$\Rightarrow r = 7$$

∴ The radius is 7 cm.

$$C = 2 \times \dfrac{22}{7} \times 7$$

$$C = 44$$

∴ Circumference $=$ 44.0 cm.

$$C = 2 \times \dfrac{22}{7} \times 10$$

$$C = \dfrac{440}{7}$$

$$C = 62.857\ldots$$

∴ Circumference $\approx$ 62.9 cm.

$$C = 2 \times \dfrac{22}{7} \times 12$$

$$C = \dfrac{528}{7}$$

$$C = 75.428\ldots$$

∴ Circumference $\approx$ 75.4 cm.

We find the arc length:

$$= \dfrac{60}{360} \times 2 \times \dfrac{22}{7} \times 3.5$$

$$= \dfrac{1}{6} \times 22$$

$$= \dfrac{11}{3}$$

∴ The arc is $\dfrac{11}{3}$ cm ≈ 3.67 cm long.

We find the arc length:

$$= \dfrac{120}{360} \times 2 \times \dfrac{22}{7} \times 6.3$$

$$= \dfrac{1}{3} \times 39.6$$

$$= 13.2$$

∴ The arc is 13.2 m long.

We find the curved part (the arc):

$$= \dfrac{75}{360} \times 2 \times \dfrac{22}{7} \times 14$$

$$= \dfrac{5}{24} \times 88$$

$$= \dfrac{55}{3} \text{ cm}$$

The two straight parts are radii, $14 + 14 = 28$ cm. So:

$$\text{Perimeter} = \dfrac{55}{3} + 28$$

$$\text{Perimeter} = \dfrac{139}{3}$$

∴ The perimeter of the sector is $\dfrac{139}{3}$ cm ≈ 46.33 cm.

The boundary is two straight sides of 80 m and two semicircles of diameter 60 m, which together make one full circle of diameter 60 m:

$$\text{Perimeter} = 2 \times 80 + \pi \times 60$$

$$\text{Perimeter} = 160 + \dfrac{22}{7} \times 60$$

$$\text{Perimeter} = 160 + \dfrac{1320}{7}$$

$$\text{Perimeter} = \dfrac{2440}{7}$$

∴ Perimeter ≈ 348.57 m.

The boundary is the outer semicircle (diameter 12), the inner semicircle (diameter 8) and two flat ends, each $\dfrac{12 - 8}{2} = 2$ cm:

$$\text{Perimeter} = \dfrac{1}{2}\pi(12) + \dfrac{1}{2}\pi(8) + 2 + 2$$

$$\text{Perimeter} = 10\pi + 4$$

$$\text{Perimeter} = \dfrac{220}{7} + 4$$

$$\text{Perimeter} = \dfrac{248}{7}$$

∴ Perimeter ≈ 35.43 cm.

The boundary is four semicircles of diameter 10 cm (two full circles):

$$\text{Perimeter} = 4 \times \dfrac{1}{2}\pi(10)$$

$$\text{Perimeter} = 20\pi$$

$$\text{Perimeter} = \dfrac{440}{7}$$

∴ Perimeter ≈ 62.86 cm.

The boundary is three semicircles of diameter 12 cm:

$$\text{Perimeter} = 3 \times \dfrac{1}{2}\pi(12)$$

$$\text{Perimeter} = 18\pi$$

$$\text{Perimeter} = \dfrac{396}{7}$$

∴ Perimeter ≈ 56.57 cm.

The figure is a $3 \times 3$ grid of squares of side 14 cm, with the four corner squares left out (a plus shape). The boundary is:

• four semicircles of diameter 14 cm, one on the outer end of each arm — together two full circles of diameter 14;
• four quarter circles of radius 14 cm, one at each corner between two arms — together one full circle of radius 14.

$$\text{Perimeter} = 2 \times \pi(14) + 2\pi(14)$$

$$\text{Perimeter} = 56\pi$$

$$\text{Perimeter} = 56 \times \dfrac{22}{7}$$

$$\text{Perimeter} = 176$$

∴ Perimeter $=$ 176 cm.

The base (28 cm) is split into four parts of 7 cm, and each part carries a small semicircle of diameter 7 cm (two below, two above). The top is a semicircle of diameter 28 cm:

$$\text{Perimeter} = \dfrac{1}{2}\pi(28) + 4 \times \dfrac{1}{2}\pi(7)$$

$$\text{Perimeter} = 14\pi + 14\pi$$

$$\text{Perimeter} = 28\pi$$

$$\text{Perimeter} = 88$$

∴ Perimeter $=$ 88 cm.

The hypotenuse is:

$$\sqrt{6^2 + 8^2} = \sqrt{100} = 10 \text{ cm}$$

The boundary is three semicircles of diameters 6, 8 and 10 cm:

$$\text{Perimeter} = \dfrac{1}{2}\pi(6 + 8 + 10)$$

$$\text{Perimeter} = 12\pi$$

$$\text{Perimeter} = \dfrac{264}{7}$$

∴ Perimeter ≈ 37.71 cm.

The top is a semicircle of diameter $4 + 4 + 4 = 12$ cm; the bottom is three semicircles of diameter 4 cm:

$$\text{Perimeter} = \dfrac{1}{2}\pi(12) + 3 \times \dfrac{1}{2}\pi(4)$$

$$\text{Perimeter} = 6\pi + 6\pi$$

$$\text{Perimeter} = 12\pi$$

$$\text{Perimeter} = \dfrac{264}{7}$$

∴ Perimeter ≈ 37.71 cm.

The outer curve is a semicircle of diameter 20 cm. The inner curve is one semicircle of diameter 10 cm above the line and another below it:

$$\text{Perimeter} = \dfrac{1}{2}\pi(20) + \dfrac{1}{2}\pi(10) + \dfrac{1}{2}\pi(10)$$

$$\text{Perimeter} = 20\pi$$

$$\text{Perimeter} = \dfrac{440}{7}$$

∴ Perimeter ≈ 62.86 cm.

In one revolution, the car moves forward by one circumference:

$$C = \pi d$$

$$C = \dfrac{22}{7} \times 56$$

$$C = 176$$

∴ The car travels 176 cm in one revolution.

$$10 \text{ km} = 10 \times 100000 \text{ cm} = 1000000 \text{ cm}$$

We find the number of revolutions:

$$= \dfrac{1000000}{176}$$

$$= 5681.8\ldots$$

∴ The tyre makes about 5682 revolutions (5681 complete revolutions).

Each arc is centred at the midpoint of a side and joins a corner of the square to its centre. Both are 7 cm from that midpoint, and the two radii make a right angle, so each arc is a quarter circle of radius 7 cm.

Each petal is bounded by 2 such arcs, and there are 4 petals, so there are 8 quarter circles (2 full circles of radius 7):

$$\text{Total perimeter} = 2 \times 2\pi(7)$$

$$\text{Total perimeter} = 28\pi$$

$$\text{Total perimeter} = 88$$

∴ The total perimeter of the petals is 88 cm.

In a regular hexagon the centre is 42 cm from every vertex, so each vertex, its neighbour and the centre form an equilateral triangle. An arc centred at a vertex, from a neighbouring vertex to the centre, has radius 42 cm and angle 60°:

$$\text{One arc} = \dfrac{60}{360} \times 2 \times \dfrac{22}{7} \times 42$$

$$\text{One arc} = \dfrac{1}{6} \times 264$$

$$\text{One arc} = 44 \text{ cm}$$

Each petal is bounded by 2 arcs, and there are 6 petals:

$$\text{Total perimeter} = 12 \times 44$$

$$\text{Total perimeter} = 528$$

∴ The total perimeter of the petals is 528 cm.

Let the radii be $r_1$ and $r_2$. The ratio of the perimeters is 5 : 4, so:

$$\dfrac{2\pi r_1}{2\pi r_2} = \dfrac{5}{4}$$

$$\Rightarrow \dfrac{r_1}{r_2} = \dfrac{5}{4}$$

∴ The ratio of their radii is 5 : 4 (the same as the ratio of the perimeters).