Chapter 11 - Circle
Practice set 42
1. Complete the table below.
| Sr. No. | Radius (r) | Diameter (d) | Circumference (c) |
|---|---|---|---|
(i) | 7 cm | .......... | .......... |
(ii) | .......... | 28 cm | .......... |
(iii) | .......... | .......... | 616 cm |
(iv) | .......... | .......... | 72.6 cm |
(i)
Given:
Radius = 7 cm
To find:
Circumference and Diameter
Solution:
Circumference = 2πr
∴ Circumference = 2 × \(\large \frac{22}{7}\) × 7
∴ Circumference = 2 × 22
∴ Circumference = 44 cm
Diameter = 2 × radius
∴ Diameter = 2 × 7
∴ Diameter = 14 cm
Ans: Circumference is 44 cm and diameter is 14 cm.
(ii)
Given:
Diameter = 28 cm
To find:
Radius and Circumference
Solution:
Radius = \(\large \frac{Diameter}{2}\)
∴ Radius = \(\large \frac{28}{2}\)
∴ Radius = 14 cm
Circumference = 2πr
∴ Circumference = 2 × \(\large \frac{22}{7}\) × 14
∴ Circumference = 2 × 22 × 2
∴ Circumference = 88 cm
Ans: Radius is 14 cm and circumference is 88 cm.
(iii)
Given:
Circumference = 616 cm
To find:
Radius and Diameter
Solution:
Circumference = 2πr
∴ 616 = 2 × \(\large \frac{22}{7}\) × r
∴ \(\large \frac{616\, ×\, 7}{2\, ×\, 22}\) = r
∴ Radius = 98 cm
Diameter = 2 × radius
∴ Diameter = 2 × 98
∴ Diameter = 196 cm
Ans: Radius is 98 cm and diameter is 196 cm.
(iv)
Given:
Circumference = 72.6 cm
To find:
Radius and Diameter
Solution:
Circumference = 2πr
∴ 72.6 = 2 × \(\large \frac{22}{7}\) × r
∴ \(\large \frac{72.6\, ×\, 7}{2\, ×\, 22}\) = r
∴ r = \(\large \frac{508.2}{44}\)
∴ Radius = 11.55 cm
Diameter = 2 × radius
∴ Diameter = 2 × 11.55
∴ Diameter = 23.1 cm
Ans: Radius is 11.55 cm and diameter is 23.1 cm.
2. If the circumference of a circle is 176 cm, find its radius.
Given:
Circumference of the circle = 176 cm
To find:
Radius of the circle
Solution:
Circumference of the circle = 2πr
∴ 176 = 2 × \(\large \frac{22}{7}\) × r
∴ \(\large \frac{176\, ×\, 7}{2\, ×\, 22}\) = r
∴ r = \(\large \frac{8\, ×\, 7}{2}\)
∴ Radius = 28 cm
Ans: Radius of the circle is 28 cm.
3. The radius of a circular garden is 56 m. What would it cost to put a 4-round fence around this garden at a rate of 40 rupees per metre?
Given:
Radius of the circular garden = 56 m
Rate of wire per meter = ₹ 40
To find:
Cost of 4-round fence around the garden
Solution:
Circumference of the circular garden = 2πr
∴ Circumference of the circular garden = 2 × \(\large \frac{22}{7}\) × 56
∴ Circumference of the circular garden = 2 × 22 × 8
∴ Circumference of the circular garden = 352 m
Now,
Length of the wire required to put 1-round fence = Circumference of the circular garden
∴ Length of wire required to put a 4-round fence = 4 × Circumference of the circular garden
∴ Length of wire required to put a 4-round fence = 4 × 352
∴ Length of wire required to put a 4-round fence = 1408 m
Rate of wire per meter = ₹ 40
∴ Cost of 4-round fence around the garden = length of wire required × rate of wire
∴ Cost of 4-round fence around the garden = 1408 × 40
∴ Cost of 4-round fence around the garden = Rs 56320
Ans: The cost to put a 4-round fence around the garden is Rs 56320.
4. The wheel of a bullock cart has a diameter of 1.4 m. How many rotations will the wheel complete as the cart travels 1.1 km?
Given:
Diameter of the wheel of the bullock cart = 1.4 m
Distance travelled by the cart = 1.1 km = 1.1 × 1000 m = 1100 m
To find:
Number of rotations the wheel completes in 1100 m
Solution:
Circumference of the wheel of the bullock cart = πd
∴ Circumference of the wheel of the bullock cart = \(\large \frac{22}{7}\) × 1.4
∴ Circumference of the wheel of the bullock cart = 22 × 0.2
∴ Circumference of the wheel of the bullock cart = 4.4 m
Distance covered in 1 rotation is the circumference of the wheel
∴ Distance covered in 1 rotation = 4.4 m
Now,
Number of rotations = \(\large \frac{Distance\, covered\, in\, 1 \,rotation}{Circumference\, of\, the\, wheel}\)
∴ Number of rotations = \(\large \frac{1100 m }{4.4 m}\)
∴ Number of rotations = 250
Ans: The wheel of the bullock cart will complete 250 rotations as the cart travels 1.1 km.
Practice set 43
1. Choose the correct option.
If arc AXB and arc AYB are corresponding arcs and m(arc AXB) = 120° then m(arc AYB) =
(i) 140°
(ii) 60°
(iii) 240°
(iv) 160°
Solution:
Measure of major arc = 360° – measure of corresponding minor arc
∴ m (arc AYB) = 360 – m (arc AXB)
∴ m (arc AYB) = 360 – 120
∴ m (arc AYB) = 240°
OPTION (iii) : 240°
2. Some arcs are shown in the circle with centre ‘O’ Write the names of the minor arcs, major arcs and semicircular arcs from among them.
Solution:
Minor arcs :
arc QXP, arc PR, arc RY, arc YQ, arc QX, arc XP, arc PRY
Major arcs :
arc PYQ, arc PQR, arc RQY, arc XPQ, arc XQP, arc XQR
Semicircular arcs :
arc QPR, arc QYR
3. In a circle with centre O, the measure of a minor arc is 110°. What is the measure of the major arc PYQ?
Given:
Measure of minor arc = 110⁰
To find:
Measure of the major arc PYQ
Solution:
Measure of major arc = 360° – measure of corresponding minor arc
∴ m (arc PYQ) = 360 – 110
∴ m (arc PYQ) = 250°
Ans: The measure of the major arc PYQ is 250°.