Mathematics Solutions Solutions for Class 7 Maths Chapter 12 Circle are provided here with simple step-by-step explanations. These solutions for Circle are extremely popular among Class 7 students for Maths Circle Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Solutions Book of Class 7 Maths Chapter 12 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Solutions Solutions. All Mathematics Solutions Solutions for class Class 7 Maths are prepared by experts and are 100% accurate.

#### Page No 77:

#### Question 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 |

#### Answer:

(i) Radius, *r* = 7 cm

Diameter, *d *= 2*r* = 2 × 7 = 14 cm

∴ Circumference, *c* = π*d*

= $\frac{22}{7}$ × 14

= 22 × 2

= 44 cm

(ii) Diameter, *d* = 28 cm

Radius, *r* = $\frac{d}{2}$ = $\frac{28}{2}$ = 14 cm

∴ Circumference, *c* = 2π*r*

= 2 × $\frac{22}{7}$ × 14

= 88 cm

(iii) Circumference, *c* = 616 cm

Now, *c* = 2π*r* (where '*r*' is the radius)

⇒616 = 2 × $\frac{22}{7}$ × *r*

⇒*r *= 616 × $\frac{1}{2}$ × $\frac{7}{22}$

⇒*r* = 98

So, radius = 98 cm

Diameter, *d* = 2*r* = 2 × 98 = 196 cm

(iv) Circumference, *c* = 72.6 cm

Now, *c* = 2π*r* (where '*r*' is the radius)

⇒72.6 = 2 × $\frac{22}{7}$ × *r*

⇒*r* = 72.6 × $\frac{1}{2}$ × $\frac{7}{22}$

⇒*r* = 11.55

So, radius = 11.55 cm

Diameter, *d *= 2*r* = 2 × 11.55 = 23.1 cm

The complete table is shown below.

Sr. No. | Radius (r) |
Diameter (d) |
Circumference (c) |

(i) | 7 cm | 14 cm | 44 cm |

(ii) | 14 cm | 28 cm | 88 cm |

(iii) | 98 cm | 196 cm | 616 cm |

(iv) | 11.55 cm | 23.1 cm | 72.6 cm |

#### Page No 77:

#### Question 2:

If the circumference of a circle is 176 cm, find its radius.

#### Answer:

Circumference, *c* = 176 cm

Now, *c* = 2π*r* (where '*r*' is the radius of circle)

⇒176 = 2 × $\frac{22}{7}$ × *r*

⇒*r* = 176 × $\frac{1}{2}$ × $\frac{7}{22}$

⇒*r *= 28

∴ Radius of the circle = 28 cm

#### Page No 77:

#### Question 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 ?

#### Answer:

Radius of the circular garden, *r *= 56 m

Circumference of the circular garden, *c* = 2π*r*

= 2 × $\frac{22}{7}$ × 56

= 352 m

∴ Length of the wire needed for one round of fencing = *c* = 352 m

Cost of one round of fencing = length of wire × cost per metre

= 352 × 40

= 14080 rupees

Cost of four round of fencing = 4 × 14080 = 56320 rupees

#### Page No 77:

#### Question 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 ?

#### Answer:

Diameter of the wheel, *d* = 1.4 m

Circumference, *c* = π*d*

= $\frac{22}{7}$ × 1.4

= 4.4 m

When the wheel completes 1 rotation, it covers a distance that is equal to its circumference.

So, number of rotations taken by the wheel to cover 4.4 m = 1

Now, the wheel covered a total distance of 1.1 km.

We know that, 1 km = 1000 m

∴ 1.1 km = 1.1 × 1000 m = 1100 m

∴ Total number of rotations taken by wheel = $\frac{\mathrm{total}\mathrm{distance}}{\mathrm{circumference}}$

= $\frac{1100}{4.4}$

= $\frac{11000}{44}$

= 250

Hence, the wheel completes 250 rotations to cover a distance of 1.1 km.

#### Page No 79:

#### Question 1:

*m*(arc AXB) = 120

^{°}then

*m*(arc AYB) = $\overline{){12}{}{e}{r}{y}{y}{t}{4}}$

^{°}(ii) 60

^{° }(iii) 240

^{°}(iv) 160

^{°}

#### Answer:

Consider that arc AXB is the minor arc and arc AYB is the corresponding major arc.

It is known that, measure of major arc = 360^{∘} − measure of the corresponding minor arc.

We have, *m*(arc AXB) = 120^{°}.

So, *m*(arc AYB) = 360^{∘} − *m*(arc AXB) = 360^{∘} − 120^{∘} = 240^{∘}

Hence, the correct answer is option (iii).

#### Page No 79:

#### Question 2:

#### Answer:

Minor arc : An arc of a circle having measure less than 180^{∘}.

Major arc : An arc of a circle having measure greater than 180^{∘}.

Semicircular arc : An arc of a circle having measure equal to 180^{∘}.

Names of minor arcs :

(i) arc PXQ

(ii) arc PR

(iii) arc RY

(iv) arc XP

(v) arc XQ

(vi) arc QY

Names of major arcs :

(i) arc PYQ

(ii) arc PQR

(iii) arc RQY

(iv) arc XQP

(v) arc QRX

Names of semicircular arcs :

(i) arc QPR

(ii) arc QYR

#### Page No 79:

#### Question 3:

In a circle with centre O, the measure of a minor arc is 110^{°}. What is the measure of the major arc PYQ?

#### Answer:

Suppose PQ is the minor arc and then *m*(arc PQ) = 110^{∘}.

We know that, measure of major arc = 360^{∘} − measure of corresponding minor arc.

∴ *m*(arc PYQ) = 360^{∘} − *m*(arc PQ)

= 360^{∘} − 110^{∘}

= 250^{∘}

Hence, the measure of major arc PYQ is 250^{∘}.

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