NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3.

- Perimeter and Area Class 7 Ex 11.1
- Perimeter and Area Class 7 Ex 11.2
- Perimeter and Area Class 7 Ex 11.4
- Perimeter and Area Class 7 MCQ

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 7 |

Subject |
Maths |

Chapter |
Chapter 11 |

Chapter Name |
Perimeter and Area |

Exercise |
Ex 11.3 |

Number of Questions Solved |
17 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3

**Question 1.**

Find the circumference of the circles with the following radus (Take π = \(\frac { 22 }{ 7 } \))

**(a)** 14 cm

**(b)** 28 mm

**(c)** 21 cm

**Solution:
**

**Question 2.**

Find the area of the following circles, given that: ( Take π = \(\frac { 22 }{ 7 } \) )

**(a)** radius = 14 mm

**(b)** diameter = 49 m

**(c)** radius = 5 cm.

**Solution:
**

**Question 3.**

If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet ( Take π = \(\frac { 22 }{ 7 } \) )

**Solution:**

Circumference of the circular sheet = 154 m

Let the radius of the circular sheet be r cm

Then, its circumference = 2nr m According to the question,

Circumference = 2πr = 154

**Question 4.**

A gardener wants to fence a circular garden of diameter 21 m. Find the length of the rope he needs to purchase if he makes 2 rounds of offense. Also find the cost of the rope, if it costs ₹ 4 per meter. ( Take π = \(\frac { 22 }{ 7 } \) )

**Solution:**

The diameter of the circular garden (r) = 21 m

Radius of the circular garden (r) = \(\frac { 21 }{ 2 } \) m

∴ Circumference of the circular garden = 2πr

= 2 × \(\frac { 22 }{ 7 } \) × \(\frac { 21 }{ 2 } \) m = 66m

⇒ Length of the rope needed to make 1 round of fence = 66 m

⇒ Length of the rope needed to make 2 rounds of fence

= 66 × 2 m = 132 m

Cost of rope per meter = ₹ 4

∴ Cost of the rope = ₹ 132 × 4 = ₹ 528.

**Question 5.**

From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take π = 3.14)

**Solution:**

Here, Outer radius, r = 4 cm

Inner radius, r = 3 cm

Area of the remaining sheet = Outer area – Inner area

= π (R^{2} – r^{2}) = 3.14 (4^{2} – 3^{2}) cm^{2}

= 3.14 (16 – 9) cm^{2}

= 3.14 × 7 cm^{2} = 21.98 cm^{2}

**Question 6.**

Saima wants to put lace on the edge of a circular table cover of a diameter of 1.5 m. Find the length of the lace required and also find its cost if one meter of the lace costs ₹ 15. (Take π = 3.14)

**Solution:**

Diameter of the table cover = 1.5 m

⇒ Radius of the table cover (r) = \(\frac { 1.5 }{ 2 } \) m

⇒ Circumference of the table cover = 2πr

= 2 × 3.14 × \(\frac { 1.5 }{ 2 } \) m = 4.71 m

⇒ Length of the lace required = 4.71 m

∵ Cost of lace per meter = ₹ 15

∴ Cost of the lace = ₹ 4.71 × 15 = ₹ 70.65

**Question 7.**

Find the perimeter of the following figure, which is a semicircle including its diameter.

**Solution:
**

**Question 8.**

Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is ₹ 15/m^{2}. (Take π = 3.14)

**Solution:**

Diameter of the table-top = 1.6 m

⇒ Radius of the table-top (r) = \(\frac { 1.6 }{ 2 } \) m = 0.8 m

∴ Area of the table-top = πr^{2}

= 3.14 × (0.8)2 m^{2}

= 3.14 × 0.64 m^{2}

= 2.0096 m^{2}

∵ Rate of polishing = ₹ 15 per m^{2}

∴ Cost of polishing the table-top = ₹ 2.0096 × 15

= ₹ 30.144

= ₹ 30.14 (approx.).

**Question 9.**

Shazli took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its side? Which figure encloses more area, the circle or the square? ( Take π = \(\frac { 22 }{ 7 } \) )

**Solution:
**

**Question 10.**

From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1 cm are removed (as shown in the following figure). Find the area of the remaining sheet. ( Take π = \(\frac { 22 }{ 7 } \) )

**Solution:
**

**Question 11.**

A circle of radius 2 cm is cut out from a square piece of an aluminium sheet of side 6 cm. What is the area of the leftover aluminium sheet? (Take π = 3.14)

**Solution:
**

**Question 12.**

The circumference of a circle is 31.4 cm. Find the radius and the area of the circle? (Take π = 3.14)

**Solution:
**

**Question 13.**

A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66 m. What is the area of this path? (π = 3.14)

**Solution:
**

**Question 14.**

A circular flower garden has an area of about 314 m^{2}. A sprinkler at the centre of the garden can cover an area that has a radius of 12 m. Will the sprinkler water the entire garden? (Taken π = 3.14)

**Solution:**

The circular area of the sprinkler = πr^{2}

= 3.14 × 12 × 12

= 3.14 × 144 = 452.16 m^{2}

Area of the circular flower garden = 314 m^{2}

Since the area of the circular flower garden is smaller than by sprinkler

Therefore, the sprinkler will water the entire garden.

**Question 15.**

Find the circumference of the inner and the outer circles as shown in the following figure? (Take π = 3.14)

**Solution:**

Radius of inner circle = 19 – 10 = 9 m

∴ Circumference of the inner circle = 2 πr = 2 × 3.14 × 9 m = 56.52 cm

The radius of the outer circle = 19 m

∴ Circumference of the outer circle = 2πr = 2 × 3.14 × 19 m = 119.32 m.

**Question 16.**

How many times a wheel of radius 28 cm must rotate to go 352 m? ( Take π = \(\frac { 22 }{ 7 } \) )

**Solution:**

**Question 17.**

The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour? (Take π = 3.14)

**Solution:**

We know that the minute hand describes one complete revolution in one hour.

∴ Distance covered by its tip = Circumference of the circle of radius 15 cm

= (2 × 3.14 × 15) cm

= 94.2 cm

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