NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.1 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.1.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 8 |

Subject |
Maths |

Chapter |
Chapter 11 |

Chapter Name |
Mensuration |

Exercise |
Ex 11.1, Ex 11.2, Ex 11.3, Ex 11.4 |

Number of Questions Solved |
5 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.1

**Question 1.**

A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?

**Solution.**

Area of the square field = a x a

= 60 m x 60 m = 3,600 \({ m }^{ 2 }\)

Perimeter of the square field = 4a

= 4 x 60 m = 240 m

∴ Perimeter of rectangular field = 240 m

⇒ 2(l + b) = 240

⇒ 2(80 + b) = 240

where b m is the breadth of the rectangular field

⇒ 80 + b = \(\frac { 240 }{ 2 } \) ⇒ 80 + bx = 120

⇒ b = 120 – 80 = 40

∴ Breadth = 40 m

∴ Area of rectangular field

= l x b = 80 m x 40 m = 3,200 \({ m }^{ 2 }\)

So, the square field (a) has a larger area

**Question 2.**

Mrs. Kaushik has a square plot ‘ with the measurement as shown in the figure. She wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a: garden around the house at the rate of ₹ 55 per \({ m }^{ 2 }\).

**Solution.**

Area of the square plot = a x a

= 25 x 25 \({ m }^{ 2 }\) = 625 \({ m }^{ 2 }\)

Area of the house = a x b

= 20 x 15 \({ m }^{ 2 }\) = 300 \({ m }^{ 2 }\)

∴ Area of the garden

= Area of the square plot – Area of the house

= 625 \({ m }^{ 2 }\) – 300 \({ m }^{ 2 }\)

= 325 \({ m }^{ 2 }\)

∵ The cost of developing the garden per square metre = ₹ 55.

∴ Total cost of developing the garden

= ₹ 325 x 55

= ₹ 17,875.

**Question 3.**

The shape of a garden is rectangular in the middle and semicircular at the ends as shown in the diagram. Find the area and the perimeter of this garden.

**Solution.**

**Question 4.**

A flooring tile has the shape of a parallelogram whose base is 24 cm and the cor-responding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m2? (If required you can split the tiles in whatever way you want to fill up the corners).

**Solution.**

Area of a flooring tile = bh

= 24 x 10 \({ cm }^{ 2 }\)

= 240 \({ cm }^{ 2 }\)

Area of the floor

= 1080 \({ m }^{ 2 }\)

= 1080 x 100 x 100 \({ cm }^{ 2 }\)

∵ m2 = 100 x 100 \({ cm }^{ 2 }\)

∴ Number of tiles required to cover the floor

=\(\frac { Area\quad of\quad the\quad floor\quad }{ Area\quad of\quad a\quad flooring\quad tile } \)

= \(\frac { 1080\times 100\times 100\quad }{ 240 } \)

= 45000.

**Question 5.**

An ant is moving around a few food pieces of different shapes scattered on the floor. For which food-piece would the ant have to take a longer round? Remember, the circumference of a circle can be obtained by using the expression c = 2πr, where r is the radius of the circle.

**Solution.**

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