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

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Textbook | NCERT |

Board | CBSE |

Category | NCERT Solutions |

Class | Class 8 |

Subject | Maths |

Chapter | Chapter 11 |

Exercise | Class 8 Chapter 11 Mensuration 11.1 |

Number of Questions Solved | 5 |

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

**NCERT TEXTBOOK EXERCISES**

**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

âˆ´ Perimeter of rectangular field = 240 m

â‡’ 2(l + b) = 240

â‡’ 2(80 + b) = 240

where b m is the breadth of the rectangular field

â‡’ 80 + b = 240/2 â‡’ 80 + bx = 120

â‡’ b = 120 â€“ 80 = 40

âˆ´ Breadth = 40 m

âˆ´ Area of rectangular field

= l x b = 80 m x 40 m = 3,200 m2

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

**Question 2. Mrs. Kaushik has a square plot â€˜ 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 â‚¹ 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}

âˆ´ Area of the garden

= Area of the square plot â€“ Area of the house

= 625 m^{2} â€“ 300 m^{2}

= 325 m^{2}

âˆµ The cost of developing the garden per square metre = â‚¹ 55.

âˆ´ Total cost of developing the garden

= â‚¹ 325 x 55

= â‚¹ 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}

âˆµ m2 = 100 x 100 cm^{2}

âˆ´ 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 = 2Ï€r, where r is the radius of the circle.**

**Solution.**

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

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