NCERT Solutions for Class 8 Maths Chapter 11 Exercise 11.1 – Mensuration, has been designed by the NCERT to test the knowledge of the student on the topic – Mensuration Introduction

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

**NCERT Solutions for Class 8 Maths Chapter 11 Exercise 11.1 – Mensuration**

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

**Sol.: **– We know that,

Area of Square = (Side)

^{2 }Area of Rectangular = Length × Breadth

As we don’t know Breadth of Rectangle, we need to find it first to calculate area.

We know that,

Perimeter of square = Perimeter of rectangular, (Mentioned in question)

Also,

Perimeter of square = Sum of all side = 4 × Side, And

Perimeter of rectangular = 2 × (Length + Breadth)

So,

4 × Side = 2 × (Length + Breadth)

4 × 60 = 2 × (80 + Breadth)

After solving this equation, we get

Breadth = [(4×60)/2] – 80

= 40m

Area of Square = (Side)

^{2}= (60)

^{2 }= 60 × 60 = 3600 m

^{2 }And, Area of Rectangle = Length × Breadth = 40 × 80 = 3200 m

^{2 }Therefore, area of Square field is larger than Rectangular field.

**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 Rs 55 per m ^{2}.**

**Sol.: **– Area of garden = Area of Total plot – Area of house

= 25 × 25 – 15 × 20

= 625 – 300

=325 m

^{2}

^{ }So, total cost of developing a garden around the house

= Total Area of garden × Cost of developing the garden per m^{2 }= 325 × 55

= ₹ 17875

**3. The shape of a garden is rectangular in the middle and semi circular at the ends as shown in the diagram. Find the area and the perimeter of this garden [Length of rectangle is 20 – (3.5 + 3.5) metres].**

**Sol.: **– As we can see in the figure,

Perimeter of the garden = Perimeter of rectangle × perimeter of semi-circle

= (2 × Length of rectangle) + (2 × π × Radius of circle)

(Here, Length of rectangle = 20 – (3.5 + 3.5) = 13 m)

(Here, Radius of circle = 7/2 = 3.5 m)

So, Perimeter of the garden = (2 x 13) + (2 x (22/7) x 3.5)

= 26 + 22

= 48 m

And, Area of the garden = Area of rectangle + 2 × Area of semi-circle

= (Length × Breadth) + 2 × (πr²/2)

= (13 × 7) + 2 × (22×3.5²/2×7)

= 91 + 38.5

=129.5 m^{2}

**4. A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m ^{2}? (If required you can split the tiles in whatever way you want to fill up the corners).**

**Sol.: **– We know that,

Area of parallelogram = Base × Height

= 24 × 10

= 240 cm

^{2}

Area of floor = 1080 m^{2 }= 1080 × 10000 m^{2 }(As 1m^{2}= 10000 cm^{2 })

= 10800000 m^{2}

^{ }So, Number of tiles required to cover the floor = (Area of floor/Area of parallelogram)

= 10800000/240

= 45000 tiles

**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, circumference of a circle can be obtained by using the expression c = 2πr, where r is the radius of the circle.**

**Sol.: **– For checking on which food-piece the ant would take longer round; we need to find Perimeter of all food-piece.

a). Perimeter of 1^{st} food-piece = circumference of semi-circle + D

= πr + D

= [(22 × 1.4)/7] + 2.8

= 7.2 cm

b). Perimeter of 2^{nd} food-piece = circumference of semi-circle + 3 sides of rectangle

= [(22 × 1.4)/7] + 1.5 +1.5 +2.8

= 10.2 cm

c). Perimeter of 3^{rd} food-piece = circumference of semi-circle + 2 sides of triangle

= [(22 × 1.4)/7] + 2 + 2

= 8.4 cm

Therefore, Perimeter of food piece (b) is largest. So, ‘Ant’ will take longest round around food piece (b).

**The next Exercise for** **NCERT Solutions for Class 8 Maths Chapter 11 Exercise 11.2 – Mensuration can be accessed by clicking here**

**Download NCERT Solutions for Class 8 Maths Chapter 11 Exercise 11.1 – Mensuration**

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