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

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.1
are the part of NCERT Solutions for Class 8 Maths. Here you can find the NCERT
Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.1**.**** **

**NCERT Solutions for Class 8 Maths Chapter 11****Mensuration**Ex 11.1**NCERT Solutions for Class 8 Maths Chapter 11****Mensuration**Ex 11.2**NCERT Solutions for Class 8 Maths Chapter 11****Mensuration**Ex 11.3**NCERT Solutions for Class 8 Maths Chapter 11****Mensuration**Ex 11.4

**Ex 11.1 Class 8 Maths 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: ****
**Perimeter
of the figure (a) = 4 × side = 4 × 60 = 240 m

Perimeter of the figure (b) = 2[l + b] = 2[80 + b] m

Perimeter of the figure (a) = Perimeter of the figure (b)

⇒ 240 = 2[80 + b]

⇒ 80 + b = 120

⇒ b = 120 – 80 = 40 m

Area of the figure (a) = (side)

^{2}= 60 × 60 = 3600 m

^{2}

Area of the figure (b) = l × b = 80 × 40 = 3200 m

^{2}

So, the area of the figure (a) is larger than the area of the figure (b).

**Ex 11.1 Class 8 Maths 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 plot = side × side = 25 m × 25 m = 625 m

^{2}

Area of the house = l × b = 20 m × 15 m = 300 m

^{2}

Area of the garden to be developed = Area of the plot – Area of the house

= 625 m^{2} – 300 m^{2} = 325 m^{2}

The total cost of developing the garden = ₹ 325
× 55 = ₹ 17,875

**Ex 11.1 Class 8 Maths 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. [Length
of rectangle is 20 – (3.5 + 3.5) metres]**Length of the rectangle = 20 – (3.5 + 3.5) = 20 – 7 = 13 m**

**Solution:**

Area of the rectangle = l × b = 13 × 7 = 91 m

^{2}

Area of two circular ends = 2(1/2 Ï€r

^{2})

= Ï€r

^{2}

= 22/7 × 7/2 × 7/2

= 77/2 m

^{2}

= 38.5 m

^{2}

Total area of the garden = Area of the rectangle + Area of two ends

= 91
m^{2} + 38.5 m^{2} = 129.5 m^{2}

Total perimeter of the garden = 13 + 13 + 2 ×
(Ï€r)

= 26 + 2(22/7 × 7/2)

= 26 + 22

= 48 m

**Ex 11.1 Class 8 Maths Question 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).

**Solution:
**Area of the floor = 1080 m

^{2}= 1080 × 10000 cm

^{2}= 10800000 cm

^{2}

[∵ 1 m^{2} = 10000 cm^{2}]

Area of 1 tile = base × height = 24 × 10 = 240 cm^{2}

Number of tiles required

**Ex 11.1 Class 8 Maths 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:**