# Mensuration Chapter 11 Class 8 Maths NCERT Textbook With Solutions PDF

NCERT Solutions for Class 8 Maths Chapter 11′ PDF Quick download link is given at the bottom of this article. You can see the PDF demo, size of the PDF, page numbers, and direct download Free PDF of ‘Ncert Class 8 Maths Chapter 11 Exercise Solution’ using the download button.

### Chapter 11:Mensuration

#### 11.1 Introduction

We have learned that for a closed plane figure, the perimeter is the distance around its boundary and its area is the region covered by it.

We found the area and perimeter of various plane figures such as triangles, rectangles, circles, etc.

We have also learned to find the area of pathways or borders in rectangular shapes.

In this chapter, we will try to solve problems related to the perimeter and area of other plane closed figures like quadrilaterals. We will also learn about the surface area and volume of solids such as cubes, cuboids, and cylinders.

#### 11.6 Solid Shapes

In your earlier classes, you have studied that two-dimensional figures can be identified as the faces of three-dimensional shapes. Observe the solids which we have discussed so far

#### 11.7.3 Cylinders

Most of the cylinders we observe are right circular cylinders. For example, a tin, round pillars, tube lights, water pipes, etc.

### NCERT Solutions Class 8 Maths Chapter 11 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?

Solution:

Side of a square = 60 m (Given)

And the length of a rectangular field, l = 80 m (Given)

According to the question,

Perimeter of rectangular field = Perimeter of square field

2(l+b) = 4×Side (using formulas)

2(80+b) = 4×60

160+2b = 240

b = 40

Breadth of the rectangle is 40 m.

Now, Area of Square field

= (side)2

= (60)2 = 3600 m2

And Area of Rectangular field

= 3200 m2

Hence, area of square field is larger.

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 m2.

Solution:

Side of a square plot = 25 m

Formula: Area of square plot =  square of a side = (side)2

= (25)2 = 625

Therefore the area of a square plot is 625 m2

Length of the house = 20 m and

Breadth of the house = 15 m

Area of the house = length×breadth

= 20×15 = 300 m2

Area of garden = Area of the square plot – Area of the house

= 625–300 = 325 m2

∵ The cost of developing the garden per sq. m is Rs. 55

Cost of developing the garden 325 sq. m = Rs. 55×325

= Rs. 17,875

Hence the total cost of developing a garden around is Rs. 17,875.