RD Sharma Class 6 ex 20.3 Solutions Chapter 20 Mensuration

In this chapter, we provide RD Sharma Class 6 ex 20.3 Solutions Chapter 20 Mensuration for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 6 ex 20.3 Solutions Chapter 20 Mensuration Maths pdf, Now you will get step by step solution to each question.

TextbookNCERT
ClassClass 6
SubjectMaths
ChapterChapter 20
Chapter NameMensuration
ExerciseEx 20.3

RD Sharma Solutions for Class 6 Chapter 20 Mensuration Ex 20.3 Download PDF

Chapter 20: Mensuration Exercise 20.3

Question: 1

The following figures are drawn on a squared paper. Count the number of squares enclosed by each figure and find its area, taking the area of each square as 1 cm2.

Squared paper

Solution:

(i) There are 16 complete squares in the given shape.

Since, Area of one square = 1 cm2

Therefore, Area of this shape = 16 × 1 = 16 cm2

(ii) There are 36 complete squares in the given shape.

Since, Area of one square = 1 cm2

Therefore, Area of 36 squares = 36 × 1 = 36 cm2

(iii) There are 15 complete and 6 half squares in the given shape.

Since, Area of one square = 1 cm2

Therefore, Area of this shape = (15 + 6 × 12) = 18 cm2

(iv) There are 20 complete and 8 half squares in the given shape.

Since, Area of one square = 1 cm2

Therefore, Area of this shape = (20 + 8 × 12) = 24 cm2

(v) There are 13 complete squares, 8 more than half squares and 7 less than half squares in the given shape.

Area of one square = 1 cm2

Area of this shape = (13 + 8 × 1) = 21 cm2

(vi) There are 8 complete squares, 6 more than half squares and 4 less than half squares in the given shape.

Area of one square = 1 cm2

Area of this shape = (8 + 6 × 1) = 14 cm2

Question: 2

On a squared paper, draw (i) a rectangle, (ii) a triangle, (iii) any irregular closed figure, Find approximate area of each by counting the number of squares complete, more than half and exactly half.

Solution:

(i) A rectangle: This contains 18 complete squares.

If we assume that the area of one complete square is 1 cm2,

Then the area o this rectangle will be 18 cm2.

Rectangle

(ii) A triangle: This triangle contains 4 complete squares, 6 more than half squares and 6 less than half squares.

If we assume that the area of one complete square is 1 cm2,

Then the area of this shape = (4 + 6 x 1) = 10 cm2

Angle

(iii) Any irregular figure: This figure consists of 10 complete squares, 1 exactly half square, 7 more than half squares and 6 less than half squares.

If we assume that the area of one complete square is 1 cm2,

Then the area of this shape = (10 + 1 x12 + 7 x 1) = 17.5 cm2

Any irregular figure

Question: 3 

Draw any circle on the graph paper, Count the squares and use them to estimate the area the area of the circular region.

Solution:

Circular

This circle on the squared paper consists of 21 complete squares, 15 more than half squares and 8 less than half squares.

Let us assume that the area of 1 square is 1 cm2.

If we neglect the less than half squares while approximating more than half square as equal to a complete square, we get:

Area of this shape = (21 + 15) = 36 cm2

Question: 4

Using tracing paper and centimeter graph paper to compare the areas of the following pairs of figures:

Tracing paper and Centimeter graph paper

Solution:

Shape of Using tracing paper

Using tracing paper, we traced both the figures on a graph paper.

This figure contains 4 complete squares, 9 more than half squares and 9 less than half squares. Let us assume that the area of one square is 1 cm2

If we neglect the less than half squares and consider the area of more than half squares as equal to area of complete square, we get:

Area of this shape = (4 + 9) = 13 cm2

This figure contains 8 complete squares, 11 more than half squares and 10 less than half squares.

Let us assume that the area of one square is 1 cm2.

If we neglect the less than half squares and consider the area of more than half squares as equal to area of complete square, we get:

Area of this shape = (8 + 11) = 19 cm2

On comparing the areas of these two shapes, we get that the area of Fig. (ii) is more than that of Fig. (i).

All Chapter RD Sharma Solutions For Class 6 Maths

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