NCERT Solutions For Class 6 Maths Chapter 10

NCERT Solutions Class 6 Maths Mensuration

Ncert Solutions For Class 6 Maths Chapter 10 PDF Free Download

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration is an important section for students preparing for class 6 maths. Several times, in exams, a majority of questions are asked from this section making it one of the most crucial chapters of class 6th maths. The NCERT solutions for class 6 Maths Chapter 10 Mensuration are prepared by our experienced faculty members to develop a strong conceptual base among students. The students can refer NCERT Solutions for class 6 maths chapter 10 pdf from the links given below. The NCERT solutions for class 6  provides a detailed explanation for the NCERT solutions. Many students may find difficulties in the concepts to remember. Some of these major concepts in this chapter of Mensuration deal with problems related to finding perimeter of a variety of different shapes. Shapes such as ‘L’ shapes, Parallelograms, Pentagons and other geometric faces. The entire chapter of measurement and mensuration deals with these concepts. Check out the NCERT Solutions for class 6 maths chapter 10 below:

NCERT Solutions For Class 6 Maths Chapter 10 Exercises

Exercise 10.1

Q.1 Find the perimeter of each of the following figures

(a)

1

Ans.

Perimeter is the distance covered along the boundary of a closed figure.

Perimeter = Sum of the distances or lengths of all the sides

= PQ + QR + RS + SP.

= 3 cm + 2 cm + 1 cm + 4 cm.

= (3 + 2 + 1 + 4) cm

= 10 cm

The perimeter of the given figure is 10 cm.

 

(b)

2

Ans.

Perimeter = Sum of the distances or lengths of all the sides

= EF + FG + GH + HE.

= 20 cm + 30 cm + 20 cm + 45 cm.

= (20 + 30 + 20 + 45) cm

= 105 cm

The perimeter of the given figure is 105 cm.

 

(c)

3

Ans.

Perimeter = Sum of the distances or lengths of all the sides.

= LM + MN + NO + OL

= 20 cm + 20 cm + 20 cm + 20 cm

= (20 + 20 + 20 + 20) cm

= 80 cm

The perimeter of the given figure is 80 cm.

 

(d)

4

Ans.

Perimeter = Sum of the distances or lengths of all the sides.

= AB + BC + CD + DE + EA

= 5 cm + 5 cm + 5 cm + 5 cm + 5 cm

= (5 + 5 + 5 + 5 + 5) cm

= 25 cm

The perimeter of the given figure is 25 cm.

 

(e)

5

Perimeter = Sum of the distances or lengths of all the sides.

We are simplifying the problem by considering a single ‘L’ shaped section and finding its perimeter. There are four such ‘L’ shaped sections and the total perimeter is four times the perimeter of a single section.

6

Perimeter of the ‘L’ shaped section = AB + BC + CD + DE + EF

= 3 + 4 + 1 + 3 + 2

= 13 cm

The perimeter of a single ‘L’ section is 13 cm

The perimeter of the entire figure = 4 x perimeter of each ‘L’ section.

= 4 x 13 cm

= 52 cm

The perimeter of the entire figure is 52 cm.

 

 

Q2. The lid of a rectangular box of sides 50 cm by 10 cm is sealed all around with a tape. What is the length of the tape required?

Ans.

7

Dimensions of the given box:

Length of the rectangular box (l) = 50 cm

Breadth of the rectangular box (b) = 10 cm

Perimeter of a rectangle = 2 x (l + b)

= 2 x (50 cm + 10 cm)

= 2 x (60 cm)

= 120 cm

The perimeter of the rectangular box = Length of the tape required to measure the box = 120 cm

 

 

Q3. A table-top measures 3 m 35 cm by 1 m 40 cm. What is the perimeter of the table-top?

Ans.

8

The length of the rectangle should be greater than the breadth (l \(>\) b)

Length of the rectangle (l) = 3 m 35 cm

=  (300 + 35) cm (∵ 1 m = 100 cm)

= 335 cm

Breadth of the rectangle (b) =  1 m 40 cm

= (100 + 40) cm (∵ 1 m = 100 cm)

= 140 cm

Perimeter of the table top = Perimeter of the rectangle

= 2 x (length + breadth)

= 2 x (l + b)

= 2 x (335 cm + 140 cm)

= 2 x (475 cm)

= 950 cm

The perimeter of the table top = 950 cm = 9.5 m

 

Q4. A photo of sides 17.8 cm by 12.7 cm is to be framed with wood. What is the length of wooden strip required?

Ans:

Length of photograph (I) = 17.8 cm

Breadth of photograph (b) = 12.7 cm

Length of required wooden strip = Perimeter of Photograph

= 2 x (/ + b)

= 2 x (17.8 + 12.7)

= 2 x 30.5 = 61 cm

 

Q5. Each side of rectangular piece of land which measures 0.8km by 0.7km is to be fenced with 5 rows of wires. How much length of wire is needed?

Ans:

Length of land (I) = 0.8 km

Breadth of land (b) = 0.7 km

Perimeter = 2 x (I + b) = 2 x (0.8 + 0.7) = 2 x 1.5 = 3.0 km

Length of wire required = 5 x 3 = 15 km

 

Q6. Find the perimeter of the following shapes:

(a) An equilateral triangle of side 5 cm.

(b) An Isosceles triangle with equal sides 6 cm each and the third side 5 cm.

(c) A triangle of sides 6 cm, 3 cm and 2 cm

Ans:

(a) Perimeter = of an equilateral triangle = 3 x Side of triangle

= (3 x 5) cm = 15 cm

(b) Perimeter = (6 cm + 3 cm + 2 cm) = 11 cm.

(c) Perimeter = (2 x 6) + 5 = 17 cm

 

Q7. A triangle has sides measuring 15 cm, 12 cm and 13 cm. Find its perimeter.

Ans:

Perimeter of the triangle = Sum of the lengths of all sides of the triangle

Perimeter = 15 + 12 + 13 = 40 cm

 

Q8. A regular hexagon of each side measuring 6 m is given. Find its perimeter.

Ans:

Perimeter of regular hexagon = 6 x Side of regular hexagon

Perimeter of the regular hexagon = 6 x 6 = 36 m

 

Q9. Find the side of the square whose perimeter is 16 m.

Ans:

Perimeter of square = 4 x Side

16 = 4 x Side

\(Side = \frac{16}{4} = 4\ cm\)

Side = 4

 

Q10. Find the side of a regular pentagon if its perimeter is 150 cm.

Ans:

Perimeter of regular pentagon = 5 x Length of side

150 = S x Side

\(Side = \frac{150}{5} = 30\ cm\)

Therefore, Side = 30 cm

 

Q11. A piece of string is 120 cm long. What will be the length of each side if the string is used to form

(a) A regular hexagon?

(b) A square?

(c) An equilateral triangle?

Ans:

(a) Perimeter = 4 x Side

120 = 4 x Side

\(Side = \frac{120}{4} = 30\ cm\)

(b) Perimeter = 3 x Side

120 = 3 x Side

\(Side = \frac{120}{3} = 40\ cm\)

(c) Perimeter = 6 x Side

120 = 6 x Side

\(Side = \frac{120}{6} = 20\ cm\)

 

Q12: What is the third side of a triangle if two of its sides are 15 cm and 12 cm, whose perimeter is 40 cm?

 Ans:

Perimeter of triangle = Sum of all sides of the triangle

40 = 15 + 12 + Side

40 = 27 + Side

Side = 40 – 27 = 13 cm

Hence, the third side of the triangle is 13 cm.

 

Q13: A square park of side 240m is to be fenced. Find the cost of fencing if the rate of fencing per meter is Rs.10.

Ans:

Length of fence required = Perimeter of the square park

= 4 x Side = 4 x 240 = 960 m

Cost for fencing 1 m of square park = Rs 10

Cost for fencing 960 m of square park = 960 x 10 = Rs 9600

 

Q14: A rectangular park of length 150 and breadth 175 is to be fenced. Find the cost if the fencing rate is Rs.15 per meter.

Ans:

Length of rectangular park (I) = 150 m

Breadth of rectangular park (b) = 175 m

Length of wire required for fencing the park = Perimeter of the park

= 2 x (/ + b) = 2 x (150 + 175) = 2 x 325 = 650 m

Cost for fencing 1 m of the park = Rs 15

Cost for fencing 650 m of the square park = 650 x 15 = Rs 9750

 

Q15. Adam runs around a square park of side 80 m. Eve runs around a rectangular park with length 70 m and breadth 40 m. Who covers a comparatively lesser distance?

Ans:

Distance covered by Adam = 4 x Side of square park

= 4 X 80 = 240 m

Distance covered by Eve = 2 x (70 + 40)

= 2 x 110 = 220 m

Therefore, Eve covers less distance.

 

Q16: What is the perimeter of each of the following figures? What do you infer from the answers?

9

Ans:

(a) Perimeter of square = 4 x 50 = 200 cm

(b) Perimeter of rectangle = 2 x (20 + 80) = 200 cm

(c) Perimeter of rectangle = 2 x (40 + 60) = 200 cm

(d) Perimeter of triangle = 60 + 60 + 80 = 200 cm

It can be inferred that all the figures have the same perimeter.

 

Q17: Ram buys 9 square slabs, each having a side of 2 m. Initially, he lays them in the form of a square.

10

(a) Looking at the figure, find out the perimeter of his arrangement.

(b) His friend Mani looked at the figure and asked him to lay them out in the shape of a cross. Looking at the second figure, find out the perimeter of the cross.

(c) Out of the two shapes, find out which one has a larger perimeter.

(d) Mani suggests that there’s a way to get an even greater perimeter. Can you guess a way of doing it?   (The paving slabs must meet along complete edges. That is, they cannot be broken.)

Ans:   

(a) Side of the square = (3 x 2) m = 6 m

Perimeter of the square = (4 x 3) m = 12 m

(b) Perimeter of the cross = 2+4+4+2+4+4+2+4+4+2+4+4 = 40m

(c) The cross formation has the greatest perimeter of all arrangement.

(d) Formation with perimeter which is greater than 40 m can’t be determined.

 

 

EXERCISE-10.2

Q1. Count the squares and find out the area of the following diagrams:

11

Ans:      

(a) The figure contains 9 fully filled squares only. Therefore, the area of this figure will be 9 square units.

(b) The figure contains 5 fully filled squares only. Therefore, the area of this figure will be S square units.

(c) The figure contains 2 fully filled squares and 4 half-filled squares. Therefore, the area of this figure will be 4 square units.

(d) The figure contains 8 fully filled squares only. Therefore, the area of this figure will be 8 square units.

(e) The figure contains 10 fully filled squares only. Therefore, the area of this figure will be 10 square units.

(f) The figure contains 2 fully filled squares and 4 half-filled squares. Therefore, the area of this figure will be 4 square units.

(g) The figure contains 4 fully filled squares and 4 half-filled squares. Therefore, the area of this figure will be 6 square units.

(h) The figure contains 5 fully filled squares only. Therefore, the area of this figure will be S square units.

(I) The figure contains 9 fully filled squares only. Therefore, the area of this figure will be 9 square units.

(j) The figure contains 2 fully filled squares and 4 half-filled squares. Therefore, the area of this figure will be 4 square units.

(k) The figure contains 4 fully filled squares and 2 half-filled squares. Therefore, the area of this figure will be 5 square units.

(l) From the given figure, it can be observed that,

Covered Area Number Area estimate ( sq units )
Fully filled squares 2 2
Half filled squares
More than half filled squares 6 6
Less than half filled squares 6 0

Total area = 6 + 2 = 8 sq units.

(m) From the above figure, it is observed that,

 

Covered Area Number Area estimate ( sq units )
Fully filled squares 5 5
Half filled squares
More than half filled squares 10 10
Less than half filled squares 9 0

Total area = 9 + 5 = 14 sq units

(n) From the above figure, it is observed that,

 

Covered Area Number Area estimate ( sq units )
Fully filled squares 8 8
Half filled squares
More than half filled squares 10 10
Less than half filled squares 9 0

Total area = 10 + 8 = 18 sq units.

 

EXERCISE-10.3

Q1. What is the area of the following rectangles whose sides’ measure:

(a) 2 cm and 5 cm

(b) 15 m and 20 m

(c) 4 km and 5 km

(d) 3 m and 60 cm

Ans:       It is known that, Area of a rectangle = Length (l) x Breadth (b)

(a) l = 2 cm

b = 5 cm

Area = l x b = 2 x 5 = 12 cm2

 

(b) l = 15 m

b= 20 m

Area = l x b= 20 x 15 = 300 m2

 

(c) l = 4 km

b= 5 km

Area = l x b = 4 x 5 = 20 km2

 

(d) l = 3 m

b = 60 cm = 0.60 m

Area = l x b=2 x 0.60 = 1.20 m2

 

Q2. Find the areas of the squares whose sides are: (a) 8 cm (b) 11 cm (c) 6 m

Ans:

It is known that, Area of a square = (Side)2

(a) Side = 8 cm Area = (8)2 =64 cm2

(b) Side = 11 cm Area = (11)2 = 121 cm2

(c) Side = 6 m Area = (6)2 = 36 m2

 

Q3. Which one has the largest and smallest area among the 3 rectangles whose dimensions measure:

(a) 5 m by 6 m

(b) 12m by 4 m

(c) 2 m by 10 m

Ans:       We know that, Area of rectangle = Length x Breadth

(a) l = 5m

b = 6m

Area = l x b = 5 x 6 = 30 m2

 

(b) l = 12m

b = 4m

Area = l x b = 12 x 4 = 48 m2

 

(c) l = 2m

b = 10m

Area = l x b = 2 x 10 = 20 m2

From the above results, we come to know that (b) has the largest area and rectangle (c) has the smallest area.

 

Q5: What is the cost of tiling a rectangular plot of land 400 m long and 300 m wide at the rate of Rs 5 per hundred sq in?

Ans:

Area of rectangular plot = 400 x 300 = 120000 m2

Cost of tiling per 100 m2 = Rs 5

Cost of tiling per 120000 m2 = \(\frac{5}{100}\times 120000 = Rs.6000\)

 

Q6: What is the area of table whose sides measure 1m 20 cm in length and 2m 50 cm in breadth?

Ans:      

Length (l) = 1 m 20cm = \((1+\frac{20}{100})\ m = 1.2\ m\)

Breadth (b) 2 m 50 cm = \((2+\frac{50}{100})\ m = 2.5\ m\)

Area = l x b = 1.2 x 2.5 = 3 m2

 

Q7: How many sq meters of carpet is needed to cover the floor of a room of sides measuring 3 m in length and 4 m 20cm in breadth?

Ans:

Length (l) = 3 m

Breadth (b) = 4 m 20 cm = \((4+\frac{20}{100})\ m = 4.2\ m\)

Area = l x b = 3 x 4.2 = 12.6 m2

 

Q8: A square carpet is laid on the floor of sides measuring 4 m in a floor of sides 4 m and 5 m. Find the remaining area of the floor that is not carpeted.

Ans:

Length (I) = 4 m

Breadth (b) = 5 m

Area of floor = l x b = 4 x 5 = 20 m2

Area covered by the carpet = (Side) 2 = (4)2 = 16 m2

Area not covered by the carpet = 20 – 16 = 4 m2

 

Q9: From a piece of land measuring 4 m long and 6 m wide, 4 sq flower beds of side 2m each are dug out. Find out the remaining area of the land.

Ans:

Area of the land = 4 x 6 = 24 m2

Area occupied by 4 flower beds = 4 x (Side)2  = 4x (2)2 = 16m2

Area of the remaining part = 24 – 16 = 8 m2

 

Q10: Split the below figures into multiple rectangles and find out their area. All dimensions are given in centimeters.

12

Ans:

(a) The above figures can be split in to multiple rectangles as follows:

13

Area of 1st rectangle = 8 x 4 = 32 cm2

                                                Area of 2nd rectangle = 12 x 2 = 24 cm2

Area of 3rd rectangle = 6 x 4 = 24 cm2

Area of 4th rectangle = 8 x 4 = 32 cm2

Total area of the whole figure = 32 + 24 + 24 + 32 = 112 cm2

(b) The above figures can be split in to multiple rectangles as follows:

Area of 1st rectangle = 6 x 2 = 12 cm2

                                                Area of 2nd rectangle = 6 x 2 = 12 cm2

Area of 3rd rectangle = 6 x 2 = 12 cm2

Total area of the whole figure = 12 + 12 + 12 = 36 cm2

 

Q11.  Split the below figures into multiple rectangles and find out their area. All dimensions are given in centimeters.

14

Ans:      

15

(a) The above figures can be split in to multiple rectangles as follows:

Area of 1st rectangle = 24 x 4 = 96 cm2

                                                Area of 2nd rectangle = 16 x 4 = 64 cm2

Total area of the whole figure = 96 + 64 = 160 cm2

(b) The above figures can be split in to multiple rectangles as follows:

16

Area of 1st rectangle = 42 x 14 = 588 cm2

                                                Area of 2nd rectangle = 14 x 14 = 196 cm2

Area of 3rd rectangle = 14 x 14 = 196 cm2

Total area of the whole figure = 588 + 196 + 196 = 980 cm2

(c) The above figures can be split in to multiple rectangles as follows:

17

Area of 1st rectangle = 10 x 2 = 20 cm2

                                                Area of 2nd rectangle = 8 x 2 = 16 cm2

Total area of the whole figure = 20 + 16 = 36 cm2

 

Q12. Tiles of sides measuring 10 cm by 4 cm each are to be fit in a rectangular region whose length and breadth are respectively:

(a) 144 cm and 100 cm

(b) 36 cm and 70 cm

Ans.       (a) Total area of the region = 144 x 100 = 14400 cm2

Area of one tile = 10 x 4 = 40 cm2

Number of tiles required = \(\frac{14400}{40}\) = 360

Therefore, 360 tiles are required.

(b) Total area of the region = 36 x 70 = 2520 cm2

Area of one tile = 40 cm2

Number of tiles required = \(\frac{2520}{40}\) = 63

Therefore, 63 tiles are required.

You can find the entire list of NCERT solutions for class 6 and this chapter deals with the different concepts of mensuration and measurements. You can also check out the NCERT solutions of class 6 maths which deals with the variety of different solutions for the different chapters of mathematics. Some of the major chapters that deals with the different chapters such as Practical Geometry, Symmetry, Ratio and Proportion, Algebra, Mensuration and Data Handling. These chapters contain numerous difficulties and problem sets which a student can learn through to solve these problems to the best of their abilities. It is important to be well versed in these topics to score the best marks in the exam. These exercises are important to students to prepare themselves for the main exams. Thus, this chapter 10 contains a variety of solutions for the different problems, to prepare themselves for the exam.