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.

### NCERT Solutions For Class 6 Maths Chapter 10 Exercises

- NCERT Solutions For Class 6 Maths Chapter 10 Mensuration Exercise 10.1
- NCERT Solutions For Class 6 Maths Chapter 10 Mensuration Exercise 10.2
- NCERT Solutions For Class 6 Maths Chapter 10 Mensuration Exercise 10.3

*Exercise 10.1 *

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

(a)

**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)

**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)

**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)

**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)

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.

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.**

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.**

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?**

**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.**

(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:**

**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 cm^{2}

(b) *l* = 15 m

*b*= 20 m

Area = *l* x *b*= 20 x 15 = 300 m^{2}

(c) *l* = 4 km

*b*= 5 km

Area = l x b = 4 x 5 = 20 km^{2}

(d) *l* = 3 m

*b* = 60 cm = 0.60 m

Area = l x b=2 x 0.60 = 1.20 m^{2}

*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 cm^{2}

(b) Side = 11 cm Area = (11)^{2 }= 121 cm^{2}

(c) Side = 6 m Area = (6)^{2} = 36 m^{2}

*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 m^{2}

(b) l = 12m

b = 4m

Area = l x b = 12 x 4 = 48 m^{2}

(c) l = 2m

b = 10m

Area = l x b = 2 x 10 = 20 m^{2}

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 m^{2}

Cost of tiling per 100 m^{2} = Rs 5

Cost of tiling per 120000 m^{2} = \(\frac{5}{100}\times 120000 = Rs.6000\)

*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 m^{2}

* *

*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 m^{2}

*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 m^{2}

Area covered by the carpet = (Side)^{ 2} = (4)^{2} = 16 m^{2}

Area not covered by the carpet = 20 – 16 = 4 m^{2}

*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 m^{2}

Area occupied by 4 flower beds = 4 x (Side)^{2} = 4x (2)^{2} = 16m^{2}

Area of the remaining part = 24 – 16 = 8 m^{2}

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

*Ans: *

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

Area of 1^{st} rectangle = 8 x 4 = 32 cm^{2}

^{ }Area of 2^{nd} rectangle = 12 x 2 = 24 cm^{2}

Area of 3^{rd} rectangle = 6 x 4 = 24 cm^{2}

Area of 4^{th} rectangle = 8 x 4 = 32 cm^{2}

Total area of the whole figure = 32 + 24 + 24 + 32 = 112 cm^{2}

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

Area of 1^{st} rectangle = 6 x 2 = 12 cm^{2}

^{ }Area of 2^{nd} rectangle = 6 x 2 = 12 cm^{2}

Area of 3^{rd} rectangle = 6 x 2 = 12 cm^{2}

Total area of the whole figure = 12 + 12 + 12 = 36 cm^{2}

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

*Ans: *

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

Area of 1^{st} rectangle = 24 x 4 = 96 cm^{2}

^{ }Area of 2^{nd} rectangle = 16 x 4 = 64 cm^{2}

Total area of the whole figure = 96 + 64 = 160 cm^{2}

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

Area of 1^{st} rectangle = 42 x 14 = 588 cm^{2}

^{ }Area of 2^{nd} rectangle = 14 x 14 = 196 cm^{2}

Area of 3^{rd} rectangle = 14 x 14 = 196 cm^{2}

Total area of the whole figure = 588 + 196 + 196 = 980 cm^{2}

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

Area of 1^{st} rectangle = 10 x 2 = 20 cm^{2}

^{ }Area of 2^{nd} rectangle = 8 x 2 = 16 cm^{2}

Total area of the whole figure = 20 + 16 = 36 cm^{2}

**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 cm^{2}

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

Therefore, 360 tiles are required.

(b) Total area of the region = 36 x 70 = 2520 cm^{2 }

Area of one tile = 40 cm^{2}

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

Therefore, 63 tiles are required.