NCERT Solutions Class 6 Maths Chapter 10

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

Ans-

(a) Perimeter of figure = 4 cm + 2 cm + 1 cm
+ 5 cm
= 12 cm
(b) Perimeter of figure = 23 cm + 35 cm + 35 cm
+ 40 cm
= 133 cm
(c) Perimeter of figure = 15 cm + 15 cm + 15 cm
+ 15 cm
= 60 cm
(d) Perimeter of figure = 4 cm + 4 cm + 4 cm
+ 4 cm + 4 cm
= 20 cm

(e) Perimeter of figure = 2.5 cm + 0.5 cm + 4 cm
+ 1 cm + 4 cm + 0.5 cm
+ 2.5 cm
= 15 cm
(f) Perimeter of figure = 4 cm + 1 cm + 3 cm
+ 2 cm+ 3 cm+ 4 cm+ 1 cm + 3 cm
+ 2 cm + 3 cm + 4 cm + 1 cm + 3 cm
+ 2 cm + 3 cm + 4 cm + 1 cm + 3 cm
+ 2 cm + 3 cm
= 52 cm

Q.2 The lid of a rectangular box of sides 40 cm by 10 cm is sealed all round with tape. What is the length of the tape required?

Ans-

Since, lid of rectangular box is like a rectangle. So,

Length of tape = 2(length + Breadth)
= 2(40 + 10)
= 100 cm
= 1 m

Q.3 A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top?

Ans-

Length of table-top = 2 m 25 cm

= 2.25 m
Breadth of table-top = 1 m 50 cm

= 1.5 m
Perimeter of table-top = 2(Length + Breadth)

= 2(2.25 + 1.50)
= 2(3.75)
= 7.50 m

Q.4 What is the length of the wooden strip required to frame a photograph of length and breadth 32 cm and 21 cm respectively?

Ans-

Length of photograph = 32 cm
Breadth of photograph = 21 cm
Perimeter of photograph = 2(Length + Breadth)
= 2(32 + 21)
= 106 cm

Q.5 A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What is the length of the wire needed?

Ans-

Length of rectangular piece of land = 0.7 km
Breadth of rectangular piece of land = 0.5 km
Perimeter of rectangular piece of land = 2(0.7 + 0.5)
= 2(1.2) km
= 2.4 km

Length of required wire = 4(2.4) km
= 9.6 km
[Since, Length of required wire = 4(Perimeter of land)]

Q.6 Find the perimeter of each of the following shapes:
(a) A triangle of sides 3 cm, 4 cm and 5 cm.
(b) An equilateral triangle of side 9 cm.
(c) An isosceles triangle with equal sides 8 cm each and third side 6 cm.

Ans-

(a) Perimeter of triangle = 3 cm + 4 cm + 5 cm
= 12 cm
(b) Perimeter of equilateral triangle
= 9 cm + 9 cm + 9 cm
= 27 cm
(c) Perimeter of isosceles triangle
= 8 cm + 8 cm + 6 cm
= 22 cm

Q.7 Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.

Ans-

Perimeter of triangle = 10 cm + 14 cm + 15 cm
= 39 cm

Q.8 Find the perimeter of a regular hexagon with each side measuring 8 m.

Ans-

Perimeter of regular hexagon = 6(Length of side)
= 6(8 cm)
= 48 cm

Q.9 Find the side of the square whose perimeter is 20 m.

Ans-

$\begin{array}{l}\mathrm{Perimeter}\text{of square}=\text{4}\times \text{side}\\ \text{\hspace{0.17em}\hspace{0.17em}Side of square}=\frac{\mathrm{Perimeter}}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}Side of square}=\frac{20\mathrm{m}}{4}\\ =5\mathrm{m}\end{array}$

Q.10 The perimeter of a regular pentagon is 100 cm. How long is its each side?

Ans-

$\begin{array}{l}\mathrm{Perimeter}\text{of regular pentagon}=\text{5}\times \text{side}\\ \text{\hspace{0.17em}Side of regular pentagon}=\frac{\mathrm{Perimeter}}{5}\\ \text{\hspace{0.17em}\hspace{0.17em}Side of regular pentagon}=\frac{100}{4}\mathrm{cm}\\ =25\mathrm{cm}\end{array}$

Q.11 A piece of string is 30 cm long. What will be the length of each side if the string is used to form :
(a) a square?
(b) an equilateral triangle?
(c) a regular hexagon?

Ans-

$\begin{array}{l}\text{Length of a piece of string}=\text{3}0\text{m}\mathrm{}\\ \left(\text{a}\right)\mathrm{Perimeter}\text{of square}=\text{Length of a piece of string}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} 4}\times \mathrm{side}\text{of square}=\text{30 m}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}side of square}=\frac{30}{4}\mathrm{m}\\ =7.5\mathrm{m}\\ \left(\text{b}\right)\mathrm{Perimeter}\text{of an equilateral triangle}\\ =\text{Length of a piece of string}\\ \text{3}\times \mathrm{side}\text{of equilateral triangle}\\ =\text{30 m}\\ \text{Side of equilateral triangle}\\ =\frac{30}{3}\mathrm{m}\\ =10\mathrm{m}\\ \left(\text{c}\right)\mathrm{Perimeter}\text{of a regular hexagon}\\ =\text{Length of a piece of string}\\ \text{6}\times \mathrm{side}\text{of regular hexagon}\\ =\text{30 m}\\ \text{Side of regular hexagon}\\ =\frac{30}{6}\mathrm{m}\\ =5\mathrm{m}\end{array}$

Q.12 Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is its third side?

Ans-

Let third side of triangle be x cm.
Perimeter of triangle = 12 cm + 14 cm + x cm
36 cm = 26 cm + x
x = 36 cm – 26 cm
= 10 cm
Thus, third side of triangle is 10 cm.

Q.13 Sweety runs around a square park of side 75m. Bulbul runs around a rectangular park with length 60 m and breadth 45 m. Who covers less distance?

Ans-

Side of square park = 75 m
Perimeter of square park = 4 (75 m)
= 300 m
Distance covered by Sweety =300 m
Length of park = 60 m
Breadth of park = 45 m
Perimeter of rectangular park = 2 (60 m + 45 m)
= 210 m
Distance covered by Bulbul = 210 m
Thus, Bulbul covers less distance.

Q.14 What is the perimeter of each of the following figures? What do you infer from the answers?

Ans-

(a) Perimeter of square = 4(25 cm)
= 100 cm
(b) Perimeter of rectangle = 2(30 cm + 20 cm)
= 2(50 cm)
= 100 cm
(c) Perimeter of rectangle = 2(40 cm + 10 cm)
= 2(50 cm)
= 100 cm
(d) Perimeter of triangle = 30 cm + 30 cm+ 40 cm
= 100 cm

Q.15

Ans-

$\begin{array}{l}\mathrm{Length}\text{of each side of square paving}=\frac{1}{2}\mathrm{m}\\ \mathrm{Number}\text{of square paving}=\text{9}\\ \text{When square paving laid in the form of square,}\\ \text{Number of square paving in each side of square form}\\ \text{3}\\ \text{So, the length of each side of new square}\\ =3\times (\mathrm{side}\mathrm{of}\mathrm{square}\mathrm{paving})\\ =3\times \frac{1}{2}\\ =\frac{3}{2}\\ \left(\mathrm{a}\right)\mathrm{Perimeter}\text{of new square}=4\times \mathrm{side}\\ =4\times \frac{3}{2}\\ =6\mathrm{m}\\ \left(\mathrm{b}\right)\mathrm{Perimeter}\text{of fig}\left(\mathrm{ii}\right)=4\times (\mathrm{side}\mathrm{having}\mathrm{single}\mathrm{paving})\\ +8\times \left(\text{side having double}\mathrm{paving}\right)\\ =4\times \frac{1}{2}+8\times \frac{1}{2}\times 2\\ =2+8\\ =10\mathrm{m}\\ \left(\mathrm{c}\right)\text{Cross figure i.e., fig}\left(\mathrm{ii}\right)\text{has greater perimeter.}\\ \left(\mathrm{d}\right)\text{Yes, if we put all paving in a straight line. It forms}\\ \text{a rectangle whose dimensions are:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Length}=9\times \frac{1}{2}\\ =4.5\mathrm{cm}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Breadth}=\frac{1}{2}\text{cm}\\ =\text{0.5\hspace{0.17em}cm}\\ \mathrm{Perimeter}\text{of rectangle}=2(\mathrm{L}+\mathrm{B})\\ =2(4.5+0.5)\\ =10\text{\hspace{0.17em}}\mathrm{cm}\end{array}$

Q.16 Find the areas of the following figures by counting square:

Ans-

(a) Number of squares in given figure = 9
So, area of given figure = 9 Sq. unit

(b) Number of squares in given figure = 5
So, area of given figure = 5 Sq. unit

(c) Number of squares in given figure = 2 + 4(1/2)

= 4
So, area of given figure = 4 Sq. unit

(d) Number of squares in given figure = 8
So, area of given figure = 8 Sq. unit

(e) Number of squares in given figure = 10
So, area of given figure = 10 Sq. unit

(f) Number of squares in given figure = 2 + 4(1/2)

= 4
So, area of given figure = 4 Sq. unit

(g) Number of squares in given figure = 4 + 4(1/2)

= 6
So, area of given figure = 6 Sq. unit

(h) Number of squares in given figure = 5
So, area of given figure = 5 Sq. unit

(i) Number of squares in given figure = 9
So, area of given figure = 9 Sq. unit

(j) Number of squares in given figure = 2 + 4(1/2)

= 4
so, area of given figure = 4 Sq. unit

(k) Number of squares in given figure = 4 + 2(1/2)

= 5
So, area of given figure = 5 Sq. unit

(l)

So, the area of given figure = 2+1+5
= 8 square cm.

(m)

So, the area of given figure = 7+1+6
=14 square cm.

(n)

So, the area of given figure = 9+2+7
=18 square cm.

Q.17 Find the areas of the rectangles whose sides are:
(a) 3 cm and 4 cm (b) 12 m and 21 m
(c) 2 km and 3 km (d) 2 m and 70 cm

Ans-

$\begin{array}{l}\because \mathrm{Area}\text{of rectangle}=\mathrm{Length}\times \mathrm{Breadth}\\ \left(\mathrm{a}\right)\mathrm{Area}\text{of rectangle}=3\mathrm{cm}\times 4\mathrm{cm}\\ =12{\mathrm{cm}}^2\\ \left(\mathrm{b}\right)\mathrm{Area}\text{of rectangle}=12\mathrm{cm}\times 21\mathrm{cm}\\ =252{\mathrm{cm}}^2\\ \left(\mathrm{c}\right)\mathrm{Area}\text{of rectangle}=2\mathrm{km}\times 3\mathrm{km}\\ =6{\mathrm{km}}^2\\ \left(\mathrm{d}\right)\mathrm{Length}\text{of rectangle}=2\mathrm{m}\\ =200\mathrm{cm}\\ \mathrm{Area}\text{of rectangle}=200\mathrm{cm}\times 70\mathrm{cm}\\ =14000{\mathrm{cm}}^2\end{array}$

Q.18 Find the areas of the squares whose sides are:

(a) 10 cm (b) 14 cm (c) 5 m

Ans-

Area of square = (side of square)²
(a) Area of square = (10)²

= 100 cm²
(b) Area of square = (14)²
= 196 cm²
(c) Area of square = (5)²

= 25 cm²

Q.19 The length and breadth of three rectangles are as given below:
(a) 9 m and 6 m
(b) 17 m and 3 m
(c) 4 m and 14 m
Which one has the largest area and which one has the smallest?

Ans-

$\begin{array}{l}\mathrm{Area}\text{of rectangle}=\mathrm{Length}\times \mathrm{Breadth}\\ \left(\mathrm{a}\right)\mathrm{Area}\text{of rectangle}=9\mathrm{m}\times 6\mathrm{m}\\ =54{\mathrm{m}}^2\\ \left(\mathrm{b}\right)\mathrm{Area}\text{of rectangle}=17\mathrm{m}\times 3\mathrm{m}\\ =51{\mathrm{m}}^2\\ \left(\mathrm{c}\right)\mathrm{Area}\text{of rectangle}=4\mathrm{m}\times 14\mathrm{m}\\ =56{\mathrm{m}}^2\\ \text{Rectangle}\left(\text{c}\right)\text{has the largest area and rectangle}\left(\mathrm{b}\right)\text{has the}\\ \text{smallest area.}\end{array}$

Q.20 The area of a rectangular garden 50 m long is 300 sq m. Find the width of the garden.

Ans-

$\begin{array}{c}\text{Area of rectangular garden}=\text{3}00\text{sq m}\\ \text{Length of rectangular garden}=50\mathrm{m}\\ \mathrm{Breadth}\text{of rectangular garden}=\frac{\mathrm{Area}\text{of garden}}{\mathrm{Length}\text{of garden}}\\ =\frac{300}{50}\\ =6\mathrm{m}\end{array}$

Q.21 A table-top measures 2 m by 1 m 50 cm. What is its area in square metres?

Ans-

Length of table top = 2 m

Breadth of table top = 1 m 50 cm

= (1 + 0.50) m

= 1.5 m

Area of table-top = Length x Breadth

= 2 m x 1.5 m

= 3 sq m

Q.22 A room is 4 m long and 3 m 50 cm wide. How many square metres of carpet is needed to cover the floor of the room?

Ans-

Length of room = 4 m

Breadth of room = 3 m 50 cm

= (3 + 0.50) m

= 3.5 m

Area of floor = Length x Breadth

= 4 m x 3.5 m

= 14 sq m
Therefore, 14 sq m carpet is required to cover the floor of the room.

Q.23 A floor is 5 m long and 4 m wide. A square carpet of sides 3 m is laid on the floor. Find the area of the floor that is not carpeted.

Ans-

$\begin{array}{c}\mathrm{Area}\text{of rectangular floor}=5\mathrm{m}\times 4\mathrm{m}\\ =20{\mathrm{m}}^2\\ \mathrm{Length}\text{of side of square carpet}=3\mathrm{m}\\ \mathrm{Area}\text{of square carpet}={\left(3\mathrm{m}\right)}^2\\ =9{\mathrm{m}}^2\\ \mathrm{Area}\text{of floor that is not carpeted}=20-9\\ =11{\mathrm{m}}^2\end{array}$

Q.24 Five square flower beds each of sides 1 m are dug on a piece of land 5 m long and 4 m wide. What is the area of the remaining part of the land?

Answer

Area of rectangular land = 5m x 4m

= 20 m²
Area of 1 flower bed = (1m)²
= 1m²
Area of 5 flower beds = 5m²
Area of remaining part of land= 20 – 5
= 15 m²

Q.25 By splitting the following figures into rectangles, find their areas (The measures are given in centimetres).

Ans-

(a)Area of given figure = 3 × 3 + 1 × 2 +3 × 3
+ 2 × 4
= 9 + 2 + 9 + 8

= 28 cm²
(b) Area of given figure = 3 × 1 + 3 × 1 +3 × 1
= 3 + 3 + 3

= 9 cm²

Q.26 Split the following shapes into rectangles and find their areas. (The measures are given in centimeters)

Ans-

(a) Area of given figure = 12 x 2 + 8 x 2

= 24 + 16

= 40 cm²
(b) Area of given figure = 7 x 7 + 7 x 21 +7 x 7
= 49 + 147 + 49

= 245 cm²

(c) Area of given figure = 1 x 5 + 4 x 1
= 5 + 4

= 9 cm²

Q.27 How many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed to fit in a rectangular region whose length and breadth are respectively:

(a) 100 cm and 144 cm

(b) 70 cm and 36 cm.

Ans-

$\begin{array}{l}\left(\mathrm{a}\right)\text{Number of bricks in wall}=\frac{\mathrm{Area}\text{of rectangular region}}{\mathrm{Area}\text{of surface of brick}}\\ =\frac{100\times 144}{12\times 5}\\ =20\times 12\\ =240\\ \left(\mathrm{b}\right)\text{Number of bricks in wall}=\frac{\mathrm{Area}\text{of rectangular region}}{\mathrm{Area}\text{of surface of brick}}\\ =\frac{70\times 36}{12\times 5}\\ =14\times 3\\ =42\end{array}$

Q.28 Find the cost of fencing a square park of side 250 m at the rate of ₹20 per metre.

Ans-

Side of square park = 250 m
Rate of fencing the park = ₹20 per metre
Perimeter of square park = 4(250 m)
= 1000 m
Cost of fencing the square park

= ₹20 x 1000 m
= ₹20,000
Thus, the cost of fencing the square park is ₹20,000

Q.29 Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of ₹12 per metre.

Ans-

Length of a rectangular park = 175 m
Breadth of a rectangular park = 125 m
Rate of fencing per metre = ₹12
Length of fencing of park = 2(L + B)
= 2(175+ 125) m
= 2(300) m
= 600 m
Cost of fencing the park = ₹12(600 m)
= ₹7200

Q.30 What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of ₹8 per hundred sq m.?

Answer

Area of rectangular plot = 500 m × 200 m

= 1, 00, 000 sq m
Rate of tilling per sq m = ₹8 per hundred
= ₹(8/100)
Cost of tilling the plot = ₹(8/100) × 1,00,000

= ₹8000