Unit 16 Congruent Figures
16.1 Congruent figures – Introduction
Let's observe the following Pair of figures
a) Do the line segments AB and PQ b) Do ∠MON and ∠XOY seem to be
seem to be equal in length? equal?
NY
A BP Q MO
O
X
c) Are ∆ABC and ∆DEF exactly d) Are these two rectangles identical
the same? to each other?
AD A DP
2.5 cm S
2.5 cm
4 cm
4 cm
1cm
1cm
B 2 cm C Q
B 3 cm C E 3 cm F 2 cm R
The above pair of figures are seemed to be the same in shape and size. They are
identical or twins. Such figures or object are said to be congruent if they have exact-
ly the same shape and size.
Let's take a rectangular sheet of paper. Fold it DD' C D D' C
along one of its diagonals and cut the folding edge A B B'
C D'
into two right angled triangular pieces. Place the BB'
triangular pieces one above another as shown in A A D
the figure. Now, answer these questions. B' B
Are the triangular pieces exactly fitted to one above another ?
Which are the angles exactly fitted ?
Which are the arms exactly fitted ?
Could you investigate some ideas about congruent triangles ?
249Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Congruent Figures
The figures which are exactly the same in shape and size are called congruent
figures.
In the adjoining figures, rectangles D 3cm C S 3cm R
ABCD and PQRS are congruent 2cm 2cm 2cm 2cm
rectangles because their shapes and A 3cm B P 3cm Q
sizes are exactly the same. In this
case, each part of rectangle ABCD can exactly be fitted over the corresponding parts
of rectangle PQRS.
On the other hand, in each pair of
figures given alongside, first figure
cannot be fitted exactly over the
second figure. Therefore, they are (pair 1) (pair 2)
not congruent figures. However, each
pair of figures are similar figures. Similar figures have the same shapes, however,
they have different sizes.
EXERCISE 16.1
General Section - Classwork
1. Which of the following pairs of figures are congruent or similar? Tell and
write "congruent or similar" in the blank space.
A RC Z B These line segments are
SW Y D .................................................
a)
RX These triangle are
C .................................................
A
These squares are
P .................................................
b)
Q
B
P
c)
Q
d) These circles are
.................................................
Vedanta Excel in Mathematics - Book 7 250 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Congruent Figures
e) These figures are
.................................................
f) These squares are
.................................................
Creative Section
2. Answer the following questions.
a) What types of figures are called congruent figures?
b) What types of figures are called similar figures?
c) Are all congruent figures similar?
d) Are all similar figures congruent?
e) Are the surfaces of wall and white-board congruent or similar?
f) Are the surfaces of the pair of door panels of your classroom congruent or
similar?
3. Let's name the congruent and non-congruent figures in the following pairs of
figures. (Example: DABC and DDEF are congruent triangles.)
a) b) P W
AD
4 cm 4 cm 4 cm 4 cm 4 cm 6 cm 2 cm 5 cm
Q 2 cm
X Y
R
B 4 cm CE 4 cm F 3 cm
K S G
4 cm 4 cm 6 cm
c) P 5 cm O d)
7 cm
10 cm
4 cm
6 cm 6 cm
3 cm
N
L 8 cm M T 7 cm RE G
e) 2 cm C H 2 cm f) S 2 cm R
D G 4 cm Y
Z
2 cm
2 cm
2 cm
2 cm
4 cm
4 cm
1 cm
1 cm
A 2 cm B E 2 cm F W 4 cm X
P 2 cm Q
251Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Congruent Figures
4. Let's write the length of unknown sides of the following pairs of congruent
figures.
a) P Z b) C G
10 cm
3 cm 2 cm 3 cm 2 cm 6 cm
6 cm
Q R X Y A 8 cm B E 8 cm F
4 cm
XY = ? BC = ?
c) CN 5 cm d) R E 2 cm G E 6 cm S
D 5 cm M
3 cm
3 cmA 5 cm 4 cm
3 cm 4 cm
4 cm
4 cm
T 6 cm I 3 cm O 2 cm
B K 5 cm L HR
KN = ? RE = ?
It's your time - Project work!
5. a) Let's take a square sheet of paper and fold it through one of its diagonals.
(i) Does the diagonal divide a square into two congruent triangles?
Again, fold the same square paper through another diagonal and cut out all
four triangles so formed.
(ii) Do two diagonals of a square divide it into four congruent triangles?
b) Take a rectangular sheet of paper and fold it through one of its diagonals.
(i) Does the diagonals divide a rectangle into two congruent triangles?
Again, fold the same rectangular paper through another diagonal and cut out
all 4 triangles so formed.
(ii) Do two diagonals of a rectangle divide it into four congruent triangles?
c) Write a short report about your findings on the activities a) and b). Then,
present in the class.
d) Let's look around your classroom. List any three different objects which have
congruent surfaces.
Vedanta Excel in Mathematics - Book 7 252 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Unit 17 Perimeter, Area and Volume
17.1 Perimeter, Area, and Volume – Looking back
1. Are the following questions related to perimeter, area, or volume? Let's write
'perimeter', 'area', or 'volume' in the blank spaces.
a) What is the length of wire required to fence a rectangular garden? ...........
b) How much paper do you need to cover a rectangular wall of your
classroom? ......
c) How much water can a stone displace from the measuring cylinder when it
is immersed into it? ...........
d) How much carpet do you need to cover the floor of a room? ...........
e) What distance do you cover by walking around a park? ...........
f) How much water can a cubical tank hold? ...........
2. Let's say and write the perimeters of these figures as quickly as possible.
a) b) 3 cm c)
b
5 cm 4 cm 2 cm 5 cm
6 cm 5 cm l
Perimeter = .............
Perimeter = ............. Perimeter = .............
3. Let's say and write the answers as quickly as possible.
a) Length of a rectangle is 5 cm and breadth is 2 cm, then perimeter is .................,
and area is .............
b) A square is 6 cm long, its perimeter is ........................, and area is ....................
c) A cube is 4 cm long, its area is ...................., and volume is .........................
17.2 Perimeter of plane figures
Triangle, rectangle, square, circle, etc. are the plane figures. The total length of the
boundary lines of plane figure is called its perimeter. Thus, perimeter is the distance
around a closed figure.
253Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Perimeter, Area and Volume
(i) Perimeter of triangle A
The perimeter of a triangle is the sum of the length of its
three sides. b cm
∴ Perimeter of ∆ABC (P) = AB + BC + CA = a + b + c c cm
Similarly, the semi-perimeter of a triangle is defined as B a cm C
half of its perimeter. It is denoted by the letter ‘s’.
∴ Semi-perimeter of ∆ABC (s) = a + b + c .
2
(ii) Perimeter of rectangle l
The opposite sides of a rectangle are equal. DC
So, the lengths AB = DC = l bb
the breadths BC = AD = b AlB
The perimeter of the rectangle ABCD
= AB + BC + CD + DA
= l + b + l + b = 2 l + 2 b = 2 (l + b)
(iii) Perimeter of square
In square, all four sides are equal. l
DC
So, AB = CD = CD = DA = l
ll
The Perimeter of square ABCD = AB + BC + CD + DA
A lB
= l+l+l+l
= 4l
∴ Perimeter of square (p) = 4l
17.3 Relation between circumference and diameter of circle
Let's draw three circles with radii 3 cm, 4 cm, and 5 cm respectively. Place three
pieces of strings along the circumference of each circle separately.
3 cm 4 cm 5 cm
Let's measure the length of each string separately with the help of a ruler.
Now, let's find the ratios of the length of circumference of each circle to its
diameter. We find that the ratio of circumference and diameter is the same for
every circle. This constant ratio is represented by the Greek letter ‘π' (Pie).
Thus, if the circumference of a circle is c and its diameter is d,
Vedanta Excel in Mathematics - Book 7 254 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Perimeter, Area and Volume
circumference =π Vedanta ICT Corner
diameter Please! Scan this QR code or
or, c browse the link given below:
or, d =π
https://www.geogebra.org/m/z5xzv6nr
c = πd
Also, diameter of a circle (d) = 2 × radius (r) = 2r
So, circumference of circle (c) = πd = π × 2r = 2πr
Perimeter of circle
The circumference of a circle is the perimeter of the circle.
∴ Perimeter of circle = πd or 2πr
Where, the approximate value of π is 3.142159..., which is approximately
equivalent to 22 .
7
Worked-out examples
Example 1: If the perimeter of a rectangular field of length 45 m is 154 m, find
its breadth.
Solution:
Here, the length of the field (l) = 45 m
perimeter of the field = 154 m
Now,p erimeter of the rectangular field = 154 m
or, 2 (l + b) = 154 m
or, 45 + b = 154 m
2
= 77 m
or, b = (77 – 45) m = 32 m
Hence, the required breadth of the field is 32 m.
Example 2: The radius of a circular pond is 42 m. Find the length of wire
required to fence it with 5 rounds. Also, find the cost of fencing at
Rs 27 per metre.
Solution:
Here, radius of the circular pond (r) = 42 m. Fencing 1 round is the perimeter of
the circular ground.
Its perimeter = 2πr Fencing 5 rounds = 5 × perimeter
= 2 × 22 × 42 m
7
= 264 m
∴ The required length of wire = 5 × 264 m = 1320 m.
Again, the required cost of fencing = 1320 × Rs 27 = Rs 35,640.
255Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Perimeter, Area and Volume
Example 3. The length of a rectangular park is two times of its breadth. The
total cost of fencing it with four rounds at the rate of Rs. 45 is
Rs 38,880. a) Find the perimeter of the park.
b) Find the length and breadth of the park.
Solution:
Let the breadth of the rectangular park be x m.
Then, the length of the park (l) = 2x m.
Rate of fencing (R) = Rs 45 per meter
Total cost of fencing (T) = Rs 38880
Now, Perimeter of park (P) = 2 (l + b)
= 2(2x + x)
= 2(3x) = 6x
∴ The required length of wire = 4 × 6x = 24x
Again,
Total cost of fencing (T) = length of wire × Rate (R)
or, 38880 = 24x × Rs 45
or, 38880 = 1080x
38880
or, x = 21x0=802 = 36 , 72
∴ l = × 36 =
Thus, the length and breadth of the park are 72m and 36m respectively.
Alternative process:
Cost of fencing with 4 rounds = Rs 38,880
Cost of fencing with 1 round = Rs 38,880 = Rs 9,720
1080
∴ Perimeter of the park = 9,720 m
45
or, 2(l + b) = 216 m
or, 2(2x + x) = 216
or, 3x = 216 = 108
2
108
or, x = 3 = 36 m
EXERCISE 17.1
General Section - Classwork
1. Tell and write the answers as quickly as possible.
a) Three sides of a triangle are x cm, y cm and z cm respectively. Its perimeter
is .......................................... and semi-perimeter is ....................................
b) If the length of a side of a square is x cm, its perimeter is ...............................
Vedanta Excel in Mathematics - Book 7 256 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Perimeter, Area and Volume
c) If the length and breadth of a rectangle are a cm and b cm respectively, its
perimeter is .........................................
d) If the radius of a circle is y cm, its perimeter is .............................
2. a) Sides of a triangle are 6 cm, 3 cm and 5 cm. Its perimeter is ..........................
and semi-perimeter is ................................
b) A squared garden is 15 m broad. Its perimeter is ............................... and
length of wire required to fence it with two rounds is ...............................
c) A rectangular ground is 20 m long and 10 m broad. Its perimeter is ...............
and length of wire required to fence it with one round is ...........................
d) If the radius of a circle is 7 cm, its perimeter is ...............
Creative Section - A
3. Let's find the perimeters of the following figures.
a) b) c)
4.1 cm 3.8 cm 3.6 cm 4.8 cm
5.6 cm 4.8 cm 9.2 cm 3.5 cm
4.6 cm
d) f) 2.5 cm 5.2cm
e)
3.5 cm 3.4 cm 1.6 cm
5.6 cm 4.7 cm
g) 1 cm
2.4 cm 2cm i)
5 cm 3cm
3 cm h) 14 cm
1 cm 3cm l) 10 cm
j) 1 cm 4cm
1 cm 2cm
1 cm
k)
3 cm 7 cm 7 cm
7 cm
6 cm 6 cm
1 cm
4. a) Find the perimeter of triangles, whose sides are
(i) 3.6 cm, 5.7 cm, 4.5 cm (ii) 6.3 cm, 8.7 cm, 5.5 cm
b) Find the perimeter of squares in which (i) l = 6.5 cm (ii) l = 7.4 cm.
c) Find the perimeter of rectangles in which
(i) l = 9.6 cm b = 6.4 cm (ii) l = 16.3 cm, b = 12.2 cm
d) Find the perimeter of circles in which (i) r = 7 cm (ii) d = 42 cm.
5. a) Find the perimeter of a triangular garden whose lengths are 15 m, 20.5 m, and
17.5 m. Also find its semi-perimeter.
257Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Perimeter, Area and Volume
b) Find the perimeter of an equilateral triangle whose one of the sides is 6.5 cm.
c) One of the two equal sides of an isosceles triangle is 5.4 cm. If the perimeter
of the triangle is 15.4 cm, find the length of remaining side of the triangle.
6. a) Chandrakala has a square garden of length 18m.
(i) Find its perimeter.
(ii) Find the length of wire required to fence it with three rounds.
(iii) If the rate of cost of fencing is Rs 25 per metre, find the total cost of
fencing.
b) Charimaya is running a race around a square track of length 75 m. Find the
distance covered by her at the end of her fifth round.
c) 540 m of wire is required to fence a square shaped of fish pond with three rounds.
(i) What is the length of wire required for one round ?
(ii) What is the perimeter of the pond ?
(iii) Find the length of the pond.
Creative Section - B
7. a) A rectangular ground is 25m long and 18m broad.
(i) Find its perimeter.
(ii) How many metres does a girl run around the ground when she
completes five rounds?
b) Mrs. Kanchhi Tamang has a vegetable garden of length 30m and breadth
18m.
(i) Find its perimeter.
(ii) Find the length of wire required to fence it with 3 rounds.
(iii) Find the total cost of fencing at the rate of Rs 20 per metre
c) Mr. Dharmendra Yadav has a rectangular mango farm having length
1.4 km and width 600 m. If he wishes to fence the farm with four rounds
over the wall by sharp pointed fencing wire at the rate of Rs 50 per meter,
find the total cost of fencing.
8. a) The radius of a circular field is 63 m. Find the perimeter of the field. Also,
22
find the length of wire required to fence it with 5 rounds. (π = 7 )
b) If the diameter of a circular pond is 84 m, find the cost of fencing around
22
it with four rounds at Rs 12 per metre. (π = 7 ) 22
7
c) If the perimeter of a circular ground is 308 m, find its radius. (π = )
d) Find the diameter of a circular field whose circumference is 132 m.
9. a) Rohit constructed a rectangular pond of length 40 m for fish farming. He
spent Rs 36,000 to fence it with four rounds at Rs 75 per metre.
(i) Find the perimeter of the pond (ii) Find the breadth of the pond.
b) The length of a rectangular land is 10 m longer than that of its breadth.
The cost of fencing around it with three rounds at Rs. 50 per metre is
Rs 13,800. Find the length and breadth of the land.
Vedanta Excel in Mathematics - Book 7 258 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Perimeter, Area and Volume
10. a) A wire is in the shape of a rectangle. Its length is 30 cm and breadth is
20 cm. If the same wire is re–bent in the shape of square, what is the meas-
ure of each side?
b) A string is in the shape of a square. Its each side is 22 cm. If the same string
is re–bent into the shape rectangle of length 24 cm, find the breadth.
11. a) The following table shows the measure of some of floors of rectangular
shapes of dog pens.
Pen Length (l) Breadth (b)
A 12 m 2m
B 8m 3m
C 6m 4m
(i) Which pen would take most fencing?
(ii) Which pen would you like to minimize the cost of fencing?
b) Suppose that your father wishes to have a rectangular kitchen garden in
the house. The following table shows the possible length and breadth of
the garden with equal areas.
Garden Length (l) Breadth (b)
A 16 m 3m
B 12 m 4m
C 8m 6m
(i) Which garden would take most fencing?
(ii) Why did your father choose the garden with least perimeter?
It's your time – Project work!
12. a) Measure the length and breadth of the whiteboard in your classroom with
the help of a measuring tape. Find the length of a wooden frame required
to enclosed the board.
b) Let's take a measuring tape and measure the total length of boundaries of
your school ground or garden of your house and discuss with your friends
about the perimeter and cost estimation of fencing at the certain rates.
17.4 Area of plane figures
The plane surface enclosed by the boundary line of a plane closed figure is known
as its area. Area is measured in square unit. For example: mm2, cm2, m2, etc. are the
units of measurement of area.
(i) Area of rectangle
In the adjoining graph, the area of each square room is
considered as 1 cm2. So, the surface enclosed by the rectangle
is 20 cm2.
∴ Area of the rectangle = 20 cm2
i.e. 5 rooms along length × 4 rooms along breadth = 20 cm2
i.e. length × breadth = 20 cm2
Thus, area of the rectangle = length × breadth = l × b
259Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Perimeter, Area and Volume
(ii) Area of square
In the given graph, area of the square is 25 cm2 = 5 cm × 5 cm
Thus, in a square, its length and breadth are equal, i.e., l = b
Thus, area of square (A) = l × b = l × l = l2
Worked-out examples
Example 1: Find the area of the following figures.
a) b)
3.5 cm
8.4 cm 14 cm
Solution: Area of rectangle (A) = l × b = 8.4 cm × 3.5 cm = 29.4 cm2
a)
b) Area of square (A1) = l2 = (14 cm)2 = 196 cm2
Example 2: The perimeter of a square field is 88 m.
a) Find its length. b) Find its area.
c) Find the cost of growing grass in the field at Rs 40 per sq. m.
Solution:
a) The perimeter of the square field = 88 m
or, 4l = 88 m
or, l = 848m = 22 m
∴ The length of the field (l) = 22 m.
b) Area of the square field (A) = l2 = (22 m)2 = 484 m2
c) The cost of growing grass in 1 m2 = Rs 40
∴ The cost of growing grass in 484 m2 = 484 × Rs 40 Total cost = Area × rate
= Rs 19,360
Example 3: A rectangular room is twice as long as its breadth and its perimeter is 48 m.
a) Find its length and breadth. b) Find its area.
c) Find the cost of carpeting its floor at Rs 75 per sq. m.
Solution:
Let, the breadth (b) of the room be x m.
So, the length (l) of the room will be 2x m
a) The perimeter of the rectangular room = 48 m
or, 2 (l + b) = 48 m
Vedanta Excel in Mathematics - Book 7 260 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Perimeter, Area and Volume
or, 2 (2x + x) = 48 m
or, 6x = 48 m
48
or, x = 6 m = 8 m
∴ The breadth of the room (b) = x = 8 m.
The length of the room (l) = 2x = 2 × 8 m = 16 m.
b) Area of the room = l × b = 16 m × 8 m = 128 m2.
c) The cost of carpeting 1 m2 = Rs 75
∴ The cost of carpeting 128 m2 = 128 × Rs 75 = Rs 9,600
EXERCISE 17.2
General Section -Classwork
1. a) Length and breadth of a rectangle are a cm and b cm respectively.
Its area is .........................
b) A square is x cm long. Its area is .........................
2. a) A rectangle is 7 cm long and 4 cm broad. Its area is .........................
b) A square is 5 cm long. Its area is .........................
Creative Section - A
3. Let's find the areas of the following figures.
a) A D b) R E
7 cm
B 9 cm C
O 12.5 cm S
4. a) Find the area of rectangles in which (i) length = 7.2 cm, breadth = 5 cm
(ii) length = 18 cm, breadth = 12.5 cm
b) Find the area of squares in which (i) length = 6 cm
(ii) breadth = 7.5 cm
Creative Section - B
5. a) A square room is 14 m long. (i) Find the area of its floor.
(ii) What is the area of carpet required to cover its floor?
(iii) Find the cost of carpeting its floor at Rs 114 per sq. m.
b) The perimeter of a square ground is 144 m.
(i) Find its length. (ii) Find its area
(iii) Find the cost of growing grasses in the ground at Rs 50 per sq. m.
c) The area of a square garden is 324 m2.
(i) Find its length. (ii) Find its perimeter.
(iii) Find the cost of fencing it with five round at Rs 40 per meter.
261Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Perimeter, Area and Volume
6. a) A rectangular park is 40 m long and 36 m broad.
(i) Find its area.
(ii) Find the cost of paving marbles all over it at Rs 90 per sq. m.
b) A rectangular room is 15 m long and 10 m broad.
(i) Find the area of carpet required to cover its floor.
(ii) Find the cost of carpeting the floor at Rs 80 per sq. m.
c) A rectangular lawn is twice as long as its breadth and its perimeter is 96 m.
(i) Find its length and breadth. (ii) Find its area.
(iii) Find the cost of growing grass all over it at Rs 45 per sq. m.
d) A rectangular field is 30 m long and its area is 750 m2.
(i) Find its breadth. (ii) Find its perimeter.
(iii) Find the cost of fencing around it with three rounds at Rs 35 per meter.
7. You have a wire of length 44 cm. You bent it one after another to form the
given shapes. (i) a square (ii) a rectangle of length 12 cm.
Calculate the area of each shape and find which shape covers more area.
17.5 Nets and skeleton models of regular solids
Let’s learn about some solids, their nets, and skeleton models. We can prepare these
skeleton models by using drinking straws or wheat straws.
(i) Tetrahedron Vertex
It is a regular solid. Face 4
1
It has four surfaces. 23
Net
Each surface is an equilateral triangle.
It has four vertices. Skeleton
It has 6 edges.
(ii) Cube
It is a regular solid and it’s also called 4
a regular hexahedron. 12 3 6
It has six surfaces. Vertex 5
Each surface is a square. Face Net
It has eight vertices.
It has 12 edges. Skeleton
(iii) Octahedron 1
It is a regular solid. 2 34567
It has eight surfaces.
Each surface is an equilateral triangle. 8
It has 6 vertices.
It has 12 edges Net
Skeleton
Vedanta Excel in Mathematics - Book 7 262 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Perimeter, Area and Volume
Let's study the following facts known from the nets and skeleton models of the
polyhedrons.
1. The faces of a regular polyhedron are exactly the same, i.e., they are congruent.
2. The line segment that joins any two faces of a regular polyhedron is called its
edge. The edges of a regular polyhedron are equal.
3. The point at which three or more than three edges meet each other is called a
vertex.
Now, study the following table to know about the number of vertices, edges, and
faces of some regular polyhedrons.
Regular No. of vertices No. of edges No. of faces F+V–E
polyhedron (V) (E) (F)
Tetrahedron 4 6 4 4+4–6=2
Hexahedron 8 12 6 6 + 8 – 12 = 2
Octahedron 6 12 8 8 + 6 – 12 = 2
Thus, in any regular polyhedron, F + V – E = 2 is true. This rule was developed by
Swiss Mathematician Euler. So, it is also called Euler’s rule.
(iv) Cylinder circular base
It is a solid object. curved surface
circular base
It has a curved surface with
two circular bases.
(v) Cone vertex
curved surface
It is a solid object.
It has a curved surface with circular base circular base
Its curved surface meet at a point called its
vertex.
EXERCISE 17.3
General Section
1. Let's tell and write the name of these solids as quickly as possible.
a) b) c) d) e)
.................... .................... .................... .................... ....................
263Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Perimeter, Area and Volume
2. Let's name the solid objects which have the following nets.
a) b) c)
............................... ............................... ...............................
d) e)
............................... ...............................
3. Let's say and write the answers as quickly as possible.
a) A tetrahedron has ............ vertices, ............ edges and ............ faces.
b) A hexahedron has ............ vertices, ............ edges and ............ faces.
c) An octahedron has ............ vertices ............ edges and ............ faces.
d) A cylinder has ............ curved surface and ............ circular bases.
e) A cone has ............ curved surface and ............ circular bases.
Creative Section
4. Draw the following polyhedrons. Write the number of their faces (F), vertices (V)
and edges (E), and show that F + V – E = 2 in each case.
a) Cube b) Cuboid c) Tetrahedron d) Octahedron
5. Draw the following nets on separate sheets of hard paper. Cut the outlines of the
nets out and fold along the dotted lines. Paste the edges of the folded faces with
glue. Name the solids you have made.
264 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Vedanta Excel in Mathematics - Book 7
Perimeter, Area and Volume
17.6 Area of solids
Cube, cuboid, cylinder, sphere, cone, pyramid, etc. are the examples of solids.
Length, breadth, and height (or thickness) are three dimensions of solid objects.
(i) Area of cuboid
A cuboid has 6 rectangular faces. Its surface area is the total l
b
sum of the area of 6 rectangular faces. h h
b
Area of top and bottom faces = l b + l b = 2l b l
Area of side faces = bh + bh = 2bh
Area of front and back faces = l h + l h = 2l h
∴ Surface area of cuboid = 2l b + 2bh + 2l h = 2 (l b + bh + l h)
(ii) Area of cube l l
A cube has 6 square faces. Each square face has area of l2. l l
∴ Surface area of a cube = 6l2
l
(i) A lidless rectangular box does not have its top face.
So, it has only 5 rectangular faces.
∴ Area of a lidless rectangular box = 2 (l b + bh + l h) – l b
(ii) A hollow rectangular box does not have top and bottom faces.
So, it has only 4 rectangular faces.
∴ Area of hollow rectangular box = 2 (l b + bh + l h) – 2l b = 2 (bh + l h)
(iii) Area of lidless cubical box = 5l2 Vedanta ICT Corner
(iv) Area of hollow cubical box = 4l2 Please! Scan this QR code or
browse the link given below:
17.7 Volume of solids https://www.geogebra.org/m/t89nau3x
The total space occupied by a solid is called its volume. Volume is 1 cm
measured in cu.mm (mm3), cu.cm (cm3), cu.m (m3), etc.
The length, breadth, and height of the cube given alongside are of 1 cm 1 cm
1 cm each. So, its volume is said to be 1 cm3.
(i) Volume of cuboid h
b
Volume of a cuboid is calculated as the product of the area of
its rectangular base and height. l×b
l
∴ Volume of cuboid = Area of base × height
=l×b×h
(ii) Volume of cube
In the case of a cube, its length, breadth and height are equal. h=l
l b=l
i.e. l = b = h
∴ Volume of cube = l × b × h
= l × l × l = l3
265Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Perimeter, Area and Volume
Worked-out examples
Example 1: A book is 25 cm long, 18 cm broad, and 2 cm thick. Find its
a) surface area b) volume
Solution:
Here, length of the book (l) = 25 cm
breadth of the book (b) = 18 cm
thickness of the book (h) = 2 cm
a) Now, the surface area of the book = 2 (l b + bh + l h)
= 2 (25 × 18 + 18 × 2 + 25 × 2) cm2
= 2 (450 + 36 + 50) cm2 = 1072 cm2.
b) Again, volume of the book = l × b × h
= 25 cm × 18 cm × 2 cm = 900 cm3
Example 2: The volume of a rectangular shoap is 72 cm3 and its height is 4 cm.
If it is placed on a table, find the area covered by it on the table.
Solution:
Here, volume of the shoap (V) = 72 cm3
height of the shoap = 4 cm
Now, volume of the shoap = Area of its base × height
∴ Area of the base × height = 72
or, Area of its base × 4 = 72
or, Area of its base = 72 = 18 cm2
4
Thus, its base (bottom) covers an area of 18 cm2 on the table.
Example 3: If the surface area of a cubical die is 96 cm2,
a) Find the length of its each edge.
b) Find its volume.
Solution:
a) Here, the surface area of the cubical dice = 96 cm2
or, 6l2 = 96 cm2
or,
l2 = 96 cm2 = 16 cm2
6
or, l = 16 cm2 = 4 cm
∴ The length of its each edge is 4 cm.
b) Again, volume of the cubical die = l3 = (4 cm3) = 64 cm3.
Vedanta Excel in Mathematics - Book 7 266 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Perimeter, Area and Volume
Example 4: A cubical room contains 343 cu. m of air.
a) Find the length of its floor.
b) Find the area of its floor.
c) Find the cost of carpeting its floor at Rs 90 per sq.m.
Solution:
a) Here, volume of the cubical room = Volume of air
∴ Volume of the cubical room = 343 m3
or, l3 = 343 m3
or, l = 3 343 cm3 = 7 m
So, the length of its floor is 7 m.
b) Now, area of the floor = l2 = (7m)2 = 49 m2
c) Again, the cost of carpeting the floor = Area × Rate = 49 × Rs 90 = Rs 4,410
Example 5: A rectangular brick is twice as long as its breath and it is 3 cm high.
If its volume is 200 cm3,
a) find its length and breadth.
b) Find its surface area.
Solution:
a) Let the breadth of the brick be x cm.
∴ The length of the brick will be 2x cm.
Now, the volume of the brick = 200 cm3
or, l × b × h = 200 cm3
or, 2x × x × 4 cm = 200 cm3
or,
2x2 = 200 cm2 = 50 cm2
or, 4
x2 = 50 cm2 = 25 cm2
2
or, x = 25 cm2= 5 cm
So, the breadth (b) = x = 5 cm and the length (l) = 2x = 2 × 5 cm = 10 cm.
b) Again, surface area of the brick = 2 (l b + bh + l h)
= 2 (10 × 5 + 5 × 4 + 10 × 4) cm2
= 2 (50 + 20 + 40) cm2 = 220 cm2
Example 6: A rectangular metallic block is 9 cm long, 6 cm broad and 4 cm
thick. If it is melted and converted into a cube, find the surface area
of the cube.
Solution:
Here, length of the block (l) = 9 cm
breadth of the block (b) = 6 cm
thickness of the block (h) = 4 cm
Now, volume of the block = l × b × h = 9 cm × 6 cm × 4 cm = 216 cm3
267Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Perimeter, Area and Volume
volume of the cube = Volume of the block
or, l3 = 216 cm3
or, l = 3 216 cm3 = 6 cm
Again, the surface area of the cube = 6l2 = 6 × (6 cm)2 = 216 cm2.
Example 7: A rectangular box completely filled with milk powder is 24 cm long,
12 cm broad, and 16 cm high. How many cubical boxes of length
8 cm are required to empty the milk powder from the box?
Solution:
Here, length of the rectangular box (l) = 24 cm
breadth of the rectangular box (b) = 12 cm
height of the rectangular box (h) = 16 cm
∴ Volume of the rectangular box = l × b × h = (24 × 12 × 16) cm3 = 4608 cm3
Also, volume of milk powder = volume of the box = 4608 cm3
Again, volume of each cubical box = (8 cm)3 = 512 cm3
Volume of milk powder
Now, the required number of cubical boxes = Volume of each cubical box
4608
= 512 = 9
Hence, 9 cubical boxes are required to empty the milk powder from the box.
Example 8: The area of the base of a rectangular water tank is 30,000 cm2. Find
the height of water level when there is 3000 litres of water in the tank.
(1l = 1000 cm3)
Solution:
Here, volume of water = 3000 l h
= 3000 × 1000 cm3
Now, volume of the part of the tank containing water = volume of water
or,A rea of its base × height = 3000 × 1000 cm3
or, 30000 cm2 × h = 3000 × 1000 cm3
or, h = 3000 × 1000 cm3
or, 30000 cm2
h = 100 cm
So, the required height of water level in the tank is 100 cm (or 1 m).
EXERCISE 17.4
General Section - Classwork
1. Let's say and write the answers as quickly as possible.
a) A cuboid is x cm long, y cm broad and z cm high. Its total surface area
is ............................ and volume is ............................
Vedanta Excel in Mathematics - Book 7 268 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Perimeter, Area and Volume
b) A cuboids is 6 cm long, 5 cm broad and 3 cm high. Its volume is .................
c) A cube is 4 cm long. Its total surface area is ........... and volume is ..............
e) A lidless cubical box is 3 cm long. Its total surface area is ..............
f) A hollow cubical box is 5 cm long. Its total surface area is ...................
g) Area of the rectangular base of a box is 15 cm2 and its height is 4 cm. Its
volume is ..............
h) Volume of a cube is 64 cm3. Its length is ...................
Creative Section - A
2. Find the total surface area and volume of these solids.
b) c) d)
a)
4 cm
2cm
5 cm
3cm
5cm
4 cm
15 cm 8 cm 4 cm
5 cm 5 cm 15 cm
3. Find the total surface area and volume of cuboids.
a) l = 8cm, b = 6cm, h = 5cm b) l = 15cm, b = 4.5 cm, h = 3cm
4. Find the volume of cuboids.
a) Area of base = 104cm2, h = 5.5 cm
b) Area of base = 170 cm2, h = 16.5 cm
5. Find the total surface area and volume of cubes.
a) l = 5cm b) b = 7cm c) h = 10 cm d) l = 12.5 cm
6. Find the total surface area of these rectangular boxes when they are (i) lidless
and (ii) hollow.
a) l = 15cm, b = 10cm, h = 6cm b) l = h = 20cm, b = 5cm
7. Find the total surface area of these cubes when they are: (i) lidless and (ii)
hollow
a) l = 12cm b) l = 15cm
8. Find the unknown measurements of the following cuboids:
a) l = 7 cm, b = 5 cm, volume = 105 cm3, find h.
269Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Perimeter, Area and Volume
b) l = 10 cm, h = 8 cm, volume = 440 cm3, find b.
c) b = 7 cm, h = 9 cm, volume = 378 cm3, find l.
d) Area of base = 60 cm2, volume = 540 cm3, find h.
9. Find the length of edge of each of the following cubes:
a) Surface area = 24 cm2 b) Surface area = 54 cm2
c) Volume = 64 cm3 d) volume = 125 cm3
10. If 1000 cm3 = 1 l, find the capacity of the following water tank in litre.
a) l = 200 cm, b = 150 cm, h = 100 cm b) l = 256 cm, b = 180 cm, h = 200 cm
c) l = 3.2 m, b = 1.6 m, h = 1.5 m d) l = 2 m, b = 1 m, h = 2.5 m
Creative Section - B
11. a) The volume of a rectangular box is 1680 cm3 and its height is 10 cm. If it is
placed on a table, find the area covered by it on the table.
b) A rectangular room contains 600 m3 of air and its height is 5 m.
(i) Find the area of its floor.
(ii) Find the cost of carpeting its floor at Rs 60 per sq.m.
c) A cubical room contains 216 m3 of air.
(i) Find the length of its floor. (ii) Find the area of its floor.
(iii) Find the cost of carpeting its floor at Rs 50 per sq. m.
d) A rectangular water tank is 150 cm high and the area of its base is 30000 cm2.
Find the capacity of the tank in litre. (1000 cm3 = 1 l)
12. a) The surface area of a cube is 24 cm2.
(i) Find the length of its edge. (ii) Find its volume.
b) The volume of a cube is 125 cm3. (ii) Find its surface area.
(i) Find the length of its edge.
c) If the surface area of a lidless cubical box is 180 cm2, find its length.
d) If the surface area of a hollow cubical box is 324 cm2, find its height.
13. a) A cuboid is twice as long as its breadth and its height is 4 cm. If the volume
of the cuboid is 288 cm3,
(i) find its length and breadth. (ii) Find its surface area.
b) The volume of a rectangular room is 490 m3. If length is two times longer
than that of its breadth and height is 5 m,
(i) find its length and breadth. (ii) Find the area of its floor.
(iii) Find the cost of carpeting its floor at Rs 45 per sq.m.
Vedanta Excel in Mathematics - Book 7 270 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Perimeter, Area and Volume
14. a) A rectangular metallic block is 9 cm long, 6 cm broad and 4 cm high. If it is
melted and converted into a cube,
(i) find the length of the edge of the cube.
(ii) Find the surface area of the cube.
b) The length, breadth, and the height of a rectangular metallic block are 18 cm,
4 cm, and 24 cm respectively. It is melted and formed into a cube. Find the
surface area of the cube.
15. a) The area of the base of a rectangular water tank is 40,000 cm2. Find the
height of water level when there is 4000 litres of water in the tank.
(1 l = 1000 cm3).
b) A rectangular water tank contains 1440 litres of water. If its length is 120 cm
and breadth is 80 cm, find the height of water level in the tank.
(1 l = 1000 cm3) .
16. a) A rectangular box completely filled with milk powder is 16 cm long, 9 cm
broad and 15 cm high. How many cubical boxes of length 6 cm are required
to empty the milk powder from the box?
b) A rectangular water tank is 80 cm long, 40 cm broad and 60 cm high. How
many rectangular jars each of them are 20 cm long, 10 cm broad and 15 cm
high are required to empty the tank full of water?
It's your time - Project work!
17. a) Let's measure the length, breadth and thickness of your maths book using a
ruler.
(i) Find the area of its base
(ii) Find its total surface area
(iii) Find its volume
b) Let's measure the length, breadth, height (or thickness) of cubical or cuboidal
objects found in your school or at home. Find the value of these objects.
c) Let's measure the length, breadth, and height of a matchbox and find its
volume. Now, place 4 more matchbox on its top and measure the height.
Again, find the volume of these combined matchboxes. Is the volume of
combined matchboxes = 5 × volume of one matchbox?
271Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Perimeter, Area and Volume
OBJECTIVE QUESTIONS
Let’s tick (√) the correct alternative.
1. The length of base of an isosceles triangle is 12 cm. If its perimeter is 32 cm, what is
the length of each of its equal side?
(A) 22 cm (B) 44 cm (C) 10 cm (D) 20 cm
2. The side lengths of a triangular kitchen garden are 30 ft, 40 ft and 50 ft., then the
length of wire required to fence it with 3 rounds is
(A) 120 ft. (B) 360 ft. (C) 210 ft. (D) 270 ft.
3. The circumference of a circle with radius ‘r’ cm is
(A) 2πr (B) πr (C) πr2 (D) 2πr2
4. In any regular polyhedral with F faces, V vertices and E edges, which of the following
relations hold?
(A) F + V – E = 2 (B) V + E – F = 2 (C) E + F – V = 2 (D) V – E – F = 2
5. A tetrahedron has 4 vertices and 4 faces, how many edges does it have?
(A) 4 (B) 6 (C) 8 (D) 10
6. A hollow cubical box is 6 cm long. Its total surface area is
(A) 216 cm2 (B) 36 cm2 (C) 72 cm2 (D) 144 cm2
7. The total surface area of a cubical die is 6 cm2, what its volume?
(A) 6 cm3 (B) 3 cm3 (C) 1 cm3 (D) 4 cm3
8. The volume of a cubical tank is 64 m3, its each side is
(A) 2 m (B) 4 m (C) 8 m (D) 16 m
9. The volume of a rectangular tank is 9 m3. How many liters of water can it hold?
(A) 90 l (B) 900 l (C) 9,000 l (D) 90,000 l
10. The property of a cone is
(A) It has a vertex (B) It has a circular base
(C) It has curved surface (D) All of the above
Vedanta Excel in Mathematics - Book 7 272 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Unit 18 Symmetry, Design and Tessellation
18.1 Symmetrical and asymmetrical shapes
A geometric shape or object is said to be symmetrical
if it can be divided into two or more identical pieces.
However, if a shape cannot be divided into the identical
pieces, it is called asymmetrical. Symmetrical Asymmetrical
18.2 Line or axis of symmetry
In the given figures, one or more
dotted line segments divide the
figures into two identical halves.
The dotted lines are called the line sl1yims amleintreyof lo1fasnydml2miseatrlyines al1reanlidnels2 oanf dsylm3 amnedtrly4
of symmetry or axis of symmetry.
Thus, the line of symmetry is the
line (or imaginary line) that passes through the centre of a symmetrical shape and
divides it into identical halves. The line of symmetry is also called the mirror line
or axis of symmetry. DC
18.3 Rotational symmetry
Let's take a rectangular sheet of paper and draw its CA B
two diagonals. The diagonals bisect each other at B
the centre of the rectangle. Now, place the tip of a A
pencil at the centre of the rectangle and rotate the B
rectangle a quarter-turn, then through the half-turn.
Here, when the rectangle is rotated about its centre
by half a turn, the result looks exactly like the original
rectangle. This is called rotational symmetry. D A C D
quarter-turn
half-turn
Thus, when a shape can be mapped onto itself by a
rotation of less than a complete turn (360°) about its centre, the shape is said to have
rotational symmetry.
18.4 Order of rotational symmetry
Study the following illustrations and investigate the idea of order of rotational
symmetry. A B
When the shape is rotated through a half-turn
it can be fitted over the original shape. It means 180°
that, when it is rotated through 2 times half-
turns, it reaches its original position. B A
∴ It has rotational symmetry of order 2.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 273 Vedanta Excel in Mathematics - Book 7
Symmetry, Design and Tessellation
When the equilateral triangle is rotated
A C 1
through a 3 of a turn (120°), it can be
120° fitted over the original shape. It means
that, when it is rotated through 3 times
BC AB p31osoitfioan.turn,
it reaches its original
∴ It has rotational symmetry of order 3.
DC D C When the square is rotated through a quarter-
90° turn ( 1 of a turn or 90°), it can be fitted over
4
AB A B the original shape. It means that, when it is
1
rotated through 4 times 4 of a turn, it reaches
its original position.
∴ It has rotational symmetry of order 4.
EXERCISE 18.1
General Section - Classwork
1. Let's say and write, how many lines of symmetry you can find for each of the
shapes below. If in doubt, trace and fold them.
2. Let's draw the lines of symmetry of the following geometrical figures by using
dotted lines.
3. Let's write True or False for the following statements.
a) If a shape has 1 turn symmetry then it has 1 turn symmetry. ...............
4 2 ...............
b) If a shape has 1 turn symmetry then it has 1 turn symmetry. ...............
2 4
c) If a shape has 3 lines of symmetry then it has rotational
symmetry of order 3.
Vedanta Excel in Mathematics - Book 7 274 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Symmetry, Design and Tessellation
d) If a shape has 4 lines of symmetry then it has rotational ...............
symmetry of order 4.
e) A shape must have line symmetry to have rotational symmetry. ...............
f) A shape is said to have rotational symmetry if the shape can
be mapped onto itself by a rotation of a complete turn (360°). ...............
Creative Section - A
4. Which of these drawings have half-turn symmetry, and which have quarter-
turn symmetry? Use tracing paper to help you to decide, if necessary.
a) b) c)
d) e) f)
1 1
4 3
5. a) Decide whether each shape has turn symmetry, turn symmetry, or
1 turn symmetry. Use tracing paper to help you decide, if necessary. Also
2
mention the order of rotational symmetry. (iv)
(i) (ii) (iii) Z
A BP S NK
D CQ RX YM L
b) Copy these figures and write the new position of vertices after the rotation
1 1 1
through 4 turn, 3 turn or 2 turn.
18.5 Tessellations
A tessellation is a covering of the plane with
congruent geometrical shapes in a repeating pattern
without leaving any gaps and without overlapping
each other. Tessellation is also known as tiling.
In a tessellation, the shapes are often polygons.
The polygon may be either an equilateral triangle,
a square, or a regular hexagon. Tessellation is done
on the surface of carpets and on the surface of
floor or wall to make the surface more attractive.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 275 Vedanta Excel in Mathematics - Book 7
Symmetry, Design and Tessellation
Thus, to make a tessellation (or tiling) –
(i) Use the sets of congruent figures (tiles).
(ii) Do not leave any gaps.
(iii) Do not have any overlaps.
18.6 Types of tessellations
(i) Regular tessellations
In this case, we use the same types
of regular polygons. The polygons
that we use may be equilateral
triangle, square, regular hexagon,
etc.
(ii) Semi-regular tessellations
In this case, we use two or more
regular polygons. In the adjoining
tessellation, regular octagons and
squares are used.
(iii) Irregular tessellations
In this case, we use irregular types
of polygons.
EXERCISE 18.2
1. Define the following terms.
a) Tessellation
b) Regular tessellation
c) Semi-regular tessellation
d) Irregular tessellation
2. What are three important rules while making a tessellation?
Vedanta Excel in Mathematics - Book 7 276 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Symmetry, Design and Tessellation
3. Let's use graph paper to copy the following tessellation patterns and complete
them.
4. Let's use graph paper and draw the following patterns of design. Colour your
designs and make them attractive.
It's your time - Project work!
5. a) Let's draw your own patterns of designs in graph papers. Colour the patterns
and demonstrate in your class.
b) Let's observe the various patterns of carpet designs. Copy the patterns in
graph papers and make attractive design by colouring.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 277 Vedanta Excel in Mathematics - Book 7
Unit 19 Scale Draw and Bearing
19.1 Scale drawing - review
Let's have discussions on the following questions.
Is it possible to draw the actual size of your house on a sheet of paper?
Is it possible to locate the actual distance between any two places on a sheet of
paper?
Is it possible to draw the actual size of coronavirus on a sheet of paper?
It is not possible to draw the actual size of very large figure (or object) or very small
figure (or object) on a sheet of paper. Similarly, it is impossible to show lengths or
distance which are too great to draw full-size. In the case of large figure, we should
reduce their actual size and for smaller figure, we should enlarge their actual size in
drawing by taking a convenient scale.
A drawing that shows a real object smaller than (a reduction) or larger than (an
enlargement) the real object is called scale drawing.
Reduction Enlargement
19.2 Scale factor
The scale of the drawing is the ratio of the size of the drawing to the actual size of
the object. For example:
(i) A distance of 5 km (ii) A length of 80 m
Scale: 1 cm to 1 km (or 1:100000) Scale: 1 cm to 10 m (or 1: 1000)
Scale drawing Scale drawing:
0 1 2 3 4 5 km 0 10 20 30 40 50 60 70 80 m
The scale of a drawing is also known as the scale factor. It is used to reduce or
enlarge a figure according to the scale ratio.
Worked-out examples
Example 1: What is the actual length which is represented by 5.4 cm on a scale
drawing with scale of 1 cm to 10 m (or 1 : 1000)?
Solution:
Here, 1 cm represents 10 m
∴ 5.4 cm represent 5.4 × 10 m = 54 m
Vedanta Excel in Mathematics - Book 7 278 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Scale Drawing and Bearing
Example 2: What is the actual distance between two places which is represented
by 4.5 cm on a map which is drawn to the scale 1 : 200000?
Solution: 200000
100
Here, the scale 1 : 200000 means that , 1 cm represents 200000 cm = m
= 2000 m = 2 km
Now, 1 cm represents 2 km
∴4.5 represents 4.5 × 2 km = 9 km
Hence, the distance between two places is 9 km.
Example 3: A rectangular garden is 16 m long and 14 m broad. If the map scale is
1:400, what are the length and the breadth of the garden on drawing?
Solution:
Here, the scale 1 : 400 means, 1 cm represents 400 cm = 4m
4 m is represented by 1 cm.
1 m is represented by 1 .
4
1
16 m is represented by 4 × 16 cm = 4 cm 3.5 cm
(14 m)
∴ The length of the garden on drawing is 4 cm.
Also, 1 m is represented by 1 cm
4
1
14 m is represented by 4 × 14 cm = 3.5 cm 4 cm
(16 m)
∴ The breadth of the room on drawing is 3.5 cm.
Example 4. The height of a tree on the picture is 3 cm. If its actual height is 24 m,
find the scale of the drawing.
Solution:
Here, height of tree on the picture = 3 cm
Actual height of tree = 24 m = 24 × 100 cm = 2400 cm
Scale of drawing = ? 3 cm
Height of tree on picture 2400 cm 1
Now, scale of drawing = Actual height of tree = = 800 = 1 : 800
∴ Required scale of drawing is 1 : 800
EXERCISE 19.1
General Section - Classwork
Tell and write the answers as quickly as possible.
1. a) When the scale is 1:100, the actual length of 3 cm is ...................... m.
b) When the scale is 1:200, the actual length of 5 cm is ...................... m.
c) When the scale is 1:300, the actual length of 4cm is ...................... m.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 279 Vedanta Excel in Mathematics - Book 7
Scale Drawing and Bearing
2. a) When the scale is 1:200, the map/drawing length of 6m is ................... cm.
b) When the scale is 1:500, the map/drawing length of 40m is ................... cm.
c) When the scale is 1: 800, the map/drawing length of 24m is ................... cm.
3. Let's study the information given in the table and fill in the blanks.
Length on map/drawing Scale Actual length
................. m
a) 2 cm 1:300 ................. m
................. km
b) 5 cm 1:2000
20 m
c) 6.5 cm 1:50000 32 km
72 m
d) ....................... cm 1:400
e) ....................... cm 1:800000
f) ....................... mm 1:9000
Creative Section - A
4. a) Sketch a tower of height 45 m according to the scale of 1 cm to 9 m.
b) The actual length and breadth of a rectangular play ground are 40 m and
30 m respectively. Draw the map of ground by taking a scale of 1 cm to 10 m.
c) A water tank is 20 m long, 10 m wide and 4 m high. Take a scale of 1 : 400.
Then, draw a map of the tank.
5. a) What is the actual length which is represented by 3.2 cm on a scale drawing
with scale of 1 cm to 4 m?
b) What is the actual height of a building which is represented by 8.5 cm on a
scale drawing with scale of 1:1000?
c) Find the actual length and breadth of a room which are represented by 6 cm
and 4 cm respectively on a scale drawing with scale of 1:200.
d) What is the actual distance between two places which is represented by
15 mm on a map which is drawn to the scale 1 mm to 5 km?
6. a) If 1 cm represents 5 m, what is the scale drawing length of 18 m?
b) The distance between two places is 50 km. What is the scale drawing
distance between these places with scale of 1 cm to 10 km?
c) A rectangular ground is 54 m long and 45 m broad. If the map scale is 1:900,
find the length and breadth on drawing.
d) The length and breadth of a rectangular garden are 24 m and 20 m respectively.
Find the scale drawing length and breadth of the garden with scale of 1:800.
Vedanta Excel in Mathematics - Book 7 280 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Scale Drawing and Bearing
Creative Section - B 3 cm 3 cm 3 cm 2 cm
7. The scale drawing dimensions of the Kitchen Bed room 3 cm
Dinning
ground floor of a house are given alongside. 2 cm room
a) Find the actual dimensions of every 1 cm Toilet Living room
Passage
room with scale of 1:200. 2 cm
Storeroom
b) Find the actual area occupied by the 3 cm 2 cm 4 cm
kitchen room.
c) Calculate the actual area occupied by the house.
It's your time - Project work!
8. a) Let's choose an appropriate scale and draw a road map of your school from
your home.
b) Let's sketch the layout of ground floor of a house with an appropriate scale to
the following measurements. (i) Sitting room - 10 m by 8 m
(ii) a bed room - 8 m by 6 m (iii) a kitchen - 6 m by 4 m
(iv) a guest room 8 m by 6 m (v) a wash room - 4 m by 3 m
19.3 Bearing
From the given compass, let's study the different North (N)
directions. In the compass NOS represents
North–South and EOW represents East-West North West (NW) North East (NE)
directions. West (W) O East (E)
Furthermore, the angle between N and E is 90°. South West (SW) South (S) South East (SE)
The direction NE lies exactly in between N and
E. So, the angle between N and NE is 45°. We
take O as the point of reference and ON (North line) as base to find direction of any
object (or place) in terms of degrees measuring clockwise from the base line (ON).
The direction is usually written using three digits. For example,
the direction of NE from O is 045°. It is called bearing of NE from O. NW N
045° NE
The direction of E from O is 090°. It is called bearing of E from O. W O E
Similarly, the direction of SW from O is 225°. It is called bearing SW 225°
of SW from O and so on. SE
S
Thus, the bearing of a point (or place) is the angle in degrees
measured from the north line in clockwise direction.
The bearing of an object (or place) is usually written in three digits. So, it is also
called three-digit bearing (or three-figure bearing).
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 281 Vedanta Excel in Mathematics - Book 7
Scale Drawing and Bearing
Worked-out examples
Example 1: Write down the bearings of A, B, and C places from O.
a) N b) N c) C N
60° A 25°
O O
120° O
B
Solution:
a) The bearing of A from O is 060°.
b) The bearing of B from O is (360° – 120°) = 240°.
c) The bearing of C from O is (360° – 25°) = 335°.
Example 2: State the bearing of point P from A.
a) N b) N c) N d) N
48° P A
P
WA E WA E WA E W 25° E
60° 40° P
P
Solution: S S S S
a) The bearing of point P from A is ∠NAP = 048°
b) The bearing of point P from A is ∠NAP = ∠NAS + ∠SAP = 180° + 60° = 240°
c) The bearing of point P from A is ∠NAP = ∠NAS – ∠PAS = 180° – 40° = 140°
d) The bearing of point P form A is ∠NAP = ∠NAW + ∠WAP = 270° + 25°= 295°
Example 3: In the adjoining diagram, if the bearing of A from B is 080°, find the
bearing of B from A. N N1
Solution:
Here, bearing of A from B is 080°. 80° A
B
Since, BN // AN1, ∠ NBA and ∠ N1AB are co-interior angles.
∴∠ N1AB = 180° – 80° [Co-interior angles are supplementary.]
= 100°
∴ Bearing of B from A = 360° – 100° = 260°.
EXERCISE 19.2 N NE
General Section NW E
1. Let's say and write the names of these eight points of the
WO SE
compass in full. SW
a) N is ....................................... NE is .......................... S
b) E is ....................................... SE is ..........................
Vedanta Excel in Mathematics - Book 7 282 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Scale Drawing and Bearing
c) S is ....................................... SW is ..........................
d) W is ....................................... NW is ..........................
2. Let's say and write how many degrees are in the angles between .
a) N and E .................... b) N and NE ....................
c) N and S .................... d) N and SE ....................
e) N and W .................... f) N and SW ....................
g) N and W .................... h) N and NW ....................
3. Let's say and write the bearings as quickly as possible.
a) N b) N c) N d) D N
75° A 80° 130°
O O
O B
C
O
300°
A from O ......... B from O ......... C from O ......... D from O .........
Creative Section - A
4. Let's study the adjoining map of MaBhaeintaddrai nagar Jumla
Nepal and answer the following Nepalgunj
questions. Jomsom
Gaighat
BBirhaatdnrIaalgpaaurrmPokhara
Butwal Kathmandu
Birgunj
a) If Kathmandu is the point of reference, write the compass directions of:
(i) Pokhara (ii) Biratnagar (iii) Mahendranagar
(iv) Ilam (v) Jumla (vi) Nepalgunj
b) If Butwal is the point of reference, write the compass directions of:
(i) Kathmandu (ii) Birgunj (iii) Baitadi
(iv) Bhadrapur (v) Jomsom (vi) Gaighat
5. Let's write down the bearings of all the planes shown below from A.
a) N b) N c) N
60° A 125°
A A
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 283 Vedanta Excel in Mathematics - Book 7
Scale Drawing and Bearing e) f) N
d) N
N
140° A 45°
AA
6. Let's find the bearing of point P from O in each the following figures.
a) N b) N c) N
P
W O 20° E WO E W O E
30°
S 55° S
d) N P N
SP
f)
e) N
P 40° P
WO E WO E W O 130° E
150°
S P S
Creative Section - B S
7. The bearings of A from B are given in the figures. Find the bearings of B from A.
a) N1 b) c) d) N1
N N N1 N N1 N
A
025° 065° A 087°
090°
B B B A B A
e) N N
f) g) h)
N
N N1 N1
N1 N1
105° 140° B A
B B 210°
A A B 330°
A
8. a) If the bearing of a place P from a point Q is 055°, find the bearing of Q from p.
b) If the bearing of Dharan from Birgunj is 084°, find the bearing from Dharan to
Birgunj.
Vedanta Excel in Mathematics - Book 7 284 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Scale Drawing and Bearing
OBJECTIVE QUESTIONS
Let’s tick (√) the correct alternative.
1. When the scale is 1:100, the actual length of 5 cm is
(A) 5 m (B) 50 m (C) 500 m (D) 5 km
2. The distance between two places is 40 km. What is the scale drawing distance
between these places with scale of 1 cm to 10 km?
(A) 4 mm (B) 4 cm (C) 40 cm (D) 4 m
3. What is the actual height of tree which is represented by 2.5 cm on a scale drawing
with scale of 1 cm to 6 m?
(A) 15 m (B) 25 m (C) 6 m (D) 8.5 m
4. The scale of a map is 1:250000. Two towns are 15 km apart. How far apart are the
towns on the map?
(A) 6 cm (B) 3.75 cm (C) 6.5 cm (D) 7.25 cm
5. The bearing of a point or place is measured with reference to
(A) East (B) South (C) West (D) North
6. The bearing of the east direction is
(A) 030o (B) 090o (C) 180o (D) 270o
7. The bearing of point A from O in the given figure is N A
N
(A) 070o (B) 290o 70°
(C) 020o (D) 110o O
8. The bearing of point P from O in the given figure is P
20°
W O
E
(A) 020o (B) 070o
(C) 160o (D) 340o N S
9. The bearing point M from O is O
40°
(A) 040o (B) 140o M
(C) 130o (D) 320o
10. The bearing of a place A from the place B is 050o, what is the bearing of B from A?
(A) 050o (B) 130o (C) 230o (D) 140o
Vedanta ICT Corner
Please! Scan this QR code or
browse the link given below:
https://www.geogebra.org/m/kpbhhy4q
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 285 Vedanta Excel in Mathematics - Book 7
Unit 20 Statistics
20.1 Statistics – Review
The modern age is the age of information and communication. In the various field
of the society, we need information in the form of numerical figures to accomplish
various economical, commercial, technical activities. Such numerical figures are
called data.
Statistics is a branch of mathematics that deals about the collection, tabulation, and
presentation of data.
20.2 Types of data and frequency table
The numerical information given below are the amounts of pocket money (in Rs)
brought by 20 students of class VII on a day.
20, 15, 20, 25, 20, 25, 10, 30, 20, 30,
25, 20, 15, 20, 30, 25, 40, 20 25, 40
The above data are not presented in the proper order. Such data are called raw data.
Now, let's arrange these data in the ascending order of the values.
10, 15, 15, 20, 20, 20, 20, 20, 20, 20,
25, 25, 25, 25, 25, 30, 30, 30, 40, 40
The above data which are presented in the proper order (either ascending or
descending order) are called arrayed data.
From an arrayed data, it is easier to observe how many times a particular figure is
repeated
Rs 10 is repeated only one time. → Frequency of Rs 10 is 1.
Rs 15 is repeated two times. → Frequency of Rs 15 is 2.
Rs 20 is repeated seven times. → Frequency of Rs 20 is 7.
Rs 25 is repeated five times. → Frequency of Rs 25 is 5.
Rs 30 is repeated three times. → Frequency of Rs 30 is 3.
Rs 40 is repeated two times. → Frequency of Rs 40 is 2.
In this way, a frequency is the number of times a figure (or value) occurs.
We can present the data and their frequencies in a table. Such table is called
frequency distribution table (or simply frequency table).
Vedanta Excel in Mathematics - Book 7 286 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Statistics
Frequency table Frequency
Pocket money (in Rs) Tally marks 1
2
10 | 7
15 || 5
20 || 3
25 2
30 |||
40 || 20
Total
Tallying is a system of showing frequencies using diagonal lines grouped in fives.
Each time five is reached, a horizontal line is drawn through the tally marks to make
a group of five. The next line starts a new group. For example,
1 is |, 2 is ||, 5 is , 7 is ||, and so on.
20.3 Grouped and continuous data
Let's consider the following marks obtained by 20 students in a Mock Test of
Mathematics.
25, 32, 45, 15, 8, 42, 26, 19, 28, 45,
36, 48, 22, 29, 18, 34, 6, 38, 41, 24
The above mentioned data are called individual data. Another way of organizing
data is to present them in a grouped form. For grouping the given data, we should
first see the smallest and the largest values. Then, we have to divide the data into
appropriate class intervals.
In the above data, smallest value is 6 and the largest value is 48. So, the class
intervals can be 0 – 10, 10 – 20,…, 40 – 50. The data arranged in this way are called
grouped data.
The frequency table of the grouped data are given below.
Marks Tally marks No of students (f)
0 – 10 || 2
10 – 20 ||| 3
20 – 30 | 6
30 – 40 4
40 – 50 |||| 5
Total 20
287Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Statistics
20.4 Line graphs
Data can also be represented by plotting the corresponding frequencies in the graph
paper. The line so obtained by joining the points is called the line graph.
While constructing a line graph, the frequencies of the items are plotted along y–axis.
The line graph given below represents the daily wags of the workers of a company.
Wages in Rs. 40 50 60 70 80 90 100
No. of workers 15 8 22 18 10 8 5
Y
25
20
No. of workers 15
10
5
0 40 50 60 70 80 90 100 X
wages (in Rs)
Example 1: The table given below shows the daily sales of a shop for six days of a week.
Day Sun Mon Tues Wed Thurs Fri
Sales (in Rs) 3000 4500 2000 4000 2500 6000
(i) Construct a line graph for the frequency table.
(ii) On which days were the sales above Rs 4,000.
Solution:
(a) 6000
5000
4000
3000
2000
1000
0
Sun Mon Tue Wed Thus Fri
b) The sales were above Rs 4,000 on Monday and Friday.
Vedanta Excel in Mathematics - Book 7 288 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Statistics
20.5 Bar graph
We have already discussed to present data in frequency tables. Alternatively, we can
also present data graphically. Different types of diagrams are used for this purpose.
Here, we shall discuss about a bar graph (or bar diagram).
(i) Simple bar graph
A simple bar graph is drawn to present a single type of data. In a bar graph, data
are represented in a series of bars that are equally wide. Bars can touch each
other or be separated by gaps of equal width. The height of the bars represent
the frequency of the data.
We should follow the rules given below while drawing a bar graph:
Choose a suitable scale to represent the whole data. Mention the frequencies in
y-axis.
Decide how wide the bars are and how much space you leave between them.
Construct the bars of the same width and at equal distance.
Worked-out examples
Example 1: The table given below shows the number of students in class 7 of
Solution: a school in five years. Represent the numbers in a bar graph.
Years (B.S.) 2075 2076 2077 2078 2079
No. of students 30 35 45 60 50
2075 2076 2077 2078 2079
(ii) Multiple bar graph
A multiple bar graph is drawn to show two or more related components of the
given data. For example, if we want to show the number of boys and girls of
a school separately, we should draw the multiple bar graph. The method of
drawing a multiple bar graph is exactly the same as a simple bar graph.
289Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Statistics
Example 2: The number of boys and girls in the primary level of a school are
given in the table below. Draw a multiple bar graph to show their
numbers.
Solution: Classes 1 2345
No. of boys 12 15 16 25 25
No. of girls 18 20 24 20 10
No. of boys and girls 35 20 24 25 25
30 15 16 20 10
5
25 23 4
Classes
20 18
15 12
10
5
0
1
EXERCISE 20.1
1. a) Define raw and arrayed data with examples.
b) Define frequency with examples.
c) Write a difference between a simple bar graph and a multiple bar graph.
2. a) The amount of pocket money (in Rs) brought by 20 students of class VII on a
day are given below. Let's construct a frequency distribution table with tally
marks.
30, 40, 25, 35, 40, 50, 40, 30, 25, 20,
40, 45, 35, 40, 30, 35, 40, 45, 30, 35
b) Daily wages (in Rs) of 30 workers in a factory are given below. Arrange the
data in proper order. Lets construct a frequency table to present the data.
850, 900, 600, 750, 800, 750, 600, 900, 800, 750,
700, 800, 750, 600, 750, 800, 900, 850, 750, 700,
800, 750, 700, 800, 850, 750, 800, 750, 700, 750.
c) In an interview of 20 married couples about their desired number of children,
the responses were as follows:
1, 2, 3, 2, 3, 1, 4, 2, 2, 1
4, 3, 1, 2, 2, 2, 1, 3, 1, 2
Let's present the above data in a frequency table using tally marks and answer
the following questions.
(i) How many couples desired a single child?
(ii) How many couples desired two children?
(iii) How many couples desired three children?
Vedanta Excel in Mathematics - Book 7 290 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Statistics
(iv) How many couples desired more than three children?
(v) What is the desirable number of children for the maximum number of
couples?
3. a) The adjoining line graph Rainfall (mm) 120
shows the average rain fall in 100
mm during 6 months of the Apr May June July Aug Sep
year 2020. 80 Months
60
(i) In which month was the 40
minimum average rainfall? 20
On which month was it 0
maximum?
................................................
(ii) How much was the average
rainfall recorded in August?
................................................
(iii) Write a paragraph about the
common trend of rainfall
during these months in your
exercise book.
b) The adjoining line graph
shows the average rain fall
in mm during 6 months of
the year 2020.
(i) On which month
was the minimum
average rainfall? On
which month was it
maximum?
(ii) How much was
the average rainfall
recorded in August?
(iii) Write a paragraph about
the common trend of
rain fall during these
months.
c) The table given below
shows the daily expense of a family.
Day Sunday Monday Tuesday Wednesday Thursday Friday
500
Expense (in Rs) 400 600 1300 1000 1200
(i) Construct a line graph for the above data.
(ii) On which days were the expenses more than Rs 1,000?
291Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Statistics
4. a) The number of students in the primary level (class 1 to 5) are given below.
Draw a simple bar graph to show the number.
Classes 12345
No. of students 45 30 50 25 15
b) The table given below shows the number of students who secured 'A' grade in
S.E.E. from your school in the last five years. Let's draw a simple bar diagram to
show the data.
Years (B.S.) 2075 2076 2077 2078 2079
No. of students 25 40 30 45 40
5. a) The table given below shows the number of boys and girls of a school from
class 6 to 10. Illustrate the numbers by drawing a multiple bar graph.
Classes 6 7 8 9 10
No. of boys 10 20 15 25 20
No. of girls 15 10 20 15 10
b) Class 7 students conducted a survey and recorded the following number
of male and female who had positive results of rapid diagnostic test for
conronavirus in the last five days. Draw a multiple bar graph to show the
numbers.
Days Sunday Monday Tuesday Wednesday Thursday
No. of male patients 7 10 12 10 7
No. of female patients 3 5 4 3 5
c) The table given below shows the S.E.E. result of a school for the last 4 years
Let's draw a multiple bar graph to show the numbers.
Year (in B.S.) 2076 2077 2078 2079
'A+' grade 5 10 15 12
'A' grade 10 15 20 27
'B+' grade 15 5 15 11
6. The multiple bar graph given below shows the number of boys and girls who
secured 'A' grade in the final examination from different classes. Answer the
following questions.
No. of boys and girls boys
girls
6 78 9
Classes
Vedanta Excel in Mathematics - Book 7 292 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Statistics
a) How many students are represented by a room in vertical column?
b) How many boys and girls secured 'A' grade in the final exam from class 6?
c) If all the students of class 7 have grade 'A' in the exam, how many students were
there in class 7?
d) If 80 students appeared in the exam from class 9, find the percentage of students
who secured 'A' grade.
e) If 30 boys appeared in the exam from class 8, find the percentage of boys who
could not secure 'A' grade.
f) If 20 students could not secure 'A' grade from class 6, find the percentage of
students who secured 'A' grade.
It’s your time - Project work!
7. a) Let’s collect and write the marks obtained by your friends in the recently
conducted mathematics exam. Group the data into the class interval of
length 10 and show them in a cumulative frequency distribution table.
b) Let’s collect and write the number of students from class 1 to 8 in your
school. Show the data in a line graph.
20.6 Average (or Mean)
Suppose Ram has Rs 10 and Hari has Rs 8.
Then, we often say that in average Ram and Hari have Rs 9.
Let's think how we get an average of Rs 9 from Rs 10 and Rs 8.
Now, let's answer what the averages of (i) 2 and 4 (ii) 5 and 9 (iii) 2, 3 and 4 are.
Thus, an average is a single number that represents the central value of a set of data.
Of course, an average indicates the quality of the given data.
An average of the given data is calculated by adding them together and dividing the
sum by the number of data.
i.e., Average = Total sum of the data
Number of data
If x is used to represent the data, n is used to represent the number of data and x is
used to represent the average or mean of the data, then
Average (x) = Σnx
Here, the symbol’ Σ' means summation of whole data.
Worked-out examples
Example 1: The marks obtained by 7 students of a class 7 in a mock test of
mathematics are given below. Find the average marks.
40, 34, 20, 22, 38, 28, 35
293Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Statistics
Solution:
Here, Σx = 40 + 34 + 20 + 22 + 38 + 28 + 35 = 217 and n = 7
∴ Average(x) = Σx = 217 = 31
n 7
Example 2: The ages (in years) of 6 children are given below. If the average
age of the children is 9 years, find the value of p.
9, 6, 8, 7, p, 13
Solution:
Here, Σx = 9 + 6 + 8 + 7 + p + 13 = 43 + p and n = 6
Σx
Now, average (x) = n
or, 9 = 43 + p
or, p = 54 6
or, 43 +
p = 54 – 43 = 11
Example 3: The average of marks obtained by Govinda in English and Maths
is 21. If the marks obtained by him in Science is also included, the
average is 24. How many marks did he obtain in Science?
Solution: Total marks in two subjects
2
Here, average marks in English and Maths =
or, 21 = x or, x = 42
2
Total marks in three subjects
Also, average marks including Science = 3
or, 24 = y or, y = 72
3
∴ Marks obtained in Science = y – x = 72 – 42 = 30.
Example 4: Ramesh obtained an average marks of 80 in Maths and Science
and an average marks of 70 in English and Science. If he obtained
75 marks in Science, find his marks in Maths and English.
Solution: average marks of Maths and Science = Total marks in two subjects
Here, 80 =
x2
or, 2
or, x = 160
∴ Total marks obtained in Maths and Science = 160
Also, total marks obtained in English and Science = 70 × 2 = 140
The marks obtained in Science = 75
∴ The marks obtained in Maths = 160 – 75 = 85
The marks obtained in English = 140 – 75 = 65.
So, his marks in Maths and English are 85 and 65 respectively.
Vedanta Excel in Mathematics - Book 7 294 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Statistics
EXERCISE 20.2
General Section - Classwork
1. Let's say and write the answers as quickly as possible.
a) Average of 4 and 6 is .................... b) Average of 3 and 5 is ....................
c) Average of 3, 4, and 5 is .................... d) Average of 5, 10, and 15 is ..................
2. a) Average of 2 and x is 3, x = ............ b) Average of 3 and y is 4, y = ............
c) Average of p and 10 is 16, p = .......... d) Average of z and 12 is 10, z = ..........
3. a) Σx = 20, n = 5, average = ........ b) Σx = 32, n = 8, average = ...........
c) n= 10, average (x) = 5, Σx = ........... d) average (x) = 6, n= 9, Σx = ...........
Creative Section - A
4. Let's find the average of the following data.
a) 20, 11, 14, 10, 15 b) 30, 40, 34, 53, 45, 32
c) 11yrs, 10 yrs, 9 yrs, 14 yrs d) 5.4cm, 4.5 cm, 7.2 cm, 6.3 cm, 8.1 cm
e) average of first five prime numbers f) average of first five multiples of 6
5. a) If the average of 7, 12, 6, 10, and p is 8, find the value of p.
b) If the average of Rs 16, Rs 20, Rs 25, and Rs y is Rs 20, find Rs y.
6. a) The average rainfall of the last week was 35 mm. If the records of daily
rainfall from Sunday to Friday were 45 mm, 22 mm, 28 mm, 36 mm, 30 mm
and 24 mm, find the rainfall on Saturday.
b) The average age of Pratik, Debashis, Sunayana, and Buddhi is 16 years. If the
age of the first three children are 7 years, 13 years, and 20 years respectively,
how old is Buddhi?
Creative Section - B
7. a) The average of daily income of a plumber from Sunday to Friday is Rs 1,300
in the last week. If he earned Rs 2,000 on Saturday, find his average of daily
income of the week.
b) The average marks obtained by Anamol in English, Maths, and Science is 70.
If he obtained 60 marks in Nepali, find his average marks in 4 subjects.
8. a) The average income on Sunday and Monday of an electrician is Rs 1,800. If
his income on Tuesday is also included, the average income will be Rs 2,100.
Find his income on Tuesday.
b) The average of the marks obtained by Abdul and Mariya in Maths is 56.
If the marks obtained by Hari is also included, the average is 60. Find the
marks obtained by Hari.
295Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Statistics
9. a) Renu obtained an average marks of 70 in Science and Maths, an average
marks of 60 in Science and English. If she obtained 55 marks in Maths, find
her marks in Science and English.
b) The average age of Kishan and Dolma is 15 years and that of Dolma and
Shashwat is 14 years. If Kishan is 17 years old, find the age of Dolma and
Shashwat.
10. a) The average temperature of a town in the last week was 32°C. If the average
temperature recorded daily from Sunday to Wednesday was 30°C and that of
recorded daily from Wednesday to Saturday was 33°C, find the temperature
of the town recorded on Wednesday.
b) The average of 9 numbers is 23. If the average of first five numbers is 20 and
that of last five numbers is 25, what is the fifth number?
c) The average age of a husband and wife was 23 years at the time of their
marriage. After 10 years, they have now a daughter of 6 years, what is the
average age of the family at present?
It's your time - Project work!
11. a) Let's make groups of 5 friends and conduct a survey inside your class to
find the amount of pocket money that your friends in each group brought
on a day.
(i) Find the average amount of pocket money of each group.
(ii) Find the average amount of pocket money of the whole students of your
class.
b) Let's conduct a survey to find the number of students of your school from
class 4 to class 8. Present the data in a frequency distribution table and find
the average (mean) number of students of these classes.
c) Let's conduct a survey to find the number of students of different ages from
class 6 to class 8 in your school. Present the data in a frequency table and
calculate the mean age. How much close your age from the mean age?
OBJECTIVE QUESTIONS
The following line graph shows the number of 35 Y
patients admitted in a hospital of first six months 30
of a year. Observe the line graph and answer the 25
questions.
1. How many patients were admitted in Asar? 20
(A) 35 (B)15 (C) 20 (D) 30 No. of patients 15
2. In which month the maximum numbers of 10
patients were admitted in the hospital?
5
(A) Bhadra (B) Asij
0 Baisakh Jesth Asar ShrawanBhadra Asoj
(C) Baishakh (D) Shrawan Months X
Vedanta Excel in Mathematics - Book 7 296 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Statistics
3. The ratio of number of patients admitted in Baishakh and Asoj is
(A) 7:4 (B) 7:3 (C) 7:5 (D) 7:6
4. How many patients were admitted in the hospital in the six months?
(A) 135 (B) 140 (C) 145 (D) 105
The given multiple bar graph represents the result of classes VI to VII. Observe the bar
graph and answer the following questions.
5. How many students secured A+ grade in class VIII?
(A) 9 (B) 8 (C) 15 (D) 10
6. Total number of students who secured B+ grade in classes VI and VII
(A) 30 (B) 15 (C) 25 (D) 55
7. Total number of students in class VII is
(A) 45 (B) 30 (C) 48 (D) 35
8. The class in which more students secured A+ grade is
(A) class VI (B) class VII (C) class VIII (D) class VI and VII
9. The ratio of number of students who secured A grade in class VI and VII is
(A) 2:3 (B) 3:2 (C) 4:5 (D) 1:4
10. What percent of students secure A grade in class VI?
(A) 50% (B) 3331% (C) 1632% (D) 6623%
Vedanta ICT Corner
Please! Scan this QR code or
browse the link given below:
https://www.geogebra.org/m/rmrsxnxr
297Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Answers
1. Set
Creative Section Exercise - 1.1
hydrogen
4. Answer the questions and show to your teacher. oxygen
5. a) {hydrogen, oxygen, nitrogen, carbondioxide } nitrogen
l carbondixoide
ap
b) {l, a, p, t, o}
to 12 d) {4, 6, 8, 9} 4
c) {1, 2, 3, 4, 5, 6, 7, 8, 9} 34 5 68
678
9
9
6. a) A = {natural numbers less than 5} b) B = {first five multiples of 5}
c) C = {prime numbers between 5 and 20} d) D = {factors of 6}
7. a) P = {11, 13, 17, 19} b) A = {f, o, t, b, a, l}
c) F = {1, 2, 3, 6, 9, 18} d) M = {6, 9}
8. a) A = {x : x is a natural number, x < 6 b) B = {y : y is a prime number less than 10}
c) C = {x : x is a square number less than 26} d) D = {f : f is a factor of 8}
9. Complete your project work and compare with your friends. Then, discuss in the class.
Creative Section Exercise - 1.2
3. Answer the questions and show to your teacher.
4. a) A = {12, 14, 15, 16, 18}, n(A) = 5 b) B = {1, 2, 3, 4, 6,12}, n(B) = 6
c) Z = {–2, –1, 0, 1, 2}, n(Z) = 5 d) W = {0}, n(W) = 1
5. a) A = {11, 13, 15, 17, 19}, finite set b) B = {11, 13, 15, ...}, infinite set
c) C = { }, empty set d) D = {15}, unit set
6. Complete your project work and compare with your friends. Then, discuss in the class.
Exercise - 1.3
Creative Section
7. Answer the questions and show to your teacher.
8. a) A = {Sunday, Saturday}, B = {Tuesday, Thursday}, equivalent sets
b) M = {f, o, l, w}, N = {w, o, l, f}, equal sets
9. a) O = {1, 3, 5, 7, 9}, S = {1, 4, 9}, overlapping sets
b) X = {2, 3, 5}, Y = {7, 11}, disjoint sets
10.
Subsets Proper subsets
a) {apple}, φ a) φ
b) {1}, {2}, {1, 2} , φ b) {1}, {2}, φ
c) {a}, {b}, {c}, {a, b} {b, c}, {a, c}, {a, b, c}, φ c) {a}, {b}, {c}, {a, b} {b, c}, {a, c}, φ
d) {1}, {2}, {3}, {4} {1, 2}, {1, 3}, {1, 4}, {2, 3}, d) {1}, {2}, {3}, {4} {1, 2}, {1, 3}, {1, 4}, {2, 3},
{2, 4}, {3, 4}, {1, 2, 3}, {2, 3, 4}, {1, 3, 4}, {2, 4}, {3, 4}, {1, 2, 3}, {2, 3, 4}, {1, 3, 4},
{1, 2, 4}, {1, 2, 3, 4}, φ {1, 2, 4}, φ
11. a) A = {1, 2, 3, 6}, B = {1, 2, 4, 8}, set of common elements = {1, 2}
b) P = {m, a, t, h}, Q = {t, h, i, n, k}, set of common elements= {t, h}
A B
12
12. a) A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, B = {1, 2, 3, 4, 6, 8, 12, 24} 0 241836
24
Set of common elements = {1, 2, 3, 4, 6, 8} P 5Q 57
4 9
8
b) P = {4, 8, 12, 16, 20}, Q = {5,10, 15, 20} 12 20 10
Set of common elements = {20} 15
16
13. a) {Mina, Shiva, Hari, Ram, Sita, Dolma} A B
Mina Ram
Shiva Sita
Hari Dolma
Vedanta Excel in Mathematics - Book 7 298 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
The words you are searching are inside this book. To get more targeted content, please make full-text search by clicking here.
Vedanta Excel in Mathematics Book 7 Final (2079)
Discover the best professional documents and content resources in AnyFlip Document Base.
Search