Vedanta Excel Maths Book 7

Triangle, Quadrilateral and Polygon

Quadrilateral 4 2 2 × 180° (4 – 2) × 180°

Pentagon 5 3 3 × 180° (5 – 2) × 180°

Hexagon 6 4 4 × 180° (6 – 2) × 180°

Heptagon 7

Octagon 8

Nonagon 9

Decagon 10

n–gon n (n – 2) × 180°

In the adjoining figure, ABCDE is a pentagon. AB, BC, CD, D
DE, and EA are the sides of the pentagon. Similarly, ∠A,
∠B, ∠C, ∠D, and ∠E are the interior angles of the pentagon. E C

A polygon is said to be regular if its all sides are equal and AB
every interior angle is equal.

The sum of the interior angles of a regular polygon is obtained by using a formula

(n – 2) × 180°, where n is the number of sides of any polygon.

So, each interior angle of a regular polygon = (n – 2) × 180° ED
n

Similarly, the angle made by the produced A y
side and its adjacent side is called the exterior angle of polygon. F C
In the adjoining hexagon, ∠x, ∠y, etc. are the exterior angles.
x
The sum of exterior angles of a regular polygon is 360°. B

So, each exterior angle = 36n0°, where n is the number of sides of polygon.

Worked-out examples

Example 1. Find out each interior angle of regular hexagon.
Solution:
Here, in a regular hexagon, number of sides (n) = 6
Each interior angle = ?

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 249 Vedanta Excel in Mathematics - Book 7

Triangle, Quadrilateral and Polygon

Now, (n – 2) × 180° (6 – 2) × 180°
n 6
Each interior angle = =

= 4 × 30° = 120q
Hence, each interior angle of regular hexagon is 120°.

Example 2 Find each exterior angle of regular octagon.
Solution:
Here, number of sides in a regular octagon (n) = 8
Each exterior angle = ?

Now, Each exterior angle = 360° = 360° = 45q
n 8

Hence, each exterior angle of a regular octagon is 45°.

EXERCISE 14.3
General Section - Classwork

1. Let's tell and write the correct answer in the blank spaces.
a) The number of sides (n) of the following polygons are:

(i) pentagon, n = ........... (ii) hexagon, n = ............

(iii) heptagon, n = ......... (iv) Octagon, n = .............

(v) nonagon, n = ........... (vi) decagon, n = ...........

b) The sum of the interior angle of a regular polygon with 'n' number of side is
obtained by the formula ................................

c) Each interior angle of a rectangular polygon with 'n' number of sides is
obtained by the formula ................................

d) Each exterior angle of a regular polygon with 'n' number of sides is obtained
by the formula ................................

Creative Section - A

2. Find the sum of the interior angles of the following polygons by using formula.

a) Triangle b) Quadrilateral c) Pentagon d) Hexagon
e) Heptagon f) Octagon g) Nonagon h) Decagon

3. Find each interior angle of the following regular polygons by using formula.

a) Triangle b) Quadrilateral c) Pentagon d) Hexagon

e Heptagon f) Octagon g) Decagon h) Dodecagon

Vedanta Excel in Mathematics - Book 7 250 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Triangle, Quadrilateral and Polygon

4. Find each exterior angle of the following regular polygons by using formula.

a) Triangle b) Quadrilateral c) Pentagon d) Hexagon

e) Heptagon f) Nonagon g) Decagon h) Dodecagon

Creative Section - B

5. Find the number of sides and name the polygons having the following sum of
interior angles.

a) 720° b) 900° c) 1080° d) 1440°

6. Find the number of sides and name the regular polygons having the following
interior angles.

a) 108° b) 120° c) 135° d) 140°

7. Find the number of sides of regular polygons having the following exterior
angles.

a) 90° b) 72° c) 60° d) 36°

8. Find the value of x° and y° from the following polygons.

a) b) c) A d) 80°
B x° y°
A A F H
C x°
B E B E D 110°
D C y° G
y° x° x° y° G
C D F
E

It's your time - Project work!

9. a) Let's make groups of five friends. Take the necessary number of straws

of equal length. Then, each friend in a group should make two model of
polygons by joining the straws with cello tape such that each group makes a
triangle, a quadrilateral, ... a decagon.

b) Now, place each model of polygon on the chart paper and trace its outline.

10. a) Let's draw the sketches of 3 sides polygon to 12 sides polygons and name

each polygon (11-sided is hendecagon and 12-sided is dodecagon).

b) How does dodecagon look like? Is it like a circle?

c) If we go on increasing the number of sides of polygons infinitely, what type
of plane shape can be obtained? Discuss in the class.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 251 Vedanta Excel in Mathematics - Book 7

Unit Congruency and Similarity

15

15.1 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
exactly the same shape and size.

The symbol # is used to denote ' is congruent to'. Thus, from above examples
(i) AB # PQ (ii) ‘MON # ‘XOY (iii) 'ABC # 'DEF (iv) ABCD # PQRS

15.2 Congruent triangles C D D' C
Let's take a rectangular sheet of paper. Fold it DD'
A B B'
along one of its diagonals and cut the folding edge C D'

into two right angled triangular pieces. Place the BB' A D
triangular pieces one above another as shown in A

the figure. Now, answer these questions.

Are the triangular pieces exactly fitted to one above another ?

Which are the angles exactly fitted ? B' B
Which are the arms exactly fitted ?

Vedanta Excel in Mathematics - Book 7 252 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Congruency and Similarity

Could you investigate some ideas about congruent triangles ?

The figures which are exactly the same in shape and size are called congruent
figures.

In the adjoining figures, 'ABC and 'PQR A P

are congruent triangles because their shapes 2.5 cm
and sizes are exactly the same. In this case, 2.5 cm
each part of 'ABC can exactly be fitted over 3 cm 85° 3 cm 85°
45° 50° 45° 50°
B 4 cm C Q 4 cm R

the corresponding parts of 'PQR.

'ABC is congruent to ' PQR is written as ' ABC # ' PQR.

15.3 Conditions of congruency of triangles
Two triangles are congruent under the following conditions.

Condition 1: Side – Side – Side (S.S.S.) axiom

If three sides of one triangle are respectively equal to three corresponding sides of
another triangle, the triangles are said to be congruent.

In 'ABC and 'PQR, AP
(i) AB = PQ (S)

(ii) BC = QR (S) B CQ R
(iii)CA = RP (S)

(iv) ? 'ABC # 'PQR (S.S.S. axiom)

(v) ? ‘A = ‘P, ‘B = ‘Q and ‘C = ‘R [Corresponding angles of congruent triangles]

The angles of triangles lying to the opposite of equal sides are called
corresponding angles. For example: sides BC = QR, the angle opposite to
BC is ‘A and the angle opposite to QR is ‘P. So, ‘A and ‘P are corresponding
angles.

Condition 2: Side-Angle-Side (S.A.S.) axiom

If two sides of one triangle are respectively equal to two sides of another triangle
and the angle made by them are also equal, the triangle are said to be congruent.

In 'ABC and 'PQR, AP
(i) AB = PQ (S) B CQ R
(ii) ‘B = ‘Q (A)
(iii)BC = QR (S)
(iv) ? 'ABC # 'PQR (S.A.S. axiom)

253Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7

Congruency and Similarity

(v) ‘C = ‘R and ‘A = ‘P [Corresponding angles of congruent triangles]
(vi) AC = PR [Corresponding sides of congruent triangles]

The sides opposite to the equal angles of congruent triangles are called
corresponding sides. For example: ‘B = ‘Q, the side opposite to ‘B is AC
and the side opposite to ‘Q is PR. So, AC and PR are corresponding sides.

Condition 3: Angle – Side – Angle (A.S.A) axiom

If two angles of one triangle are respectively equal to two angles of another triangle
and the adjacent sides of the angles are also equal, the triangles are said to be
congruent.

In 'ABC and 'PQR, AP

(i) ‘B = ‘Q (A)

(ii) BC = QR (S) B CQ R
(iii)‘C = ‘R (A)

(iv) ? 'ABC # 'PQR (A.S.A. axiom)

(v) AC = PR and AB = PQ [Corresponding sides of congruent triangles]

(vi) ‘A = ‘P [Corresponding angles of congruent triangles]

Condition 4: Right angle-Hypotenuse-Side (R.H.S.) axiom

If hypotenuse and one of two other sides of a right angled triangle are respectively
equal to the hypotenuse and a side of other right angled triangle, the triangles are
said to be congruent.

In rt. ‘ed 'sABC and PQR, AP

(i) ‘B = ‘Q (R)

(ii) AC = PR (H) B CQ R
(iii) AB = PQ (S)

(iv) ? 'ABC # 'PQR (R.H.S. axiom)

(v) BC = QR [Corresponding sides of congruent triangles]

(vi) ‘C = ‘R and ‘A = ‘P [Corresponding angles of congruent triangles]

15.4 Similar triangles

The figures which are exactly the same in shape P
75°
but their sizes may be different are known as A 60°

similar figures. In the given figure, 'ABC and 75°
60° 45°
'PQR are similar triangles because they have equal B CQ 45° R
angles and hence the same shape.

Thus, if three angles of one triangle are respectively equal to three angles of
another triangle, they are said to be similar.

'ABC is similar to 'PQR is written as 'ABC ~ 'PQR. The symbol ~ is used to
denote ‘is similar to’.

Vedanta Excel in Mathematics - Book 7 254 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Congruency and Similarity

Here, AB and PQ, BC and QR, CA and RP are the corresponding sides of the similar
triangles.

The corresponding sides of similar triangles are always proportional, i.e., the ratios
of the corresponding sides are equal.

? AB = BC = CA .
PQ QR RP

Worked-out examples

Example 1: Use necessary axiom and show that 'ABC # 'PQR. Also, write their

corresponding sides and angles.

Solution: AP

In 'ABC and 'PQR,

(i) ‘B = ‘Q = 60q (A)

(ii) BC = QR = 3.5 cm (S) 60° 75° 60° 75°
(iii)‘C = ‘R = 75q (A) Q R
(iv) ?'ABC # 'PQR [A.S.A. axiom] B 3.5 cm C 3.5 cm

(v) AB = PQ and AC = PR [Corresponding sides of congruent triangles]

(vi) ‘A = ‘P [Corresponding angles of congruent triangles]

Example 2: In the figure, 'DEF ~ 'XYZ. Find the length of YZ.
Solution:

Here, 'DEF ~ 'XYZ D X
60° 60°
? DE = EF 8 cm Y 40° 80° Z
XY YZ 6 cm
8 4
or, 6 = YZ 40° 80°
4 cm
or, 8YZ = 24 E F

or, YZ = 24 = 3 cm
8

EXERCISE 15.1
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 B These line segments are
D .........................
a)

C

A

P These triangle are
.........................
b)
RC
Q
B

255Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7

Congruency and Similarity

P SW Z These squares are
RX Y .......................
c)
These circles are
Q .........................

d)

e) These figures are
.........................

f) These squares are
.........................

2. The following pairs of triangles are congruent. Let's tell and write the necessary
axioms of congruency.
a) b)

....................... axiom ....................... axiom

c) d)

....................... axiom ....................... axiom

3. From these pairs of congruent triangles, tell and write the corresponding sides
and corresponding angles as quickly as possible.

A P Corresponding angle of ‘A is .........................
Corresponding angle of ‘R is .........................
a)

Corresponding angle of ‘B is .........................

B CQ R

b) F X Corresponding side of FD is .........................
Corresponding side of YZ is .........................
D EZ Y Corresponding angle of ‘F is .........................

Vedanta Excel in Mathematics - Book 7 256 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Congruency and Similarity

4. Let's tell and write the equal ratios of corresponding sides of these pairs of
similar triangles.
X

a) A b) F

P

B CQ R D EY Z
................................................
................................................

Creative Section - A

5. a) Define the meaning of congruent figures and similar figures.

b) Are all congruent figures similar?

c) Are all similar figures congruent?

d) Draw the sketch of two congruent triangles and name them. Write the name
of corresponding sides and corresponding angles.

e) Write the necessary conditions (or axioms) to be two triangles congruent.

6. Lets use the necessary axioms and show that the following pair of triangles are

congruent. Also write their corresponding sides and angles.

a) A D b) P X

B 50°4 cm C E 50° F Q 110° 35° R Y 110° 35° Z
3.5 cm 4 cm3.5 cm 2 cm 2 cm
3.6 cm
c) K 3.6 cmRd)EO
5 cm
5 cm 3.2 cm
3.2 cm

L 4 cm M S 4 cm T F 4.5 cm G P 4.5 cm Q

7. Let's use the necessary axioms and show that the following pairs of triangles

are congruent. Also, write their corresponding sides and angles.

a) A P b) X ZD F

B CQ R Y E
R C
c) K d) Q

L MS T O PB A
257Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7

Congruency and Similarity

8. In the following pairs of similar triangles, find the unknown sizes of the sides.

a) A P b) F Z
30°
50° 50° 30°
8 cm
6 cm 6 cm
a cm
B 55° 75° C Q 55° 75° R 115° 115°
4 cm x cm 35°
D 6 cm E X 35° Y
3 cm G
c) d)
12 cm S x cm K E
R 50° 45° U 50° 45° V 80° y cm 80° 6 cm

85° 10 cm 85° L 60° 40° M 60° 40°
T 5 cm 15 cm 9 cm
F
W

Creative Section - B A C
9. a) In the figure alongside, use A.S.A. axiom and prove that
B D
'ABO # 'CDO. S R
P
b) In the adjoining figure, use S.A.S. axiom and prove that Q
'PQR # 'PRS. A

c) In the figure alongside, AD A BC and AB = AC,
prove that ∆ABD = ∆ ACD by using RHS axiom.

d) In the adjoining figure, ABCD is a square. B DC
DC
Prove that 'ABC # 'ADC by using
AB
(i) S.A.S. axiom (ii) R.H.S. axiom

(iii) S.S.S. axiom (iv) A.S.A. axiom separately.

It's your time - Project work!

10. 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 finds on the activities a) and b). Then, present
in the class.

Vedanta Excel in Mathematics - Book 7 258 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Unit Construction

16

16.1 Construction of perpendicular bisector of a line segment

In the adjoining figure, the line segment PQ is perpendicular to P

AB and it divides AB into two equal parts. It means PQ bisects B
AB perpendicularly. Therefore, PQ is the perpendicular bisector A

of AB. Thus, a line segment which is perpendicular to a given line

segment and bisects the line segment is called the perpendicular Q
bisector of the given line segment.

Let's draw a line segment AB and construct its perpendicular bisector.

Steps of construction

(i) Draw a line segment AB.

(ii) With centre at A and radius more than half of AB, draw
two arcs above and below AB.

(iii) With centre at B and the same radius, draw two arcs
intersecting the first two arcs at P and Q respectively.

(iv) Join P and Q. PQ intersects AB at C. Measure AC, BC, and
‘PCA (or ‘PCB).

Here, AC = BC and ‘PCA = 90q.

? PQ is the perpendicular bisector of AB B R

16.2 Transferring angles D T P
OC A QS
The process of constructing an angle equal to
the given angle at the given vertex is called
transferring angles.

Construct an angle at Q equal to ‘AOB.

Steps of construction

(i) From the point Q, draw a line segment QP.

(ii) With centre at O and a suitable radius draw an arc to intersect OA at C and OB
at D.

(iii) With centre at Q, draw the same arc to intersect QP at S.

(iv) Measure the length of CD with the help of compasses and draw an arc of the
same length from S to cut previous arc at T.

(v) Join Q and T and produce it to R.

Thus ‘PQR = ‘AOB is drawn.

259Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7

Construction

16.3 Construction of different angles DSC

Construct 60q, 120q and 90q with the help of compasses.

Steps of construction RQ

(i) Draw a line segment AB. B

AP

(ii) With centre at A and a suitable radius, draw an arc to cut AB at P.

(iii) From P, draw an arc with the same radius to cut the first arc at Q. Join A, Q, and
produce to C.

Here, ‘BAC = 60q.

(iv) From Q draw an arc with the same radius to cut the first arc at R. Join A, R, and
produce to D.

Here, ‘BAD = 120q.

(v) From Q and R draw two arcs with the same radius to intersect each other at S.

(vi) Join A and S.

Here, ‘BAS = 90q. D E C
Construct 75q with the help of compasses. R

Steps of construction QP B
(i) Draw a line segment AB. A
(ii) At A, construct ‘BAC = 60q and ‘BAD = 90q.

(iii) From P and Q, draw two arcs with the same radius to intersect each other at R.

(iv) Join A, R and produce to E.

Here, ‘BAE = 75q. C

Construct 30q and bisect it. Q RD
Steps of construction S TE
(i) Draw a line segment AB.
(ii) At A, construct ‘BAC = 60q. AP B

(iii) From P and Q, draw two arcs to intersect each other at R.

(iv) Join A, R and produce to D.

Here, ‘BAD = 30q.

(v) From P and S draw two arcs to intersect each other at T.

(vi) Join A, T and produce to E.

Here, ‘BAE = 1 ‘BAD = 1 u 30q = 15q.
2 2

EXERCISE 16.1

1. Let's draw the line segments of the following length and construct their

perpendicular bisectors.

a) AB = 5.4 cm b) PQ = 6.8 cm c) MN = 7.2 cm

Vedanta Excel in Mathematics - Book 7 260 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Construction

2. Let's transfer each of the following angles at the given vertices.

a) C b) Z S c) R L

P

BA XY Q P

3. Let's construct the following angles with the help of compasses.

a) 60q b) 120q c) 90q d) 30q e) 45q
f) 75q g) 105q h) 150q i) 15q 1
j) 22 2 q

4. a) Draw a line segment AB = 5 cm. Construct ‘BAC = ‘ABC = 60q. Now, draw

the perpendicular bisectors of each side of ' ABC. Do all the perpendicular

bisectors intersect at the same point?

b) Draw a line segment PQ = 6.2 cm. Construct ‘QPR = 60q and ‘PQR = 45q.
Now, draw the angular bisectors of each angle of 'PQR. Do all the angular
bisectors intersect at the same point?

16.4 Construction of triangles

Let’s learn to construct the various types of triangles under the following conditions.

C

1. When the lengths of 3 sides are given

Construct a triangle ABC in which AB = 5.5 cm, 3.8 cm 4.6 cm

BC = 4.6 cm and CA = 3.8 cm

Steps of construction

(i) Draw a line segment AB = 5.5 cm. A 5.5 cm B

(ii) With centre at B and radius 4.6 cm, draw an arc.

(iii) With centre A and radius 3.8 cm, draw another arc to intersect the first arc

at C.

(iv) Join A, C and B, C.

Thus, ABC is the required triangle.

2. When the lengths of two sides and angle made by them are given

Construct a triangle ABC in which

AB = 5.2 cm, AC = 4.5 cm and ‘A = 45q. X
C
Steps of construction
(i) Draw a line segment AB = 5.2 cm 120° 90° 60°
(ii) At A, construct ‘BAX = 45q

(iii) With centre at A and radius 4.5 cm, draw 45° 5.2 cm B
an arc to cut AX at C. A

(iv) Join B and C.

Thus, ABC is the required triangle.

261Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7

Construction

3. When two angles and their common adjacent side are given

Construct a triangle ABC in which AB = 4.8 cm, ‘A = 60q and ‘B = 30q.

Steps of construction

(i) Draw a line segment AB = 4.8 cm

(ii) At A and B construct ‘BAX = 60q and ‘ABY = 30q respectively. AX and

BY intersect each other at C. YC X

Thus, ABC is the required triangle.

60° 30° B
A 4.8 cm

4. When lengths of hypotenuse and a side of a right angled triangle are given

Construct a right angled triangle ABC in which a side AB = 4.5 cm, and

hypotenuse AC = 6 cm. X
Steps of construction

(i) Draw a line segment AB = 4.5 cm C
(ii) At B, construct ‘ABX = 90q 6 cm
(iii) With centre A and radius 6 cm, draw an arc to

intersect BX at C. 90°
B
(iv) Join A and C. A 4.5 cm

Thus, ABC is the required right-angled triangle.

EXERCISE 16.2

1. Let's construct triangle ABC under the following conditions.

a) AB = 4 cm, BC = 4.5 cm, CA = 5 cm

b) BC = 5.5 cm, CA = 4.8 cm, AB = 4.2 cm

c) AC = 6.5 cm, AB = 5.4 cm, BC = 4.8 cm

2. Let's construct triangle PQR in which

a) PQ = 4.7 cm, ‘P = 60q, PR = 5.3 cm

b) QR = 5.6 cm, ‘Q = 45q, RP = 6.2 cm

c) PR = 4.5 cm, ‘R = 75q, PQ = 5.8 cm

3. Let's construct triangle XYZ in which

a) XY = 3.8 cm, ‘X = 60q, ‘Y = 45q

b) YZ = 4.6 cm, ‘Y = 120q, ‘Z = 30q

c) XZ = 5.3 cm, ‘X = 75q, ‘Z = 45q

4. Let's construct the right angled triangle ABC in which

a) AB = 4.5 cm, hypotenuse AC = 5.7 cm

b) AB = 3 cm, hypotenuse BC = 5 cm
c) BC = 5 cm, hypotenuse AB = 6.5 cm

Vedanta Excel in Mathematics - Book 7 262 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Construction

16.5 Construction of parallelograms
Let’s construct parallelograms under the following given conditions.

1. When two adjacent sides and angle made X
C
by them are given D 4.5 cm

Construct a parallelogram ABCD in

which AB = 4.5 cm, BC = 3.5 cm, and 3.5 cm
3.5 cm
‘ABC = 60q.

Steps of construction 60° B

(i) Draw a line segment AB = 4.5 cm. A 4.5 cm
(ii) At B, construct ‘ABX = 60q.

(iii) With the centre at B and radius BC = 3.5 cm draw an arc to intersect BX

at C.

(iv) With centre at C and radius CD = AB = 4.5 cm, draw an arc.

(v) With centre at A and radius AD = BC = 3.5 cm, draw another arc to cut

the previous arc at D.

(vi) Join A, D and C, D.

Thus, ABCD is the required parallelogram.

2. When base, diagonal, and angle made by the diagonal with base are given

Construct a parallelogram ABCD in which base AB = 5 cm, diagonal AC = 6.3

cm and ‘BAC = 30q.

Steps of construction D 5 cm X
(i) Draw base AB = 5 cm 6.3 cm C
(ii) At A, construct ‘BAX = 30q

(iii) With centre at A and radius AC = 6.3 cm,

draw an arc to cut AX at C. A 30° 5 cm B
(iv) With centre at C and radius CD = AB = 5

cm, draw an arc.

(v) With centre at A and radius AD = BC, draw another arc to cut the previous

arc at D.

(vi) Join A, D, and B, C.

Thus, ABCD is the required parallelogram.

3. When the length of a diagonals and an angle XD

made by them are given

Construct a parallelogram ABCD in which

diagonals AC = 5 cm, BD = 5.8 cm, and they A C
BY
bisect each other making an angle of 30q. O

Steps of construction

(i) Draw a diagonal AC = 5 cm,

263Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7

Construction

(ii) Draw the perpendicular bisector of AC and mark its mid-point O.

(iii) At O, construct ‘AOX = 30q and produce XO to Y.

(iv) Here, O is also the mid-point of BD. With centre at O and radius
1
OB =OD = 2.9 cm ( 2 of BD) cut OY at B and OX at D.

v) Join A and D, B and C, A and B, and C and D.

Thus, ABCD is the required parallelogram.

16.6 Construction of squares
Let's construct squares under the following conditions.

1. When a side of a square is given X Y
Construct a square ABCD in which AB = 4 cm. D C

Steps of construction 90° 4 cm 90°
A B
(i) Draw a line segment AB = 4 cm.
(ii) At A and B construct

‘BAX = 90q and ‘ABY = 90q respectively.
(iii) With centres at A and B and radius 4 cm, cut

AX at D and BY at C.
(iv) Join C, D.

Thus, ABCD is the required square.

2. When the length of a diagonal is given

Construct a square ABCD in which diagonal BD = 5 cm.

Steps of construction Y X
(i) Draw the diagonal BD = 5 cm. B 45° A
(ii) Diagonals bisect the angles of a square.
5 cm 45°
So, construct ‘DBX = ‘BDY = 45q. D
(iii) Let, BX and DY intersect at A.
(iv) With centres at B and D and radius equal

to AB (or AD), draw two arcs intersecting
at C.
(v) Join B, C and D, C.
Thus, ABCD is the required square.

C

Vedanta Excel in Mathematics - Book 7 264 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Construction

16.7 Construction of rectangles Y X
D C
1. When two adjacent sides of a rectangle are given:
Construct a rectangle ABCD in which AB = 4.5 90° 4.5 cm 90°
cm and BC = 3.5 cm. A B

Steps of construction:
(i) Draw a line segment AB = 4.5 cm.
(ii) At A and B, construct ‘ABX = ‘BAY = 90q.
(iii) With the centres A and B and radius 3.5 cm,

draw two arcs to cut AY at D and BX at C.
(iv) Join C and D.

Thus, ABCD is the required rectangle.

2. When the diagonals and an angle made by them are given:

Construct a rectangle ABCD in which diagonals AC = BD = 5 cm and they
bisect each other making 60q angle.

Steps of construction. X
D
(i) Draw AC = 5 cm and draw its perpendicular
60°
bisector to find its mid-point O. O

(ii) At O, construct ‘AOX = 60q and produce XO to

Y. 1 C
2
(iii) With centre at O and radius 2.5 cm ( of BD) A

draw two arcs to cut OX at D and OY at B.

(iv) Join A and D, B and C, A and B, and C and D. BY
Thus, ABCD is the required rectangle.

16.8 Construction of rhombus

1. When a side and angle made by two adjacent sides are given.

Construct a rhombus ABCD in which AB = 4.5 cm and ‘ ABC = 45q.

Steps of construction: 4.5 cm C B
4.5 cm 4.5 cm
(i) Draw AB = 4.5 cm
(ii) At B, construct ‘ABX = 45q. D A 4.5 cm

(iii) With centre at B and radius
4.5 cm, cut BX at C.

(iv) With centres at A and C and
radius 4.5 cm, draw two arcs
intersecting each other at D.

(v) Join A, D and C, D.

Thus ABCD is the required
rhombus.

265Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7

Construction

2. When two diagonals are given: X
D
Construct a rhombus ABCD in which diagonals AC = 5.4
cm and BD = 4.2 cm. OC
B
Steps of construction:
Y
(i) Draw AC = 5.4 cm and draw its perpendicular A

bisector XY. Mark the mid-point of AC as O.

(ii) With centre at O and radius 2.1 cm ( 12 of BD), draw
two arcs to cut OX at D and OY at B.

(iii) Join A and B, B and C, C and D, and D and A.

Thus ABCD is the required rhombus.

16.9 Construction of kite
1. When two unequal adjacent sides and angle between them are given

Construct a kite ABCD in which AB = 4.5 cm, AD = 3 cm and ‘BAD = 120q.

Steps of construction X 3 cm C
(i) Draw AB = 4.5 cm. D
(ii) At A, construct ‘BAX = 120q.
3 cm
(iii) With centre at A and radius 3 cm, cut AX at D.
120° B
(iv) With centre at D and radius 3 cm, draw arc. 4.5 cm
With centre at B and radius 4.5 cm, draw
another arc to cut previous arc at C.

(v) Join C, D and B, C.

Thus, ABCD is the required kite.

2. When two unequal adjacent sides and angle between two equal adjacent sides

are given X C

Construct a kite ABCD in which AB = 3 cm, BC = 5 cm 5 cm
and ‘BAD = 90q.

Steps of construction 3 cm
5 cm
(i) Draw AB = 3 cm.

(ii) At A, construct ‘BAX = 90q. A 90° 3 cm B
(iii) With centre at A and radius 3 cm, draw an arc to

cut AX at D.

(iv) With centres at D and B and radius 5 cm, draw two arcs intersecting each
other at C.

(v) Join C, D and B, C.
Thus, ABCD is the required kite.

Vedanta Excel in Mathematics - Book 7 266 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Construction

3. When two diagonals are given X
D
Construct a kite ABCD in which diagonals 2 cm
AC = 5.5 cm and BD = 4 cm. 90°
O 5.5cm
Steps of construction:
B
(i) Draw AC = 5.5 cm Y C
(ii) Mark any point O on AC and construct A

‘AOX = 90q. Produce XO to Y. Dcman(d21OoYf 2 cm

(iii) With centre at O and radius 2
BD) draw two arcs to cut OX at

at B.

(iv) Join A and B, A and D, B and C, and C and D.

Thus, ABCD is the required kite.

EXERCISE 16.3

1. Let's construct the parallelogram ABCD in which
a) AB = 4 cm, BC = 3 cm and ‘ABC = 60q
b) AB = 5 cm, AD = 3.5 cm and ‘BAD = 30q
c) Base AB = 4.5 cm, diagonal AC = 6.5 cm and ‘BAC = 30q
d) Base AB = 4 cm, diagonal BD = 6 cm and ‘ABD = 45q.

e) Diagonals AC = 4.8 cm, BD = 5.8 cm and they bisect each other making an
angle of 30q.

f) Diagonals AC = 5 cm, BD = 6 cm and they bisect each other making an
angle of 60q.

2. Let's construct the squares PQRS in which

a) PQ = 3.5 cm b) QR = 4.5 cm

c) Diagonal BD = 5.5 cm d) Diagonal AC = 5 cm

3. Let's construct the rectangles ABCD in which-

a) AB = 5 cm and BC = 3.5 cm b) BC = 5.4 and CD = 4 cm

c) Diagonals AC = BD = 6 cm and they bisect each other making an angle of 45q.

d) Diagonals AC = BD = 5.6 cm and they bisect each other making an angle of 60q.

4. Let's construct the rhombus PQRS in which

a) PQ = 4 cm and ‘PQR = 30q b) PQ = 4.5 cm and ‘QPS = 60q

c) Diagonals PR = QS = 5 cm d) Diagonals PR = QS = 6 cm

5. Let's construct the kite ABCD in which

a) AB = 5 cm, AD = 3 cm and ‘BAD = 120q

b) BC = 5.5 cm, CD = 3.2 cm and ‘BCD = 105q

c) AB = 6 cm, AD = 3.5 cm and ‘BAD = 90q

d) BC = 5.6 cm, AB = 4 cm and ‘ABC = 75q

e) Diagonal AC = 5 cm and BD = 3 cm

f) Diagonals AC = 6 cm and BD = 4 cm

267Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7

Unit Circle

17

17.1 Circle and its different parts – review

A circle is a plane figure bounded by a curved line and every point
of the line is equidistance from a fixed point called the centre of the
circle.

Circumference of a circle
The curved boundary line of a circle is called its circumference.
The length of the circumference represents the perimeter of the
circle.

Radius of a circle B
The straight line drawn from the centre of a circle to a point on
its circumference is called the radius of the circle. OA
In the figure given alongside, OA is radius of the circle. Radius (r)
Radii of the same circle are always equal. OA = OB = r

Chord of a circle O Chord
The straight line segment that joins any two points on the A B
circumference of a circle is called the chord of a circle.
In the given figure, AB is the chord of the circle.

Diameter of a circle

The chord that passes through the centre of a circle is called
the diameter of the circle. Diameter is the longest chord of any
circle.
COD
In the figure, CD is the diameter of the circle. The length of the Diameter

diameter of the circle is two times its radius.

? Diameter = 2 u radius.

Sector of a circle O
The region inside a circle bounded by its two radii (plural of AB
radius is radii) is called the sector.
In the adjoining figure, the shaded region AOB is the sector.

Arc of a circle A
The part of a curve between two given points on the curve of a Arc O
circle is called an arc.
In the figure alongside, AB is the arc. B

Vedanta Excel in Mathematics - Book 7 268 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Circle

Segment of a circle O B
The region bounded by an arc and its corresponding chord is A
called the segment of a circle.
In the given figure, the shaded region represents a segment. B
AC
Semi-circle
Half part of a circle is called a semi-circle. A diameter divides a O
circle into two halves and each half is the semi-circle.

In the given figure, ABC is a semi-circle.

EXERCISE 17.1
General Section - Classwork

1. Let's tell and write the names of different parts of these circles.

a) b) O c) Q
OR
O P Q
AB N
M P
A CB
centre is .......... diameter is .......... sector is ................

radii is .......... chord are .............. segment is ...............

2. Let's tell and write the answers as quickly as possible.

a) If the radius of a circle is 2cm, its diameter is ...................

b) If the radius of a circle is 3.5cm, its diameter is ...................

c) If the diameter of a circle is 6cm, its radius is .......................

d) If the diameter of a circle is 9cm, it radius is ........................

Creative Section

3. a) Define diameter and radius of a circle with a diagram.
b) What is the relation between a diameter and a radius of a circle?
c) Define: (i) chord (ii) sector, and (iii) segment of a circle with an
appropriate diagram.
d) How do you distinguish between sector and segment of a circle? Write with
a diagram.
e) In what condition does a chord of a circle become a diameter?

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 269 Vedanta Excel in Mathematics - Book 7

Circle

4. Let's copy the following figures. Name the centres, radii, chords, diameters,
sectors, arcs, segments and semi-circles.

a) Q R P b) X c) X

O P B MAN
A

Y YY

5. Let's copy the following figures. Name the angles formed in the circumference
of each circle by chords. Also name the arc opposite to each angle.

a) B b) Q c) L N d) E

P KOM A C
P O
AO C O
Y

XR BD

6. Let's copy the figures. Name the angles formed at the centre of each circle by
radii. Also, name the arc opposite to each angle.

a) b) C Z F d) M

O Y c) E

P Q KS
N
A B
X D

It's your time : Project Work!

7. a) Let's draw two circle on a chart paper and cut out each circle.
(i) Fold one circular paper into half and again fold into half. How many
sectors did you make. Colour one of the sectors.
(ii) Are there two diameters and four radii passing through the centre of
this folded circular paper?
(iii) Take another circular paper and fold it to get a sector. Colour the sector.

b) If a circular piece of paper without marking the centre is given to you, how
do you find its centre just by folding? Let's try and find the centre.

c) Let's draw three circles with radius 3 cm, 4 cm, and 5 cm respectively.
Now, find the diameter of each circle by using the relation: d = 2r.

Then measure the length of a diameter of each circle by using a ruler and
justify that d = 2r.

Vedanta Excel in Mathematics - Book 7 270 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Unit Perimeter, Area and Volume

18

18.1 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 tell 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 tell 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 .........................

18.2 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.

271Approved 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 l
In square, all four sides are equal. DC
So, AB = CD = CD = DA = l
The Perimeter of square ABCD = AB BC CD DA ll

= l l l l A lB

= 4l
? Perimeter of square (p) = 4l

(iv) Perimeter 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 ‘S' (Pie).
Thus, if the circumference of a circle is c and its diameter is d,

Vedanta Excel in Mathematics - Book 7 272 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Perimeter, Area and Volume

circumference =S
diameter
c
or, d =S

or, c = Sd

Also, diameter of a circle (d) = 2 u radius (r) = 2r

So, circumference (or perimeter) of circle (c) = Sd = S × 2r = 2Sr

? Perimeter of circle = Sd or 2Sr

Where, the approximate value of S 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,perimeter of the rectangular field = 154 m

or, 2 (l + b) = 154 m

or, 45 + b = 154 m = 77 m
2

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.

Its perimeter = 2Sr Fencing 1 round is the perimeter of

=2u 22 u 42 m = 264 m the circular ground.
7 Fencing 5 rounds = 5 × perimeter

? The required length of wire = 5 u 264 m = 1320 m.

Again, the required cost of fencing = 1320 u Rs 27 = Rs 35,640.

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.

273Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7

Perimeter, Area and Volume

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

or, x = 38880 = 36 ,
1080

? l = 2x = 2 × 36 = 72

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 18.1
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 7 274 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.

275Approved 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 7 276 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.

18.3 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 u 4 rooms along breadth = 20 cm2
i.e. length u breadth = 20 cm2

Thus, area of the rectangle = length u breadth = l u b

277Approved 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

(iii) Area of triangle
Let’s do the following activities to find the formula of area of triangles.

Take a triangular shape of piece of paper and fold it as shown in the figures.

AA A

X 1 h Y F h
2 E

h P 1 h 1 h
2 2

BDC BDC B DC

Now, cut the folded edges AP, XP and YP. Place the edge AX along XB and the

edge AY along YC. Thus, a rectangle BCEF is formed by this arrangement.

Here, area of the triangle ABC = area of rectangle BCEF

= length u breadth

= BC u CE A

= base u height h
1 1 B bC
=bu 2 h = 2 bh

Thus, area of triangle = 1 base u height = 1 b u h
2 2

(iv) Area of parallelogram

Again, let’s do the following activities to find the formula of area of parallelograms.
Take a parallelogram shape of piece of paper and fold it as shown in the figures.

D EC D EC FD E

hh h

A bB A bB AbB

Cut the folded edge BE. Now place the edge BC of 'BCE along the edge AD. Thus,

a rectangle ABEF is formed by this arrangement.

Here, area of the parallelogram ABCD = Area of the rectangle ABEF

= length u breadth D C
= AB u BE h B
= base u height
=buh

Thus, area of parallelogram = base u height Ab
=buh

Vedanta Excel in Mathematics - Book 7 278 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Perimeter, Area and Volume

(iv) Area of circle
Now, let’s do the following activities to find the formula of the area of circles.

Take circular piece of paper. Cut it into equal pieces as small as possible and
arrange the pieces as shown in the figures.

DC

(breadth)

A 1 of 2πr B
2

(length)

Thus, a rectangle ABCD is formed by this arrangement.

The length of rectangle ABCD (l) = 1 u circumference = 1 u 2Sr = Sr
2 2
The breadth of rectangle ABCD (b) = r

Here, area of the circle= Area of the rectangle ABCD

= length u breadth Or
= Sr u r

= Sr2

Thus, area of circle = Sr2

Worked-out examples

Example 1: Find the area of the following figures. d)
a) b) c)
3.5 cm r = 7cm
6 cm

8.4 cm 15 cm 14 cm

Solution: Area of rectangle (A) = l × b = 8.4 cm × 3.5 cm = 29.4 cm2
a)
Area of triangle (A) = 1 b × h = 1 × 15 cm × 6 cm = 45 cm2
b) 2 2
22 22
c) Area of circle (A) = Sr2 = 7 × 72 = 7 × 7 × 7 = 154 cm2
d)
Area of square (A1) = l2 = (14 cm)2 = 196 cm2

Radius of semi–circle (r) = d = 14 = 7 cm
2 7

Area of semi–circle (A2) = 1 Sr2 = 1 × 22 × 72 = 77 cm2
2 2 7

? Area of the figure (A) = A1 + A2 = 196 + 77 = 273 cm2

279Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7

Perimeter, Area and Volume

Example 2: Find the area of the shaded regions in the following figures.
a) b) c)

6 cm
10 cm
9 cm
8 cm
9cm 3 cm
6 cm

16 cm 14 cm 20 cm

Solution: = length u breadth = 16 cm u 10 cm = 160 cm2
a) Area of bigger rectangle = length u breadth = 9 cm u 6 cm = 54 cm2

Area of smaller rectangle

? Area of the shaded region =Area of bigger rectangle – Area of smaller rectangle

= 160 cm2 – 54 cm2 = 106 cm2

b) Area of rectangle = l u b = 14 cm u 9 cm = 126 cm2
Area of parallelogram = b u h = 6 cm u 3 cm = 18 cm2

? Area of the shaded region = Area of rectangle – Area of parallelogram
= 126 cm2 – 18 cm2 = 108 cm2

c) Area of parallelogram = b u h = 20 cm u 8 cm = 160 cm2

Area of triangle = 1 b u h = 1 u 20 cm u 8 cm = 80 cm2
2 2

? Area of the shaded region = Area of parallelogram – Area of triangle

= 160 cm2 – 80 cm2 = 80 cm2.

Example 3: 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 Total cost = Area × rate
? The cost of growing grass in 484 m2 = 484 u Rs 40
= Rs 19,360

Example 4: 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:

Vedanta Excel in Mathematics - Book 7 280 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Perimeter, Area and Volume

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

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 u 8 m = 16 m.

b) Area of the room = l u b = 16 m u 8 m = 128 m2.

c) The cost of carpeting 1 m2 = Rs 75
? The cost of carpeting 128 m2 = 128 u Rs 75 = Rs 9,600

Example 5: If the perimeter of a circular ground is 484 m,
a) Find its radius. b) Find its area.

Solution:
a) The perimeter of a circular ground = 484 m

or, 2Sr = 484 m

or, 2u 22 u r = 484 m
7
484 × 7
or, r = 2 × 22 m = 77 m

b) Now, area of the circular ground = Sr2 = 22 u 77 m u 77 m = 18,634 m2.
7

EXERCISE 18.2

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 ......................
c) Base and height of a triangle are p cm and q cm respectively.

Its area is .......................
d) Base and height of a parallelogram are x cm and y cm respectively.

Its area is ....................

e) Radius of a circle is r cm. 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 .....................
c) Base of a triangle is 8 cm and height is 5 cm. Its area is ..................

d) Base of a parallelogram is 10 cm and height is 6 cm. Its area is ....................

281Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7

Perimeter, Area and Volume

Creative Section - A

3. Let's find the areas of the following figures.

a) b) c) d)

A R EP MP S
D
6.3 cm
7 cm 5 cm

B 9 cm C R 14 cm E Q 10 cm RT
12.5 cm S
e) X O
E g) h)

f) A

9 cm O 7 cm A O B
12 cm 28 cm

Y 16 cm Z F 16 cm G

4. Let's find the areas of the following figures. c)
a) b)
10 cm
10 cm 10 cm 9 cm 9 cm

16 cm 8 cm 7 cm 20 cm 7 cm 8 cm

d) e) f)

10 cm 14 cm
21 cm

18 cm 8 cm 25 cm 30.5 cm

5. Let's find the area of the shaded regions in the following figures.

a) b) c) d)

15 cm

10 cm 5 cm 12cm 5cm 25 cm 9cm
5 cm 16 cm
21 cm 6cm
10 cm 14 cm
g)
8 cm 25 cm h)

e) f)

AD

12 cm 10cm 15 cm
3 cm 6cm
6 cm 9 cm
15 cm
B E 18 cm C 20 cm 28 cm

6. a) Find the area of rectangles in which (i) length = 7.2 cm, breadth = 5 cm
(ii) length = 18 cm, breadth = 12.5 cm

Vedanta Excel in Mathematics - Book 7 282 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Perimeter, Area and Volume

b) Find the area of parallelograms in which (i) base = 12 cm, height = 6.5 cm

(ii) base = 25 cm, height = 8.4 cm

c) Find the area of triangles in which (i) base = 6 cm, height = 545 cm

(ii) base = 20 cm, height = 8.4 cm

d) Find the area of circles in which (i) radius = 14 cm,

(ii) diameter = 35 cm.

Creative Section - B

7. 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.

8. 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.

9. a) If the perimeter of a circle is 44 cm,

(i) Find its radius. (ii) Find its area.

b) A carpenter makes a table with the circular top with circumference

220 cm. (i) Find its radius. (ii) Find its area.

c) The circumference of a circular field is 308 m.
(i) Find its diameter and radius.
(ii) Find the cost of paving stones all over it at Rs 70 per sq. m.

283Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7

Perimeter, Area and Volume

10. a) Pooja has a flower garden of circular shape of diameter 28 m. A sprinkler
at the centre of the garden can spray the water in 240 m2. Can the sprinkler
spray water in the entire garden?

b) 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, and (iii) a circle.
Calculate the area of each shape and find which shape covers more area.

18.4 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
It has four surfaces. 1
23
Each surface is an equilateral triangle. Net

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
Each surface is a square. Face 5

It has eight vertices. Net

It has 12 edges. 1 Skeleton
2 34567 Skeleton
(iii) Octahedron
8
It is a regular solid. Net

It has eight surfaces.

Each surface is an equilateral triangle.

It has 6 vertices.

It has 12 edges

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.

Vedanta Excel in Mathematics - Book 7 284 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Perimeter, Area and Volume

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

It is a solid object. curved surface
circular base
It has a curved surface with circular base

Its curved surface meet at a point called its
vertex.

EXERCISE 18.3
General Section
1. Let's tell and write the name of these solids as quickly as possible.

a) b) c) d) e)

.................... .................... .................... .................... ....................
2. Let's name the solid objects which have the following nets.

a) b) c)

............................... ............................... ...............................
d) e)

............................... ...............................

285Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7

Perimeter, Area and Volume

3. Let's tell 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.

18.5 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
sum of the area of 6 rectangular faces. b

Area of top and bottom faces = l b + l b = 2l b h h
b
Area of side faces = bh + bh = 2bh l
l
Area of front and back faces = l h + l h = 2l h l

? Surface area of cuboid = 2l b + 2bh + 2l h = 2 (l b + bh + l h)

(ii) Area of cube l
A cube has 6 square faces. Each square face has area of l2. l
? Surface area of a cube = 6l2

(i) A lidless rectangular box does not have its top face. l

Vedanta Excel in Mathematics - Book 7 286 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Perimeter, Area and Volume

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
(iv) Area of hollow cubical box = 4l2

18.6 Volume of solids

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 lub h
l b
Volume of a cuboid is calculated as the product of the area of
its rectangular base and height. h=l
l b=l
? Volume of cuboid = Area of base u height
=lubuh

(ii) Volume of cube
In the case of a cube, its length, breadth and height are equal.
i.e. l = b = h
? Volume of cube = l u b u h
= l u l u l = l3

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 u 18 + 18 u 2 + 25 u 2) cm2
= 2 (450 + 36 + 50) cm2 = 1072 cm2.

b) Again, volume of the book = l u b u h
= 25 cm u 18 cm u 2 cm
= 900 cm3

287Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7

Perimeter, Area and Volume

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 u height

? Area of the base u height = 72

or, Area of its base u 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
96
or, l2 = 6 cm2 = 16 cm2

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.

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 u Rate = 49 u 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.

Vedanta Excel in Mathematics - Book 7 288 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Perimeter, Area and Volume

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 u b u h = 200 cm3

or, 2x u x u 4 cm = 200 cm3

or, 2x2 = 200 cm2 = 50 cm2
4

or, 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 u 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 u b u h = 9 cm u 6 cm u 4 cm = 216 cm3
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 u (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 u b u h = (24 u 12 u 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

289Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7

Perimeter, Area and Volume

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 u 1000 cm3

Now, volume of the part of the tank containing water = volume of water

or,Area of its base u height = 3000 u 1000 cm3

or, 30000 cm2 u h = 3000 u 1000 cm3

or, h = 3000 × 1000 cm3
30000 cm2

or, h = 100 cm

So, the required height of water level in the tank is 100 cm (or 1 m).

EXERCISE 18.4
General Section - Classwork

1. Let's tell 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 ............................

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 ...................

Vedanta Excel in Mathematics - Book 7 290 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Perimeter, Area and Volume

Creative Section - A
2. Find the total surface area and volume of these solids.

a) b) c) d)

2cm 4 cm
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.

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

291Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7

Perimeter, Area and Volume

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.
(i) Find the length of its edge.
(ii) Find its surface area.

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 7 292 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?

293Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7

Unit Symmetry, Design and Tessellation

19

19.1 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

19.2 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 la1reanlidnels2 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

19.3 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.

19.4 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.

Vedanta Excel in Mathematics - Book 7 294 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Symmetry, Design and Tessellation

When the equilateral triangle is rotated
A C 1
BC A through a 3 of a turn (120q), it can be

120° fitted over the original shape. It means

that, when it is rotated through 3 times
B p13osoitfioan.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 90q , it can be fitted over
B 4
A 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 19.1

General Section - Classwork

1. Let's tell 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. 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.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 295 Vedanta Excel in Mathematics - Book 7

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 (360q).

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)

5. a) Decide whether each shape has 1 turn symmetry, 1 turn symmetry, or
symmetry. Use 4 3
1
2 turn tracing paper to help you decide, if necessary. Also

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.

19.5 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. Thus, to make a tessellation (or tiling) –

Vedanta Excel in Mathematics - Book 7 296 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Symmetry, Design and Tessellation

(i) Use the sets of congruent figures (tiles).
(ii) Do not leave any gaps.
(iii)Do not have any overlaps.

19.6 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 19.2

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?

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 297 Vedanta Excel in Mathematics - Book 7

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.

Vedanta Excel in Mathematics - Book 7 298 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur