Vedanta Excel in Mathematics Book - 8 Final (2078)

Geometry: Triangle

EXERCISE 15.1

General Section – Classwork

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

a) If a°, b°, and c° are the angles, of a triangle, a° + b° + c° = ............................

b) If x°, 40°, and 60° are the angles of a triangle, x° = ............................

c) If x°, y°, and z° are the angles of an equilateral triangle, x° = y° = z° = .............

d) If m° and n° are the base angles of an isosceles right-angled triangle m° = n°

= ........................

e) If one of the acute angles of a right-angled triangle is 50°, another

acute angle is ..............

f) x° is the exterior angle of a triangle, and a° and b° are its opposite interior

angles x° = ............................

Creative Section - A

2. a) If xq, 50q and 30q are the angles of a triangle, find xq.

b) If xq, 2xq and 30q are the angles of a triangle, find xq and 2xq.

c) If 3xq, 4xq and 5xq are the angles of a triangle, find these angles.

d) If the angles of a triangle are in the ratio 2:3:5, find them.

e) If two acute angles of a right angled triangle are in the ratio 1:2, find them.

f) If xq is the exterior angle and 70q and 50q are its opposite interior angle of a
triangle find xq.

3. Find the unknown sizes of angles in the following figures. d) F

a) b) P c) M

Ax x+7° 52°

x 30° C Q 3x 40° R K x x+5° L Dx yE
B 60°

e) X f) E g) A h) P 86° T
b z By Q

Y aZ y x 130° Cx 36° D y x
F GH S 124°

RM

4. Find the unknown sizes of angles in the following figures.

a) b) Q c) d)
A A CP
58° X R
45° y40x° z x
xz P 2x
P 65° a Q B O 85° 87°
Bb cC y
O
52° Y
y
R
Q S
D

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Geometry: Triangle

e) f) g) h)
KB AC
A EP S
z 42° 75°
x x Oy G F x 30° E

60° 65° y 65° x 72° y
B yz Q z A BD

CD R L

i) A j) P k) l)
X PS
72° q T
A y 57°
P Y 100° d eA Q
B 54° r pp 40° Z 140° x 80° z
48° 52° 50° R
QB QQ
Q x25°yC P acb O
R R B P ab

Creative Section - B z z
2
5. a) In the adjoining figure, show that a = 90q –

Ax yB

AD

b) In the given figure, show that AB//CD 46° 88°
B C
GB
A

c) In the figure alongside, GP bisects ‘AGH and HP bisects P

‘GHC. Show that ‘GPH = 1 right angle. C H D
C
DP

d) In the adjoining figure, AP and BP are the angular A B
bisectors of ‘BAD and ‘ABC respectively. Show that A E
‘APB = 1 right angle.

e) In the given figure, ABC is an equilateral triangle. B CD

If AB//CE, prove that ‘ACE = ‘ECD.

6. a) Let's draw two equilateral triangles ABC of different size. Verify experimentally

that each angle of an equilateral triangle is 60°.

b) Let's draw two isosceles triangle PQR of different measurements. From the vertex

P, draw a straight line to meet the mid-point of QR at S. Verify experimentally

that PS is perpendicular to QR.

c) Let's draw two isosceles triangles XYZ of different measurements. From the

vertex X draw XP perpendicular to YZ at P. Verify experimentally that YP = PZ.

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Geometry: Triangle

15.3 Congruent triangles DD' C D D' C

Let's take a rectangular sheet of paper. Fold it A BB' A B B'
along its diagonal. Cut the folded edge. Place two
triangles, one above another as shown in the C D'
figure alongside. Here, the two triangular A
pieces can be exactly fitted to one above B
another. In such a case, the triangular
pieces are said to congruent. The parts (sides and
angles) fitting one above another are called the
corresponding parts.

Thus, 'ABD and 'B'CD' are congruent B'
triangles. It is written as 'ABD # 'B'CD'.
D
Corresponding sides of AB, AD and

BD are CD', B'C and B'D' respectively.

Similarly, the corresponding angles of ‘A, ‘B,

and ‘D are ‘C, ‘D', and ‘B' respectively. In

congruent triangles corresponding sides are equal
and corresponding angles are also equal.

15.4 Experiment on the conditions of congruency of triangles
by construction.

Construction 1: ' PQR is a given triangle. R R'

(i) Measure the length of PQ with the help
of a compass and draw P'Q' = PQ.

(ii) With the centres at P' and Q' and the radii
equal to PR and QR respectively draw
two arcs to intersect each other at R'.
P Q P' Q'

(iii) Join P', R', and Q'R'.

Here, three sides of 'PQR are equal to three sides of 'P'Q'R'. So, 'PQR and 'P'Q'R are
congruent triangles.

Construction 2: ' ABC is a given triangle.

(i) Measure AB with the help of a compass C X
and draw A'B' = AB. C'

(ii) Measure ‘A with the help of a compass
and draw the same angle at A' by drawing
A'X.

(iii) Measure the length of AC by a compassed
and cut the same length A'C' on A'X.
A B A' B'

(iv) Join B' and C'.

Here two side of ΔABC are equal to two sides of A'B'C and angle made by them ‘A and
‘A' are also equal. So, ΔABC and ΔA'B'C' are congruent triangles.

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Geometry: Triangle

Construction 3: ' XYZ is a given triangle. Z Z'

(i) Measure XY with the help of a
compass and draw X'Y' = XY.

(ii) Measure ‘X and ‘Y by a compass X Y X' Y'
and draw the same angles at X' and
Y' respectively such that two straight
line segments intersect at Z'.

Here, two angles of 'XYZ are equal to two angles of 'X'Y'Z'. Also, the sides between
these angles are also equal. So, 'XYZ and 'X'Y'Z' are congruent triangles.

Construction 4: 'PQR is a given right angled triangle, right angled at Q.

(i) Measure QR by a compass and X
P'
draw Q'R' = QR. P

(ii) Construct an angle of 90° at Q'

such that ‘ R'Q'X = 90°.

(iii) Take an arc equal to PR and cut
Q'X from R' at P'.

(iv) Join P' and R'.

Here, both the triangles are right Q R Q' R'

angled triangles. Hypotenuse of

'PQR is equal to the hypotenuse of 'P'Q'R'. Also a side QR of 'PQR is equal to the side

Q'R' of 'P'Q'R'. So, 'PQR and 'P'Q'R' are congruent triangles.

15.5 Conditions of congruency of triangles

There are 3 sides and 3 angles in a triangle. Two triangles can be congruent if 3 parts
(out of 6 parts) of one triangle are equal to 3 corresponding parts of another triangle
under the following conditions. These conditions are used as axioms.

(i) S.S.S. axiom

When three sides of a triangle are respectively equal to three corresponding sides
of another triangle, they are said to be congruent.

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

b) BC = QR (S)

c) CA = RP (S)

d) ? 'ABC # 'PQR (S.S.S. axiom) B CQ R

Corresponding parts of congruent triangles are also equal.

? ‘A = ‘P, ‘B = ‘Q and ‘C = ‘R

Note: The parts opposite to the equal parts of congruent triangles are called the
corresponding parts.

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Geometry: Triangle

(ii) S.A.S. axiom

When two sides of one triangle and angle made by them are respectively equal to

the corresponding sides and angle of another triangle, they are said to be congruent

triangles.

In 'ABC and 'PQR, AP

a) AB = PQ (S)

b) ‘B = ‘Q (A) B CQ R
c) BC = QR (S)

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

Now, ‘C = ‘R and ‘A = ‘P [Corresponding angles of congruent triangles]

Also, CA = RP [Corresponding sides of congruent triangles]

(iii) A.S.A. axiom

When two angles and their adjacent side of one triangle are respectively equal
to the corresponding angles and sides of another triangle, they are said to be
congruent triangles.

In 'ABC and 'PQR, AP
a) ‘B = ‘Q (A)

b) BC = QR (S)

c) ‘C = ‘R (A) B CQ R
d) 'ABC # 'PQR (A.S.A. axiom).

Now, AC = PR and AB = PQ [Corresponding sides of congruent triangles]

Also, ‘A = ‘P [Corresponding angles of congruent triangles]

(iv) R.H.S. axiom

In two right angled triangles, when the hypotenuse and one of the two remaining
sides are respectively equal, they are said to be congruent triangles.

In right angled 's ABC and PQR, A P R
a) ‘B = ‘Q (R) B CQ
b) AC = PR (H)
c) BC = QR (S)
d) ? 'ABC # 'PQR (R.H.S. axiom)

Now, ‘C = ‘R and ‘A = ‘P [Corresponding angles of congruent triangles.]
Also, AB = PQ [Corresponding sides of congruent triangles.]

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Geometry: Triangle

Worked-out examples

Example 1: Find the value of x in the following pairs of congruent triangles. Also

(i) write the sizes of corresponding angles and corresponding sides.

B A P (ii) D
40°
Solution: 80° 55°(x+0.5)cm (2x-1)cm
3.5cm 45° E (x+0.5)cm F YX
(3x-1.5)cm
C4.2cm 50° 4.5cm
R 45° Z

(x+1)cm Q

(i) Here, 'ABC # 'PQR
‘A = ‘P = 55q In 'ABC, ‘A = 180° – (80° + 45°) = 55°

? BC = QR
x + 1 = 3.5

or, x = 3.5 – 1= 2.5 cm
Now, ‘C = ‘R = 45q
? AB = PQ = (x + 0.5) cm = (2.5 + 0.5 cm) = 3 cm

Also, ‘B = ‘Q = 80q In 'PQR, ‘Q = 180° – (45° + 55°) = 80°
? AC = PR = 4.2 cm

(ii) Here, 'DEF # 'XYZ

‘E = ‘Y = 90q

? DF = XZ

or, 3x – 1.5 = 4.5

or, 3x = 4.5 + 1.5 = 6
or, 6
Now, x = ‘3 X = 2 cm
‘D = = 40q

? EF = YZ = (x + 0.5) cm = (2 + 0.5) cm = 2.5 cm

Also, ‘F = ‘Z = 50q

? DE = XY = (2x – 1) cm = (2 u 2 – 1) cm = 3 cm

Example 2: In the figure alongside, AB = CD and AB // DC, show that

a) AD = BC b) AD // BC D C
B
Solution:
Proved
In 'ABC and 'ACD,

(i) AB = CD [Given]

(ii) ‘BAC = ‘ACD [Being alternate angles] A

(iii) AC = AC [Common side]

(iv) ? 'ABC # 'ACD [S.A.S. axiom]

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

(vi) ‘ACB = ‘CAD [Corresponding angles of congruent triangles]

(vii) AD // BC [Being equal alternate angles]

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Geometry: Triangle

EXERCISE 15.2
General Section

1. From these pairs of congruent triangles. Let's tell and write the necessary conditions
of congruency.

a) b)

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

c) d)

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

2. From these pairs of congruent triangles. Let's tell and write the corresponding sides
and corresponding angles.

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

Corresponding angle of ‘R is ....................

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

B CQ R

b) G Z Corresponding side of FG is .......................
Corresponding angle of ‘Z is ....................
Corresponding angle of ‘F is ....................

E FX Y

c) D A C

Corresponding side of DF is .......................

Corresponding side of DE is .......................

E FB Corresponding angle of ‘B is ....................
C S
d) R Corresponding side of AC is ......................
B Corresponding angle of ‘C is ....................
A Q Corresponding angle of ‘R is ....................

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Geometry: Triangle

3. What are the necessary conditions to be A D
EF
included in the following conditions so that
'ABC # ' DEF?

a) AB =DE, ‘B = ‘E .......................

b) AB = DE, BC = EF, ....................... BC

c) ‘A = ‘D, ‘B = ‘E .......................

Creative Section - A

4. 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 D

2.5cm 3cm2.5cm 3.5cm
3.8cm
B 4.2cm3.8cm 6.4cm CQ 4.2cm R 120° 120°
c) K d) 3cm E
Y 3.5cm ZF
T W G

3.6cm 60°

L 3.6cm MS 6.4cm R 40° 80° 40°
X 4.7cm Y E 4.7cm F

5. Use the necessary axioms and show that the following pairs of triangles are
congruent. Also, write their corresponding sides and angles.

a) Q b) W X D
A
55°

R P 60°

B C Y E 55° 65°
c) F

E Z d) A D

FG C F
Y XB E

6. Mention the necessary axioms to show the following pairs of triangles congruent.
Also, find the value of x in each case.

a) A b) c) d)

x A C A xD A E
x
5 cm
6.5 cm 50° 45° x
B D CB BC
DB C D

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Geometry: Triangle

7. Find the value of x in the following pairs of congruent triangles. Also, write the

sizes of corresponding angles and sides.

a) A P b) W 20(x°+3.5)cm D
45° 115°
(x+315.3°)cm 85° x cm 60° 3.2cm 45° (2x–1.5)cm
R X x cm Y F 2.5cm E
35°

B 5.5cm C Q

c) M d) (x+2)cm30° CE
Q B 86°
(x+0.8)cm 3.1cm 30° 6.5cm
(x+3.2)cm (x+2.3)cm
(2x–2)cm F

R 38° 52° 64° 3.1cm
L
5.2cm P K D
B
Creative Section - B A
A

8. a). In the figure alongside, AB// CD and AB = CD. Prove C O D
S R
that
i '$2% # 'COD

ii) AO = OD and BO = OC.

b). In the adjoining figure, PQ = SR and PQ // SR, show PQ
that A

i PS = QR
ii) PS // QR

c) In the given figure, ABC is an isosceles triangle in A B PC
which AB = AC. If AP ⊥ BC, prove that Q

i) 'APB # 'APC OD

(ii) ‘B = ‘C
(iii) BP = PC

d). In the adjoining figure, AP ⊥ CD, BQ ⊥ CD and CP
OP = OQ. Prove that AP = BQ.

It's your time - Project work! B

9. Let's take a rectangular chart paper. Fold it diagonally into two halves and cut out

the halves using scissors. Now, put one half on the another half and identify the

corresponding angles and corresponding sides of two congruent triangles.

15.6 Similar triangles P

Although the adjoining triangles are not equal in A 75°
size, they have exactly the same shape. Because of 75°
their similar shape, they are similar triangles.
45° 60° 45° 60°
Thus, if two triangles are equiangular, they are the B CQ R
similar triangles.

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Geometry: Triangle

The similar triangles ABC and PQR is written as 'ABC ~ 'PQR. The symbol ~ is used

to denote the similarity of triangles. Furthermore, if two triangles are similar, the ratios

of their corresponding sides are equal. It means the corresponding sides of similar

triangles are proportional.

? AB = BC = AC
PQ QR PR

Worked-out examples P
X

Example 1: In the given figure, ∆PQR ~ ∆XYZ, find the 5 cm7 cm
Solution sizes of ‘R and ‘Y. Also find the length of 10 cm
side PQ.
Q 65° Y 95° Z
R

Here, ‘R = ‘Z = 95q [Corresponding angles of similar triangles are equal.]
[Corresponding sides of similar triangles are proportional]
‘Y = ‘Q =65q

Again, 'PQR ~ 'XYZ

? PQ = PR
XY XZ

or, PQ = 10 =2
7 5

or, PQ = 7 u 2 = 14 cm

Example 2: In the given figure, ∆ABC ~ ∆PQC. Find the value of x and y.
Solution
A

Here, 'ABC ~ 'PQC y cm P 20 cm
? AB and PQ are corresponding sides and their ratio is AB. 5 cm 10 cm

PQ B Q x cm C
AC and PC are corresponding sides and their ratio is AC. 24 cm
PC

BC and QC are corresponding sides and their ratio is BC .
QC
AB BC AC
Now, PQ = QC = PC [Corresponding sides of similar triangles are proportional]

or, y = 24 = 20
5 x 10

Taking the first and the third equal ratios,
y 20
5 = 10 = 2

or, y = 5 u 2 = 10 cm

Also, taking the second and the third equal ratios,

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Geometry: Triangle

24 = 20 =2
x 10

or, 2x = 24

or, x = 24 = 12 cm
2

? x = 12 and y = 10 cm.

EXERCISE 15.3
General Section – Classwork

1. Let's say and write the corresponding sides of these similar triangles. Also write the
equal ratios of the corresponding sides.

a) C Corresponding side of BC is ..............
A
F

Corresponding side of AC is ..............

Corresponding side of AB is ..............

BD E Equal ratios of corresponding sides are ........................

b) P A Corresponding side of QR is .................
Corresponding side of PQ is .................
QB Corresponding side of PR is .................
R Equal ratios of corresponding sides are ......................

Creative Section A P
75°
2. a) In the given figure 'ABC ~ 'PQR. Q
If ‘C = ‘R = 50° and AC = 12 cm, 9 cm x cm50°
find the length of side PQ and also 8 cmR
find the sizes of ‘A and ‘Q. 8 cm
B 55° 50°12 cmC

b) In the adjoining figure, 'DEF ~ 'XYZ. D X
If DE = 10 cm, EF = 6 cm, DF = 8 cm, E 6 cm F Y Z
YZ = 12 cm , ‘E = ‘Y and ‘D = ‘X, find the 10 cm
value of x and y. y cm

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Geometry: Triangle A
X
3. a) In the given figure, 'ABC ~ 'XCY, BC = 9 cm,
BY = 3 cm and XY = 4 cm.
Find the length of AB.

B Y C
D A
b) In the given figure, 'CDE ~ 'ABC. Find the value x
of x with suitable reasons. E 7 cm10 cm B
8 cm
C C
6 cm B

c) In the adjoining figure, 'ABC ~ 'ADE. AD = 15 cm, A 15 cm D 25 cm
DE = 10 cm, BC = 25 cm and AC = 20 cm. Find length 20 cm E 10 cm
of AB and AE.

d) In the figure alongside, ‘BAD = ‘ACD and 4 cmA
'ABD ~ 'ABC. Find the length of BD. 10 cm

e) In the given figure, 'PQR ~ 'XQY. If ‘Q = 40q, BC
‘R = 60q, PR:XY = 5 :3 and PQ = 15 cm, D
find ‘QXY and the length of QX.
8 cm
P

X

40° 60°
Q YR

A

f) In the given figure, 'ABC ~ 'AXC, AX A BC, 3 cm 4 cm
AB = 3 cm and AC = 4 cm. BX
C
Find the length of CX. 5 cm

15.7 Conditions of similarity of triangles

Two triangles can be similar under the following conditions:
(i) If two angles of one triangle are respectively equal to two angles of another triangle,

they are said to be similar.

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Geometry: Triangle

In 'ABC and ∆PQR A P
a) ‘ABC = ‘PQR B CQ R
b) ‘ACB = ‘PRQ
c) ? 'ABC ~ 'PQR.

Note: When two angles of one triangle are equal to two angles of another triangle,
their remaining angles are equal.

(ii) When the ratios of the corresponding sides of two triangles are equal, i.e. the

corresponding sides of triangles are proportional, the triangles are said to be

similar. K

In 'KLM and 'XYZ, X

KL = LM = MK
XY YZ ZX

? 'KLM ~ 'XYZ L MY Z
Here,

a) KL is the opposite side of ‘M and ‘M = ‘Z; the opposite side of ‘Z is XY.

So, KL and XY are corresponding sides.

b) LM is the opposite side of ‘K and ‘K = ‘X; the opposite side of ‘X is YZ.
So, LM and YZ are corresponding sides.

c) MK is the opposite side of ‘L and ‘L = ‘Y; the opposite side of ‘Y is ZX.
So, MK and ZX are the corresponding sides.

(iii) If any two corresponding sides of triangles are proportional and the angles
included by them are equal, the triangles are said to be similar.

In 'DEF and 'RST, D
a) DE = EF R

RS ST

b) ‘DEF = ‘RST E FS T
c) ? 'DEF ~ 'RST.

Worked-out examples

Example 1: In the adjoining figure, ‘PXY = ‘PQR = 40°. Show that
a) ∆PQR ~ ∆PXY
P

b) XY PY
QR = PR
Solution X 40°
40° Y
a) In 'PQR and 'PXY, Q R

(i) ‘PQR = ‘PXY [Both of them are 40q]

(ii) ‘QPR = ‘XPY [Being common angles]

(iii) 'PQR ~ 'PXY [Two pairs of angles of two triangles are equal]

b) XY = PY [Corresponding sides of similar triangles are proportional]
QR PR

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Geometry: Triangle

Example 2: In the given figure, ABC is a right angled A
D
triangle where ‘A = 90°. If AD A BC, prove
that

a) 'ABC ~ 'ADC B C
b) AC2 = BC.CD

Solution

a) In 'ABC and 'ADC,

(i) ‘BAC = ‘ADC [Both of them are right angle (90q)]

(ii) ‘ACB = ‘ACD [Being common angle]

(iii) ‘ABC = ‘DAC [Remaining angles of the triangles]

(iv) ? 'ABC ~ 'ADC

b) AC = BC
CD AC

or, AC2 = BC.CD Proved

EXERCISE 15.4

General Section – Classwork

1. Find out the pairs of similar triangles from the following figures.

AP X E

7 cm 20° 1.8 cm 20°
5 cm 120° 70° 120°
R Y 2.1 cm Z F
B 4 cm C Q G

M D 8 cm 10 cm TW
150°
ES
15 cm
8 cm

2.4 cm
70° 10 cm 14 cm 150°
2.8 cm R X 12 cm
K LF Y

(i) ....................... and ............ are similar (ii) ...................... and ............are similar

(iii) ....................... and ............ are similar (iv) ...................... and ............are similar

Creative Section - A A Q
P 50° C
2. a) In the adjoining figure, ‘ABC = ‘APQ = 50q. Show that B 50°
(i) 'ABC ~ 'APQ
(ii) BC = AC .
PQ AQ

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Geometry: Triangle

b) In the given figure, ‘DEF = ‘DHG = 50q. Prove that D

(i) 'DHG ~ 'DEF 50° H F
(ii) DG = GH G
E 50°
DF EF

c) In the figure alongside, PQ//WY. Show that W
(i) 'WXY ~ 'PXQ P
(ii) WY.PX = WX.PQ X

QY

d) In the given figure, AB//CD. Prove that AB
(i) 'AOB ~ 'COD
(ii) AO.OC = BO.OD. O
C
D

e) In the adjoining figure, ABC is a right angled triangle A

where ‘A = 90q. If AD A BC, prove that D
6 cm
(i) 'ABC ~ 'ABD (ii) 'ABC ~ 'ACD CE

(iii) 'ABD ~ 'ACD (iv) AB2 = BC.BD B C
(v) AC2 = BC.CD (vi) AC.BD = AB.AD

Creative Section - B A

3. a) In the adjoining figure, ‘BED = ‘BAC prove that x cm 4 cm D 6 cm
(i) 'ABC ~ 'BDE
(ii) Find the value of x. 8 cm B
9 cm

6 cm A
3 cm
b) In the figure given alongside, prove that 8 cm
(i) 'ABC ~ 'BDC D
(ii) Find the length of AC
B 4 cm C

c) In the given figure, PQ A QR, ST A QR. Show that P8m
(i) 'PQR ~ 'SRT S
(ii) Find the length of the flag. R 6m T 6m Q

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Unit Geometry: Quadrilateral and
Regular Polygon
16

16.1 Quadrilaterals - Looking back

Classroom - Exercise R
Q
Let’s say and write the answers as quickly as possible. S C
1. a) Name of the given quadrilateral is ........................... P B

b) Vertices of the given quadrilateral are ...........................
c) Sides of the given quadrilateral are ...........................
d) Angles of the given quadrilateral are ...........................

2. a) In the quadrilateral PQRS,‘P + ‘Q + ‘R + ‘S = ...........................

b) If ‘P + ‘Q + ‘S = 300°, then ‘R = ...........................

The plane figure bounded by four line segments is called a D

quadrilateral. In the given figure, ABCD is a quadrilateral. AB, BC,

CD, and DA are the sides of the quadrilateral ABCD. Similarly,

‘A, ‘B, ‘C, and ‘D are the angles of the quadrilateral. A

In any quadrilateral, the total sum of its angle is 360q. ? ‘A + ‘B + ‘C + ‘D = 360q.

16.2 Some special types of quadrilaterals D C
B
On the basis of certain special properties, there are some

special types of quadrilaterals. Parallelogram, rectangle, square,

rhombus, etc. are some special types of quadrilaterals. A

1. Parallelogram

In the adjoining figure, quadrilateral ABCD is a parallelogram.
It’s opposite sides are equal and parallel. A quadrilateral can be a parallelogram, if
it satisfies any one of the following conditions.

(i) Its two pairs of opposite sides are parallel
(ii) Its two pairs of opposite angles are equal

(iii) Any two of its opposite sides are equal and parallel
(iv) Its diagonals bisect each other.

Experiment 12: Experimental verification of opposite sides of parallelograms.

Step 1: Draw three parallelograms ABCD of different sizes with the help of set-squares.

DC D

DC C

A B AB A
Fig (i) Fig (ii) B

Fig (iii)

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Step 2: Measure the sides of each parallelogram. Write the measurements in the table.

Fig. Length of Length of Result
opposite sides opposite sides
AB = CD and AD = BC
AB CD AD BC AB = CD and AD = BC
(i) AD = CD and AD = BC

(ii)

(iii)

Conclusion: The opposite sides of a parallelogram are equal.

Experiment 13: Experimental verification of opposite angles of parallelograms.

Step 1: Draw three parallelograms ABCD of different sizes with the help of set
squares.

DC B

DC
AC

A Fig (i) B A B
Fig (ii)
Fig (iii) D

Step 2: Measure the angles of each parallelogram with the help of protractor. Write the
measurements in the table.

Fig. Size of Size of

opposite angles opposite angles Result

‘A ‘C ‘B ‘D
(i) ‘A = ‘C and ‘B = ‘D

(ii) ‘A = ‘C and ‘B = ‘D
(iii) ‘A = ‘C and ‘B = ‘D

Conclusion: The opposite angles of a parallelogram are equal.

Experiment 14: Experimental verification of diagonals of parallelograms.
Step 1: Draw three parallelograms ABCD of different sizes with the help of set

squares.

DC DC DC
O O
O
AB AB AB
Fig (i) Fig (ii)
Fig (iii)

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Step 2: Draw diagonals in each parallelogram.
Step 3: Measure the lengths of AO, OC, BO, and OD in each parallelogram. Write the

measurements in the table.

Fig. Parts of AC Parts of BD Result
AO OC BO OD
AO = OC and BO = OD
(i) AO = OC and BO = OD
(ii) AO = OC and BO = OD
(iii)

Conclusion: Diagonals of parallelogram bisect each other.

2. Rectangle D 90° C
A B
The adjoining quadrilateral is a rectangle. A rectangle
is a parallelogram in which each of its angle is 90q. The
diagonals of rectangle are equal.

Experiment 15: All the angles of a rectangle are equal and are right angle.

Step 1: Draw three rectangles ABCD of different measurements.

DC
DC

D C

AB A fig (ii) B A fig(iii) B
fig (i)

Step 2: Measure ‘A, ‘B, ‘C, and ‘D with the help of a protractor and write the
measurements in the table.

Fig. No. ‘A ‘B ‘C ‘D Result
(i)

(ii) ‘A = ‘B = ‘C = ‘D = 90°
(iii)

Conclusion: All the angles of a rectangle are equal and are right angle.

Experiment 16: Experimental verification of diagonals of rectangles.
Step 1: Draw three rectangles ABCD of different sizes by using set squares.

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D

D C DC

O O A C
A O
B AB
Fig (i) Fig (ii)
B
Step 2: Draw diagonals in each rectangle. Fig (iii)

Step 3: Measure the lengths of AC and BD in each rectangle. Write the measurements in
the table.

Fig. AC BD Result
(i) AC = BD
(ii) AC = BD

(iii) AC = BD

Conclusion: The diagonals of a rectangle are equal.

3. Square C
B
The quadrilateral given alongside is a square. It is a rectangle in D
which its adjacent sides are equal. In other words, all sides of a
square are equal.

The diagonals of a square bisect each other perpendicularly. A

Experiment 17: Experimental verification of a diagonals of a squares

Step 1: Draw three squares ABCD of different measurements by using set squares.

DC DC DC

AOB AO B O
Fig (i) Fig (ii)
AB
Step 2: Draw diagonals in each square. Fig (iii)

Step 3: Measure AO, OC, BO, OD and ‘AOB, ‘BOC, ‘COD, ‘AOD. Write the
measurements in the table.

Fig. AO OC BO OD ‘AOB ‘BOC ‘COD ‘AOD Result
(i) AO = OC, BO = OD,

‘AOB = ‘BOC = ‘COD = ‘AOD
(ii) AO = OC, BO = OD,

‘AOB = ‘BOC = ‘COD = ‘AOD

(iii) AO = OC, BO = OD,

‘AOB = ‘BOC = ‘COD = ‘AOD

Conclusion: Diagonals of a square bisect each other perpendicularly.

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Experiment 18: The diagonals of a square bisect the vertical angles.
Step 1: Draw a square ABCD. Draw diagonals AC and BD.
DC

AB
Step 2: Measure the vertical angles and the angles formed due to the division of each

vertical angle by the diagonals. Write the measurements in the table.

Vertical angles Other angles formed by diagonals at Result

‘BAD = each vertex ‘CAD = ‘BAC = ‘CAD
‘ABC = ‘BAC = ‘CBD = ‘ABD = ‘CBD
‘BCD = ‘ABD = ‘ACD = ‘BCA = ‘ACD
‘ADC = ‘BCA = ‘BDC = ‘ADB = ‘BDC
‘ADB =

Conclusion: The diagonals of a square bisect the vertical angles.

Worked-out examples

Example 1: In the following figures, find the unknown sizes of angles.

Solution: (i) (ii) (iii)

xy c b
37°
115° z

yx z w 110° a d
73°

(i) x = 115q [Being opposite angles of a parallelogram]
y + 115q = 180q [Being the sum of co-interior angles]
[Being opposite angles of a parallelogram]
or, y = 180q – 115q = 65q
z = y = 65q [Being alternate angles]
[Being corresponding angles]
? x = 115q = z = 65q [Being the sum a straight angle]

(ii) y = 110q [Being opposite angles of a parallelogram]
z = 110q

w + 110q = 180q
or, w = 180q – 110q
or, w = 70q

x = w = 70q
? w = x = 70q and y = z = 110q

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(iii) a + 73q + 37q = 180q [Being the sum of the angles of a triangle]

or, a = 180q – 110q

or, a = 70q

a = b = 70q [Being opposite angles of a parallelogram]

c = 73q [Being alternate angles]

d = 37q [Being alternate angles]

?a = b = 70q, c = 73q and d = 37q.

Example 2: In the adjoining parallelogram, find the length of AB.

Solution:

In the parallelogram ABCD, (4x–1) cm

AB = CD [Being opposite sides of the parallelogram] D C

or, 2x + 3 = 4x – 1

or, 2x – 4x = – 1 – 3

or, – 2x = – 4 AB
=24 =2x2+ (2x+3) cm
or, x =
? Length of AB 3 = 2 u 2 + 3 = 4 + 3 = 7 cm.

Example 3: In the adjoining quadrilateral ABCD, AB // CD and BC // AD.

Prove that

a) ∆ABC # ∆ACD DC
b) AB = CD

c) AD = BC

Solution: AB

a) In 'ABC and 'ACD,

(i) ‘BAC = ‘ACD (A) [Being alternate angles]

(ii) AC = AC (S) [Being common side]

(iii) ‘ACB = ‘CAD (A) [Being alternate angles]
(iv) ?'ABC # 'ACD [A.S.A. axiom]

b) AB = CD [Corresponding sides of congruent triangles]

c) AD = BC [Corresponding sides of congruent triangles]

EXERCISE 16.1
General Section - Classwork

1. Let’s say and write the unknown angles as quickly as possible.

a) b)

yx z 120°

45° z xy

x = .............., y = .................. z = ............. x = ............., y = .............., z= .......

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c) a d) r 120°
c q

b d 110° sp

a = ................., b = ..................... p = ................, q = .....................

c = ................., d = ..................... r = ................., s = .....................
2. a) In the adjoining rectangle,
DC
if AO = 2 cm, BD = ....................
2cm O B
b) In the given square, if A R
SQ = 6 cm , PO = ......................
S

O

c) In the given parallelogram, if PQ
GP = 2.5 cm and HF = 3 cm, HG

EG = ............... and PF = ....................... P

Creative Section - A EF

3. Let’s calculate the unknown sizes of angles in the following figures.

a) b) 120° c) y d) b
y xy z 110° a

55° z xz x w 105° cd

e) f) g) 5x+5 y h)
x y 4x wy yb
30°

55° z 5x z 4x-5 z a x
35°

i) j) k) l) a
80° 45° 110° x c 40°
x
xz 150° b xy
zy y 30°

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m) n) o) p)
56° b
yx 40° z 3x

z 30° x d c x y
y 70° a 40° 150°

4. Find the values of x. Then, find the lengths of the sides of the following

parallelograms.

a) b) (x+4) cm R c) d) H G
D (2x–3) cm S Y
C Z (3x–5) cm

AB P Q (4x–1) cm EF
(x+7) cm (3x–2) cm (2x+3) cm C

(x+5) cm B
W X

Creative Section -B D

5. a) In the adjoining quadrilateral ABCD, ‘A = ‘C and

‘B = ‘D. Prove that, quadrilateral ABCD is a

parallelogram. A
Hint: Sum of the angles of quad. ABCD is 360q

S R

b) In the adjoining quadrilateral PQRS, PQ // SR and Q
PS // QR. Prove that
C
(i) 'PQR # 'PRS P O
(ii) PQ = SR
B
(iii) PS = QR

c) In the given figure, ABCD is a parallelogram. Diagonals D
AC and BD are intersecting at O. Prove that

(i) 'AOB # 'COD A
(ii) AO = OC and BO = OD.

d) In the adjoining figure, ABCD is a rectangle. Prove that D C
(i) 'BAD # 'ABC A B
(ii) AC = BD

SR

e) In the given figure, PQRS is a square. Prove that O
(i) 'POQ # 'QOR PQ
(ii) QO A PR.

6. a) Draw two parallelograms ABCD of different measurements. Then explore
experimentally the relationship between their opposite sides.

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b) Draw two rectangles PQRS of different measurements and explore experimentally
the relationship between their diagonals.

c) Explore experimentally the relationship between the diagonals of square. (Two
figures of different measurement are necessary)

It’s your time - Project work
7. a) Let’s take a rectangular sheet of paper and fold it diagonally. Then, explore the

relationship between the diagonals of a rectangle.
b) Let’s fold a rectangular sheet of paper to make a square. Cut out the square.

Now, fold the square sheet of paper diagonally. Then, explore the relationship
between the diagonals of a square.
8. a) Let’s take a rectangular sheet of paper and fold it as shown in the diagram to
make a parallelogram. Cut out the parallelogram.

Now, fold the parallelogram diagonally and explore the relationship between
the diagonals of a parallelogram.

16.3 Regular polygons

A polygon is a closed plane figure bounded by three or more than three line segments.
Triangles, quadrilateral, pentagon, hexagon, etc. are a few examples of polygons.

A polygon having all sides and angles equal is called a regular polygon. Equilateral
triangle, square, regular pentagon, etc. are a few examples of regular polygons.

Equilateral triangle Square Regular pentagon Regular hexagon

Interior angles of a polygon B A
The angles formed by the sides of a polygon inside the polygon C
are known as interior angles of the polygon. In the given figure,
‘ABC, ‘BAC, and ‘ACB are the interior angles of the 'ABC.

Exterior angles of a polygon C
When a side of a polygon is produced, it substends an angle with D BE
another side of the polygon outside the polygon. The angle so
formed is called an exterior angle of the polygon. In the given
figure, side AB of the quadrilateral ABCD is produced to E. So, A
‘CBE is the exterior angle of the quadrilateral.

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16.4 Sum of the interior angles of a polygon A

(i) The figure alongside is a triangle. The sum of the interior C
angles of the triangle is 180q. C

? ‘A + ‘B + ‘C = 180q. B B
(ii) The figure alongside is a quadrilateral. D

The number of sides of the quadrilateral (n) = 4 A

The sum of the angles of a quadrilateral = 360q
= 2 u 180q
= (n – 2) u 180q

On the basis of this knowledge, we can generalise a formula to find out the sum of the
angles of any polygon having ‘n’ number of sides.

Polygons No. of sides (n) No. of triangles Sum of the interior angles of
Triangle
formed polygon

3 1 1 u 180q = (3 – 2) u 180q

Quadrilateral

4 2 2 u 180q = (4 – 2) u 180q

Pentagon

5 3 3 u 180q = (5 – 2) u 180q

Hexagon

…… 6 4 4 u 180q = (6 – 4) u 180q
n – gon …… …… ……

n n–2 (n – 2) u 180q

Thus, the sum of the interior angles of a polygon = (n – 2) u 180q
In the case of a regular polygon, its n number of interior angles are equal.

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? The sum of ‘n’ number of interior angles = (n – 2) u 180q

? The size of each interior angle = (n – 2) u 180q
n

Thus, each interior angle of a regular polygon = (n – 2) u 180q
n

16.5 Sum of the exterior angles of a polygon x

The figure alongside is a quadrilateral. The sides of the quadrilateral d c w
are produced only to one direction. ya b

Let, a, b, c, and d are the interior angles of the quadrilateral. Also, z
w, x, y, and z are the exterior angles of the quadrilateral.

Here, the sum of the interior angles of the quadrilateral is

a + b + c + d = (4 – 2) u 180q = 2 u 180q

Also,a + z = 180q, b + w = 180q, c + x = 180q and d + y = 180q
?(a + z) + (b + w) + (c + x) + (d + y) = 4 u 180q
or, (a + b + c + d) + (w + x + y + z) = 4 u 180q
or, 2 u 180q + (w + x + y + z) = 4 u 180q
or, w + x + y + z = 4 u 180q – 2 u 180q = 2 u 180q = 360q

Thus, the sum of the exterior angle of any polygon = 360q

In the case of a regular polygon, its n number of exterior angles are equal.

? The sum of n number of exterior angles = 360q

? The size of each exterior angle 360°
=n

360°
Thus, each exterior angle of a regular polygon = n .

Worked-out examples

Example 1: Find the sum of interior angles of an octagon.
Solution:
In an octagon, number of sides (n) = 8
Now, the sum of interior angles of an octagon = (n – 2) u 180q

= (8 – 2) u 180q = 6 u 180q = 1080q

Example 2: Find the interior and exterior angle of a regular hexagon.

Solution:

In a regular hexagon, number of sides (n) = 6 180q
(n – 2) u 180q (6 – 2) u
Now, interior angle of the regular hexagon = n = 6

= 4 u 30q = 120q

360° 360°
Again, the exterior angle of the regular hexagon = n = 6 = 60q

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Example 3: In the adjoining hexagon, find the size of x°. 70° 105°
Solution:
Here, the sum of the interior angles of hexagon= (n – 2) u 180q 70°

xq + 130q + (180q – 70q) + 105q + (180q – 70q) + 130q = (6–2)u180q 130°
or, xq + 130q + 110q + 105q + 110q + 130q = 4 u 180q x° 130°

or, xq + 585q = 720q

or, xq = 720q – 585q = 135q

EXERCISE 16.2

General Section - Classwork

1. Let’s say and write the correct answer in the blank spaces.

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

b) The sum of the exterior angle of any regular polygon is ................................

c) Each interior angle of a regular 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 ....................................................

e) 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 = ..................

Creative Section - A

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

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

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

3. Find the size of interior angles of the following regular polygons by using formula.

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

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

4. Find the size of exterior angles of the following regular polygons by using formula.

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

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

5. Find the number of sides of polygons whose sum of interior angles are given below.

a) 540° b) 720° c) 900° d) 1080° e) 1440°

6. The size each interior angle of regular polygons are given below. Find the number of
sides of the polygons.

a) 90° b) 108° c) 120° d) 135° e) 140°

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7. Find the number of sides of regular polygon whose measure of each exterior angles
are given below.

a) 72° b) 60° c) 45° d) 40° e) 36°

8. Find the unknown sizes of angles of the following polygons.

a) x b) c) 87°
105° 60° x x 105°

100° 75° 95°

80°
95°

d) e) 80° f) x
70° 64°
x 30°
25° x 20° 130° 130°
65°
75° 110°

g) h) i) x x
72° 165° 160°

150° 150° (x+10)°
x
135° x 160° 60°
140° 50°
b)
Creative Section - B

9. Find the value of x° in each case.
a)

x
x

c) d)

x x

It’s your time - Project work!

10. a) Let’s draw six equilateral triangles of the same measurement of length in a chart
paper. Cut out each triangle and arrange them taking different number of triangles
to form the following polygons. (i) Rhombus (ii) Trapezium (iii) Hexagon

b) How many rhombus did you form?
c) How many trapezium did you form?
d) How many hexagon did you form?

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Unit Geometry: Construction

17

17.1 Construction of angles - Looking back

Classroom - Exercise

1. Let’s study the following construction. Tell and write the measurement of angles
being constructed in the diagrams.
a) b) c) d)

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

17.2 Construction of rectangle
1. When the diagonals and an angle made by them are given

Construct a rectangle ABCD in which diagonals X
AC = BD = 5cm and they bisect each other making an D
angle of 60°.

Steps of construction.

(i) Draw AC = 5 cm and draw its perpendicular A 60° C
bisector to find its mid-point O. O BY

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

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

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

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

2. When a side and an angle made by the diagonal with the side are given.

Construct a rectangle ABCD in whch AB = 5.4 cm and ‘BAC = 30°.

Steps of construction D X
(i) Draw AB = 5.4 cm. C

(ii) At A, construct ‘BAX = 30°

(iii) At B, construct an angle of 90°.

It intersects AX at C.

(iv) With centre at C and radius equal

to the length of AB draw an arc. 30° B
A
(v) With centre at A and radius equal
to the length of BC draw another

arc to cut the previous arc at D.

(vi) Join A, D and C, D. Thus, ABCD is the required rectangle.

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Geometry: Construction

3. When a diagonal and angle made by it with a side are given

Construct a rectangle PQRS in which diagonal X

PR = 6 cm and ‘RPS = 60° YS

Steps of construction 60° 30° R
(i) Draw a diagonal PR = 6 cm P 30° 6 cm Z
(ii) Construct ‘RPX = 60° at P.
(iii) Construct ‘PRY= 30° at R. Q

RY intersects PX at S
(iv) Construct ‘RPZ = 30° at P.
(v) Draw an arc with radius equals to RS from

P to cut PZ at Q.
(vi) Join R, Q.

PQRS is the required rectangle.

17.3 Construction of regular polygons
Construction of a regular pentagon

1. Construct a pentagon with the length of each side 4.5 cm.

In a pentagon, n = 5. So,

each interior angle of pentagon = n–2 × 180° = 5 – 2 × 180° = 108°
n 5

Steps of Construction QD
4.5cm
(i) Draw AB = 4.5 cm.

(ii) At B, draw ‘ABP = 108° with the help

of a protractor. P

(iii) Draw an arc of radius 4.5 cm from B to E 108° C
cut BP at C.

(iv) At C, draw BCQ = 108° with the help of R

the protractor. 4.5cm

(v) Draw an arc of radius 4.5 cm from C to

cut CQ at D. 108°
B
(vi) At D, draw ‘CDR = 108° with the help A 4.5cm

of the protractor.

(vii) Draw an arc of radius 4.5 cm from D to cut DR at E.

(viii) Join E, A

ABCDE is the required regular pentagon.

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Geometry: Construction Geometry: Construction

2. Construct a regular pentagon inside a circle with diameter 6 cm.

Steps of construction R

(i) Draw a line segment PQ = 6 cm and construct

its perpendicular bisector RS that intersects PQ A
at O.

(ii) With the centre at O and the radius OP = OQ,

(iii) draw a circle that intersects RS at A and X. B E
Construct the perpendicular bisector of OQ that TQ

cuts OQ at T. P UO

(iv) From T draw an arc of radius equal to TA to cut

OP at U.

(v) From U take an arc of radius equal to UA and C D
starting from A draw arcs to cut the circumference X
of the circle at B, C, D, and E respectively.

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

Now, ABCDE is the required regular pentagon. S

Construction of a regular hexagon R E Q
1. Construct a regular hexagon with the length of S 120°

each side 3.5 cm. 3.5cm P

In a hexagon, n = 6.

So, each interior angle of hexagon
n – 2 6–2
= n × 180° = 6 × 180° = 120°. 120° C

Steps of construction 3.5cm

At every vertex, construct an angle of 120°.

Then, follow the similar process as in case of A 3.5cm B
construction of regular pentagon.

2. Construct a regular hexagon of a side 4.5 cm inside a circle.

Steps of construction E D
O
(i) Draw a straight line segment AB = 4.5 cm.
B
(ii) From A and B draw two arcs each of radius
equal to AB to interest at O.

(iii) With the centre at O and radius equal to C

OA draw a circle. F

(iv) Starting from B draw arcs each of radius

equal to OA to cut the circumference of

the circle at C, D, E, and F respectively.

(v) Join B, C, D, E, F, and A.

Now, ABCDEF is the required regular hexagon. A

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Geometry: Construction

Construction of a regular octagon

Construct a regular octagon with the length of each side 3 cm.

In an octagon, n = 8. So, each interior angle

of octagon S F R Q
135° E
n– 2 8 – 2
n 8 135°
= × 180° = × 180° = 135° 135°
135° 3 cm
G 135° 3 cm135° D

Steps of construction

At every vertex, construct an angle of 135° T H 135° C
and follow the process as like in the case of
construction of regular pentagon. U 135° 3 cm
A 3 cm B

1. Construct a regular octagon inside a circle with diameter 4.8 cm.

Steps of construction

(i) Draw a straight line segment AE = 4.8 cm C

and construct its perpendicular bisector. The

perpendicular bisector intersects AE to O. B

(ii) With centre at O and radius equal to OA,

construct a circle such that the circle cuts the

perpendicular bisector at C and G. A E
F
(iii) Construct the angular bisector of ‘AOC that H
cuts the circumference of the circle at B and F.
G
(iv) Construct the angular bisector of ‘EOC that
cuts the circumference of the circle at D and H.

(v) Join A, B, C, D, E, F, G, H, and A.

Now, ABCDEFGH is the required regular octagon.

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Geometry: Construction

EXERCISE 17.1

Creative Section
1. Construct a rectangle ABCD in which

a) Diagonal AC = BD = 6 cm and they bisect each other making an angle of 45°.
b) Diagonal AC = BD = 8.5 cm and they bisect each other making an angle of 60°.
c) Diagonal AC = 5.5 cm and ‘BAC = 60q.
d) Diagonal BD = 6 cm and ‘ABD = 30q.
2. Construct rectangle ABCD in which

a) AB = 5 cm and ‘BAC = 30°
b) BC = 5.4 cm and ‘BCA = 45°
c) The length of a side is 4.6 cm and a diagonal makes an angle of 60° with this

side.
d) The length of side BC is 8 cm and diagonal BD = 10 cm.
3. a) Construct a regular pentagon ABCDE in which

AB = BC = CD = DE = EA = 4 cm
b) Construct a regular pentagon in a circle with diameter 5 cm.
4. a) Construct a regular hexagon ABCDEF in which

AB = BC = CD = DE = EF = FA = 4.5cm.
b) Construct a regular hexagon of a side of 5.5 cm inside a circle.
5. a) Construct a regular octagon ABCDEFGH in which

AB = BC = CD = DE = EF = FG = GH = 5 cm.
b) Construct a regular octagon in a circle with diameter 5 cm.
It’s your time - Project work
6. a) Let’s draw a rectangle of your own measurements of sides.
b)Let’s draw a square of your own measurements of sides.
c) Let’s draw a regular pentagon of your own measurements of sides.
d)Let’s draw a regular hexagon of your own measurements of sides.

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Unit Coordinates

18

18.1 Coordinates - Looking back
Classroom - Exercise

1. Let's study the adjoining graph, tell and write the answers as quickly as possible.

a) The straight line XOX' Y
is called ....................................
A
B

b) The straight line YOY'

is called .................................... X' O DX

c) The point O is C

called .................................... Y'
d) The coordinates of the points A, B, C

and D are ........................, ........................., ........................ and .........................

e) The coordinates of the origin are .........................

In the graph given alongside, XOX' and YOY' are known as the coordinate axes. Here,

XOX' is called the x-axis (abscissa) and YOY' is called the y-axis (ordinate). The point O

is called the origin. In the graph, the coordinates of the point A is (5, 3), B is (–4, 2), C is

(–3, –5), and D is (2, –3). Y

Furthermore, the intersected x-axis and X' QP X
y-axis divide the coordinate plane into
4 regions. These regions are called the (O0, 0)
quadrants. S

The region XOY is called the first R
quadrant. Y'

The region X’OY is called the second
quadrant.

The region X’OY’ is called the third
quadrant.

The region XOY’ is called the fourth
quadrant.

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Coordinates

18.2 Pythagoras Theorem PerpendicularA Hypotenuse
B
The adjoining triangle is a right angled triangle ABC right angled at B. Base C
The side AC opposite to the right angle is called its hypotenuse. The
sides AB and BC are called its perpendicular and base respectively.

Pythagoras was a Greek Mathematician who lived around 2500 years
ago. He discovered a significant fact about right-angled triangles
known as Pythagoras Theorem.

Pythagoras Theorem states that in a right angled triangle, the square of the hypotenuse
is equal to the sum of the squares of perpendicular and base.

? (hypotenuse)2 = (perpendicular)2 + (base)2 F
h2 = p2 + b2

Study the following illustration and try to understand the G I
Pythagoras Theorem. B
A
In the given figure, 'ABC is a right angled triangle, H
right angled at B. Here AC = 5 cm is the hypotenuse,
AB = 4 cm is perpendicular and BC= 3 cm is base. C

Here, area of the square ACDE = 25 sq. cm.

Area of square BCHI = 9 sq. cm. ED
Area of square ABFG = 16 sq. cm.

Thus, area of square ACDE = area of square BCHI + area of square ABFG

i.e. square of hypotenuse = square of perpendicular + square of base

h2 = p2 + b2.

18.3 Pythagorean Triples

Suppose the hypotenuse (h) of a right angled triangle be 5, perpendicular (p) be 4, and
base (b) be 3. Then, according to Pythagoras theorem,

h2 = p2 + b2
or, 52 = 42 + 32
or, 25 = 16 + 9
or, 25 = 25
Thus, the relation h2 = p2 + b2 is satisfied by the particular values of h, p and b. Here,
5, 4, and 3 are called the Pythagorean Triples.
When a set of Pythagorean triple is known, we can find other sets of triples multiplying
each number of triple by the same natural number. For example,

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Coordinates

3, 4 and 5 is a triple then,
3 × 2 = 6, 4 × 2 = 8 and 5 × 2 = 10
So, the other set of triples is 6, 8, and 10.
Similarly, 15 (3 × 5), 20 (4 × 5), and 25 (5 × 5) are also the triples.

Experiment 19: Experimental verification of Pythagoras Theorem.
Step 1: Draw three right angled triangles ABC of different sizes and right angled at B.

A
B BA

BC C A C Fig (iii)
Fig (i) Fig (ii)

Step 2: Measure the lengths of the sides of each triangle and write the measurements in

the table.

Fig No. AB BC CA AB2 BC2 CA2 AB2 + BC2 Results

(i) CA2 = AB2 + BC2

(ii) CA2 = AB2 + BC2

(iii) CA2 = AB2 + BC2

Conclusion: The square of the hypotenuse of a right-angled triangle is equal to the sum
of the squares of perpendicular and base.

Worked-out examples

Example 1: In the adjoining right angled triangle ABC, calculate the length of AC.

Solution = BC = 8 cm A
Here, base (b)

perpendicular (p) = AB = 6 cm 6 cm

hypotenuse (h) = AC = ?

Now, by using the Pythagoras theorem, B 8 cm C

h2 = p2 + b2

or, h2 = 62 + 82

or, h2 = 36 + 64

or, h2 = 100

or, h = 100 = 10 cm

? The required length of AC is 10 cm.

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Coordinates

Example 2: In the given figure, a ladder 13 m long rest against a vertical wall. If the
height of the wall at which the upper end of the ladder is supported is
12 m, find the distance of the foot of the ladder from the wall.

Solution:
Let PR be the length of the ladder, PQ be the height of the wall, and QR be the distance
between the wall and foot of the ladder.

Here, hypotenuse (h) = PR = 13 m

perpendicular (p) = PQ = 12 m P

base (b) = QR

Now, by using Pythagoras theorem, 13 m 12 m
h2 = p2 + b2

or, 132 = 122 + b2

or, 169 = 144 + b2 RQ

or, b2 = 169 – 144 = 25
or, b = 25 = 5 m.

So, the required distance is 5 m.

EXERCISE 18.1
General Section – Classwork

1. Let's say and write the answers as quickly as possible.

a) p = 1, b = 2, h = .................. b) p = 2, b = 3, h = ......................

c) p = 3, b = 4, h = ................. d) h= 3, p = 1, b = ........................

e) h = 5, b = 3, p = ................. f) h = 5, p = 4, b = ........................

2. Let's say and write the distance between origin and the given points.

a) (0,0) and (1, 2) distance = ................ b) (0,0) and (2, 3) distance = .................

c) (0,0) and (3, 1) distance = .................d) (0,0) and (2, 1) distance = .................

Creative Section - A

3. Find the unknown lengths of sides of the following right angled triangles.

a) b) c) 12 cm d) M
A RX K
Q
12 cm
4 cm 10 cm ZL9 cm
6 cm

B 3 cm P Y
C 13 cm

4. Using Pythagoras theorem, find whether the following sides of triangles are the
sides of right angled triangles.

a) 3 cm, 4 cm, 5 cm b) 6 cm, 8 cm, 10 cm c) 9 cm, 12 cm, 16 cm

d) 7 cm, 10 cm, 14 cm e) 5 cm, 5 3 cm, 10 cm f) 3 3 cm, 4 cm, 8 cm

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Coordinates

5. Test and identify whether the following sets of numbers are pythagorean tripples or
not.

a) 6, 8, 10 b) 9, 15, 20 c) 7, 24, 25 B

6. a) In the adjoining figure, ABC is a triangle. If 15 cm 9 cm 15 cm

BD A AC, AB = BC = 15 cm, BD = 9 cm, find the

length of AC. A D C

b) Find the length of diagonal of the adjoining 12 cm
rectangle.
16 cm
c) If the length and breadth of a rectangle are 12 cm and 13 cm
5 cm respectively, what will be the length of its diagonal?
x cm
Creative Section -B S

7. a) In the adjoining figure, find the values of x and y. y cm R
3 cm

P 4 cm Q

B

b) Find the length of x and y in the adjoining A 13 cm x cmE 5 3cm C
12 cm D y cm
diagram.

A

8. a) A ladder 10 m long rests against a vertical wall 6 m 10 m 6m
above the ground. At what distance does the ladder C
touch the ground from the bottom of the wall?

B

A

b) The vertical height of a tree is 18 m. The uppermost part

of the tree is broken by wind 5 m above the ground. At

what distance does the uppermost part of the tree touch

the ground from the foot of the tree? B C

c) In the figure alongside, ABCD is a rectangular ground B C

of length 40 m and breadth 30 m. Ramesh reached at 30 m
C walking from A to B and B to C. But Shyam reached

to C walking from A to C directly. Who walked the A 40 m D
shorter distance and by how much?

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Coordinates

d) In the figure alongside, AB is an electric pole. When an A
electric wire falls, its one end touches the ground at P,
5 m away from the base of the pole. If the length of the wire 13 m
from the top of the pole to the ground is 13 m, complete
the following problems.

(i) Find the height of the pole. P 5m B 9m Q

(ii) If the other end of the wire touches the grand at Q, 9
m away from the base to the opposite side of the pole, find the length of
the falling wire.

18.4 Distance between two points

Let the coordinates of the point A be (x1, y1) and that of B be (x2, y2).

Draw perpendiculars AP from A and Y

BQ from B to the x - axis. Also, draw

perpendicular AR from A to BQ.

Here, OP = x1 PA = QR = y1 B(x2, y2)

OQ =x2 QB = y2 y2-y1

PQ = AR = OQ – OP = x2 – x1 A(x1,y1) x2-x1 R

BR = QB – QR = y2 – y1 X' X

Thus, AR = x2 – x1 and BR = y2 – y1 OP Q B
(10, 9)
Now, from right angled-triangle ARB, by Y'

Pythagoras theorem,

h2 = p2 + b2

AB2 = BR2 + AR2

AB2 = (y2 – y1)2 + (x2 – x1)2

AB = (x2 – x1)2 + (y2 – y1)2 is the distance between the points A and B.

Example 1 : Find the distance between the point p (4, 1) and Q (10, 9) .

Solution : x1 = 4 and y1 = 1 A
In A (4, 1), x2 = 10 and y2 = 9 (4, 1)
In B (10, 9),

Now, by using the distance formula,

AB = (x2 – x1)2 + (y2 – y1)2 = (10 – 4)2 + (9 – 1)2
= 62 + 82 = 36 + 64 =
100 = 10 units .

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Coordinates

Example 2: A point G is on the x-axis 8 units right from the origin. Another point H
is on the y-axis 6 units above the origin. Find the distance between the
points G and H.

Solution: Y

Here, the coordinates of the point G is (8, 0) H
and that of H is (0, 6)

So, x1 = 8, x2 = 0, y1 = 0, y2 = 6 X' O(0, 0) GX
Now, by using distance formula, Y'
GH = (x2 – x1)2 + (y2 – y1)2

= (0 – 8)2 + (6 – 0)2
= 64 + 36
= 100 = 10 units

? The distance between the point G and H is 10 units.

Example 3: The centre of a circle is O (5, 8) and A (12, 9) is any point in its
circumference. Find the diameter of the circle. Does another point
B (9, 4) lie in the circumference of the circle?

Solution:
Here, O (5, 8) is the centre of the circle. A (12, 9) is a point in its circumference.

So, x1 = 5, x2 = 12, y1 = 8 and y2 = 9. A (12, 9)
Now, by using distance formula, O (5, 8)
OA = (x2 – x1)2 + (y2 – y1)2

= (12 – 5)2 + (9 – 8)2
= 72 + 12
= 50 = 25 × 2 = 5 2 units

? The radius of circle (OA) = 5 2 units

? The diameter of the circle = 2(OA) = 2 × 5 2 = 10 2 units.

Again, to find the distance between O (5, 8) and B (9, 4)
OB = (9 – 5)2 + (4 – 8)2 = 42 + (–4)2 = 32 = 16 × 2 = 4 2 units

? OA ≠ OB.

So, OB is not the radius of the given circle. The point B (9, 4) does not lie in the
circumference of the circle.

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Example 4: Show that the points A (5, 3), B (1, 2), C (2, –2), and D (6, –1) are the

vertices of a square.

Solution: D(6, –1) C(2, –2)

Let, A (5, 3), B (1, 2), C (2, –2), and D (6, –1) are the vertices

of a quadrilateral ABCD. A(5, 3) B(1, 2)
Now, by using the distance formula,

AB = (x2 – x1)2 + (y2 – y1)2 = (1 – 5)2 + (2 – 3)2 = (–4)2 + (–1)2 = 17 units

BC = (x2 – x1)2 + (y2 – y1)2 = (2 – 1)2 + (–2 – 2)2 = 12 + (–4)2 = 17 units

CD = (x2 – x1)2 + (y2 – y1)2 = (6 – 2)2 + (–1+ 2)2 = 42 + 12 = 17 units

DA = (x2 – x1)2 + (y2 – y1)2 = (5 – 6)2 + (3+ 1)2 = (–1)2 + 42 = 17 units

Also, diagonal AC = (x2 – x1)2 + (y2 – y1)2 = (2– 5)2 + (–2 – 3)2 = 34 units

And, diagonal BD = (x2 – x1)2 + (y2 – y1)2 = (6– 1)2 + (–1 – 2)2 = 34 units

Thus, in quadrilateral ABCD,
The sides AB = BC = CD = DA = 17 units
Also, the diagonals AC = BD = 34 units

Here, all sides of the quadrilateral ABCD are equal and it's diagonals are also equal.

So, the quadrilateral ABCD is a square and the given points are the vertices of the
square.

Example 5: Show that the points (4, 3), (3, 2), and C (2, 1) are collinear.

Solution: A (4, 3) B (3, 2) C (2, 1)
Let the points be A (4, 3), B (3, 2), and C (2, 1).

Now, by using distance formula,
AB = (x2 – x1)2 + (y2 – y1)2 = (3 – 4)2 + (2 – 3)2 = (–1)2 + (–1)2 = 2 units
BC = (x2 – x1)2 + (y2 – y1)2 = (2 – 3)2 + (1 – 2)2 = (–1)2 + (–1)2 = 2 units
AC = (x2 – x1)2 + (y2 – y1)2 = (2 – 4)2 + (1 – 3)2 = (–2)2 + (–2)2 = 8 = 2 2 units
Here, AB + BC = 2 + 2 = 2 2 = AC

Hence, the given points are collinear.

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Coordinates

Example 6: If the distance between the points (a, 0) and (0, 3) is 5 units, find the
value of a.
Solution: Q (0, 3)

Let P (a, 0) and Q (0, 3) be the two given points. P (a, 0)
Here, PQ = 5 units. Squaring both sides
Now, by using distance formula,
52 = ( a2 + 9)2
PQ = (x2 – x1)2 + (y2 – y1)2 or, 25 = a2 + 9
or, 5 = (0 – a)2 + (3 – 0)2 or, a2 = 25 – 9
or, 5 = (–a)2 + 32 or, a2 = 16
or, 5 = a2 + 9
or, a = 16 = 4

So, the required value of a is 4.

EXERCISE 18.2
General Section

1. Say and write the distance between the given points.
a) (m1, n1) and (m2, n2) = ..........................
b) (0,0) and (3, 4) distance = ..........................
c) (0,0) and (4, 3) distance = ..........................
d) (0,0) and (1, 3) distance = ..........................
e) (0,0) and (2, 3) distance = ..........................

Creative Section - A

2. Find the distance between the following points.

a) (1, 2) and (3, 4) b) (5, 3) and (1, 6) c) (–2, 1) and (4, 3)

d) (4, 5) and (–3, 6) e) (–5, –4) and (–13, 2)

f) (5 + 3, 2 – 3) and (7 + 3, 2 + 3)

3. a) A point P is on the x-axis, 6 units right from the origin and another point Q is
on the y-axis, 8 units above the origin. Find the distance between P and Q.

b) A point A lies on the x-axis 9 units left from the origin and another point B lies
on the y-axis 12 units below the origin. Find the distance between A and B.

c) Find the distance between origin and a point P (–5, 5).

Creative Section - B

4. a) When the map of Nepal is presented in a coordinate plane, the coordinates of
Kathamndu is found to be (5, 3) and that of Pokhara (4, 2 3). Find the map
distance between these places. If 1 unit represents 100 km, find the actual
distance of Pokhara from Kathmandu.

b) Biratnagar lies at the point (1, 5) and Dharan lies at (3, 9) when the map
of Nepal is placed on the graph of coordinate plane. If 1 unit represents
4.5 5 km, find the actual distance between these two places.

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Y

c) The adjoining figure is the Tikapur
map of Nepal presented in the Bhairahawa
coordinate axes of the graph.
Find the distance between X
Tikapur and Bhairahawa. If
1 unit represents 18 5 km,
find the actual distance of
Tikapur to Bhairahawa.

O

Y

d) From the given graph of Hetauda
map of Nepal, find the Damak
actual distance between
Hetauda and Damak, if 1 unit
represents 4.1 85 km.

OX

5. a) The centre of a circle lies at O (4, 6) and A (–5, 18) is any point in its
circumference.

(i) Find the radius of the circle.

(ii) Find the diameter of the circle.

(iii) Show that B (13, –6) also lies in the circumference of the circle.

b) P (1, 3) is the centre of a circle and a point Q (7, 11) is any point in its
circumference. Find the diameter of the circle. Does another point (–5, 10) lie
in the circumference of the circle?

6. a) Show that the points A (3, 4), B (7, 8), and C (11, 4) are the vertices of an
Isosceles triangle.

b) Show that the points P (–6, 2), Q (1, 7), and R (6, 3) are the vertices of a scalene
triangle.

c) Show that the points A (2, –2), B (8, 4), C (5, 7), and D (–1, 1) are the vertices
of a rectangle ABCD.

d) Show that the points P (1, 1), Q (4, 4), R (4, 8), and S (1, 5) are the vertices of
a parallelogram.

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e) Show that the points A (3, 1), B (8, 1), C (8, 6), and D (3, 6) are the vertices of
a square.

f) Show that the points (4, 6), (–1, 5), (–2, 0), and (3, 1) are the vertices of a
rhombus.

g) Show that the points P (1, 1), Q (–1, –1), and R (– 3, 3) are the vertices of an
equilateral triangle.

7. Show that the following points are collinear.

a) (1, 2), (4, 5), (8, 9) b) (–3, 7), (0, 4), (2, 2)

c) (–1, –1), (2, 3), (8, 11) d) (4, –3), (2, 1), (–1, 7)

8. a) If the distance between the points (a, 3) and (8, 3) is 7 units, find the
value of a.

b) If the distance between the points (0, 5) and (a, 0) is 13 units, find the value

of a. Y
It's your time - Project work!

9. a) Let's draw three straight D A
10. a) lines of length in whole X
number of centimetres (say X' O
3 cm, 4 cm, 5 cm, ... ) in B
a squared graph paper by
using a ruler. Now, write the
coordinates of the joining
points of each straight line
and find the lengths by using
distance formula.

Are the lengths given by C Q
distance formula and the P
lengths measured by a ruler
equal? Y'

Let's draw the following plane shapes in the separate squared graph papers.

(i) An equilateral triangle (ii) An isosceles triangle

(iii) A square (iv) A rectangle

Then write the coordinates of the vertices of these shapes and find the length
of sides of each plane shape by using distance formula. Also find the length of
diagonals in the case of square and rectangle.

b) Let's draw a circle with centre at origin in a squared graph paper by using a
pencil compass. Then find the length of it's any 3 radii and a diameter. Show
that radii of the same circle are equal and diameter is two times the radius of
the circle.

c) Let's draw two right angled triangles in a squared graph paper by using a
ruler. Write the coordinates of the vertices of each triangle. Find the
length of perpendicular, base and hypotenuse by using distance formula.
Show that h2 = p2 + b2 in each of the right-angled triangles.

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Unit Circle

19

19.1 Perimeter of circle - Looking back
Classroom - Exercise

Study the given figures. Tell and write the answers as quickly as possible.

1. In the circle given alongside, C B
a) O is called the ............................ Q
b) OA is called the ............................ AO
c) AB is called the ............................ P
d) PQ is called the ............................

e) If OB = 2 cm, OC = ............................, AB = ............................ B

A

2. In the adjoining circle,

a) The shaded region AOB is called ............................ O

b) The shaded region PRQ is called ............................ P Q

R

Draw three circles with radii 2 cm, 3 cm and 4 cm respectively. Place three pieces of
thread along the circumference of each circle.

2 cm 3 cm 4 cm

Now, measure the length of each thread separately by placing them on a ruler. The
length of each thread gives the circumference of the circle on which it is placed.
Now, find the ratios of the length of a circumference of each circle to its diameter.

You will find that the ratio circumference is almost the same for every circle. This
diameter

constant ratio is represented by the Greek letter ‘S’ (pie).

Thus, if c be the circumference and d be the diameter of a circle,
circumference
then, diameter =S

or, c =S
d
or, c = Sd

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Circle

Again, diameter of a circle (d) = 2 u radius (r).
So, circumference (or perimeter) of a circle (c) = 2Sr
?Perimeter of circle = Sd or 2Sr
The approximate value of S is 3.142159..., which is approximately equivalent to 272.

19.2 Area of circle

Let’s do the following activities to find the formula of area of circles.
Take a circular piece of paper. Cut it into equal pieces as small as possible and arrange
the pieces as shown in the figure.

DC

A B r (breadth)
1
2 of 2Sr

(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 circle = Area of the rectangle ABCD.
= length u breadth
= Sr u r
= Sr2

Thus, area of circle = Sr2

Again, radius of cirlce (r) = d
2

So, area of circle = Sr2 = S ( d )2 = Sd2
2 4

Worked-out examples

Example 1: If the diameter of a circle is 7 cm find its (i) perimeter and (ii) area.

Solution: = diameter (d) = 7 cm = 3.5 cm
Here, the radius of the circle (r) 2 2

(i) Now, the perimeter of the circle = 2Sr

= 2 × 22 × 3.5
7

= 22 cm

(ii) Also, the area of the circle = Sr2

= 22 × 3.5 × 3.5
7

= 38.5 cm2

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Circle

Example 2: When a boy completes 5 rounds around a circular pond, he covers a

distance of 440 m.

Solution: (i) Find the perimeter of the pond
(ii) Find the diameter of the pond. (π = 272)

(i) Here, the distance covered in 5 rounds = 440 m

The distance covered in 1 round = 440 m = 88 m
5
? The perimeter of the pond = 88 m

(ii) We know that the perimeter of a circle = Sd

? Sd = 88 m
22
or, 7 u d = 88 m

or, d = 88 × 7 m = 28 m
22
So, the perimeter of the pond is 88 m and its diameter is 28 m.

Example 3: If the perimeter of the base of a cylindrical tank is 308 cm, find the
diameter and area of the base of the tank.

Solution:

Here, the perimeter of the base = 308 cm

or, Sd = 308 cm

or, 22 × d = 308 cm
7

or, d = 98 cm

So, the diameter of the base of the tank is 98 cm.

Again, area of the base = Sd2
4

= 22 × 1 × 98 × 98 = 7546 cm2.
7 4

So, the area of the base of the tank is 7546 cm2.

Example 4: Find the area of the shaded region in the following figures.

a) b) c)

7 cm 8 cm 452°8cm
O 14 cm

10 cm

Solution: 14 cm

a) Here, the radius of the circle (r) = 7 cm

? Area of the circle (A1) = Sr2 = 22 × 7 × 7 cm2 = 154 cm2
7

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Circle

The length of the rectangle (l) = 10 cm

The breadth of the rectangle (b) = 8 cm

? Area of the rectangle (A2) = l × b = 10 cm × 8 cm = 80 cm2
Now, the area of the shaded region = A1 – A2
= 154 cm2 – 80 cm2 = 74 cm2

b) Here, the length of the side of the square (l) = 14 cm

? Area of the square (A1) = l2 = (14 cm)2 = 196 cm2
Also, the diameter of the semi-circle = 14 cm

Then, the radius of the semi-circle (r) = 14 cm = 7 cm.
2
1 1 22
? Area of a semi-circle = 2 Sr2 = 2 × 7 × 7 × 7 cm2

= 77 cm2.

Now, Area of two semi-circles (A2) = 2 × 77 cm2 = 154 cm2.
the area of the shaded region = A1 – A2
= 196 cm2 – 154 cm2 = 42 cm2

c) We know that 360° makes 1 (whole) circle.

1° makes 1 part of a circle.
360
1 1
45° makes 360 × 45 = 8 part of a circle.

So, the area of the sector (A) = 1 of the area of the circle.
8
Here, the radius of the circle (r) = 28 cm

Angle at the centre of a shaded sector (T) = 45°

Area (A) = ?

Now, Area of sector (A) = 1 Sr2
8
1 22
= 8 × 7 × 28 × 28 cm2

= 308 cm2

Hence, the area of shaded sector is 308 cm2.

EXERCISE 19.1
General Section – Classwork

Let’s say and write the answers as quickly as possible.

1. a) The ratio of circumference of any circle to its diameter is represented by a
symbol ................................

b) The approximate value of S is ................................

c) If r be the radius of a circle, its perimeter = ...............................

d) If x be the radius of a circle, its area = ................................

e) If d be the diameter of a circle, its perimeter = ................................

f) If y be the diameter of a circle, its area = ................................

Vedanta Excel in Mathematics - Book 8 246 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Circle

2. a) In the figure, the area of the bigger circle is 510 cm2 and that
of the smaller circle is 350 cm2. The area of the shaded region
is ................................

b) In the given figure, the area of the square is 225 cm2 and
that of the circle is 154 cm2. The area of the shaded region
is ................................

Creative Section - A

3. Find the perimeter and area of the following circles.

a) b)

7 cm 28 cm

4. a) Find the perimeter of a circle with radius 3.5 cm.
b) Find the area of a circle with radius 14 cm.
c) Find the perimeter and area of a circle with diameter 14 m.

5. a) If the radius of a circular pond is 70 m, find its
(i) perimeter and (ii) area

b) If the diameter of a circular floor of a room is 35 m find its
(i) perimeter (ii) area

6. a) Find the length of wire required to fence a circular park of diameter 280 m
with 3 rounds.

b) The radius of a circular play ground is 77 m. Calculate the distance covered
by a runner in 5 complete rounds around the ground.

c) The radius of a vegetable garden is 63 m. If 1188 m of wire is given to fence
around it, how many rounds of fencing can be made?

Creative Section - B

7. a) When a marathon runner completes 3 round around a circular track he
covers a distance of 66 km. Find the radius of the track.

b) If 3520 ft of wire is required to fence a circular garden with 4 rounds, find
the diameter of the garden.

c) If a wheel of a bicycle covers 440 m in 200 rotations, find the diameter of
the wheel.

8. a) The perimeter of a circle is 132 cm, calculate (i) its radius and (ii) the area.
b) The perimeter of a circular park is 880 m. Find the area of the park.
c) The area of the base of a cylindrical tank is 38.5 m2. Find the perimeter of
the base.
d) The area of a circular pond is 1386 m2. Find the length of wire required to
fence around it with 1 round.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 247 Vedanta Excel in Mathematics - Book 8

Circle

9. a) Samriddhi bent a rope to form a circle. If she got the circumference of the
circle is 30 cm more than its diameter. What was the length of radius?

b) The rope required to measure the diameter of a circular pond is 90 m
shorter than measuring the circumference. Find the diameter of the pond.

10. Find the area of the shaded regions in the following figures.

a) b) c) 10cm

14 cm 42 cm r=3cm

14 cm e) f)

d) 14 cm

9 cm

7 cm 8 cm

g) h) 14 cm 14 cm

i)

12 cm 28 cm 21 cm
60°

12 cm 28 cm

It’s your time: Project work!

11. a) Let’s take a piece of wire and measure its length. Then bent it to form
a circle and measure its diameter. Calculate the circumference by using
formula C = Sd. Write the conclusion on length of wire you had before and
the circumference of the circle that you just calculated using formula.

b) Let’s draw three circles of different length of radii in a chart paper. Then
use thread to measure their circumference and complete the table given
below.

Fig. no. Diameter (d) Circumference (C) C Remarks
(i) d

(ii)

(iii)

Draw a conclusion and present in your class and weekly seminar of your
school.

Vedanta Excel in Mathematics - Book 8 248 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur