Vedanta Excel in Mathematcs Book -10 Final (2078)

Geometry – Area of Triangles and Quadrilateral
o) The areas of two parallelograms are equal. The altitude of one parallelogram is

4 cm and the base of the other is 6 cm. If the base of the first parallelogram is 9 cm,
find the height of the second parallelogram.

Creative section A DF E

3. a) In the adjoining figure, ABCD and C
BCEF are parallelograms, AE // BC. S
Prove that ABCD = BCEF.

B
P

b) In the figure alongside, QR // PS.
Prove that ' PQR = ' QRS.

c) In the given figure, DX // AB. Prove that QR
D CX

area of ' ABX = 1 area of ABCD.
2

d) In the given figure, ABCD is a parallelogram. AB
X is any point within it. Prove that the sum of DC
the areas of ' XCD and ' XAB is equal to half of
the area of ABCD. X
AB

M N

e) In the adjoining figure, triangles AMB and ANB O B
are standing on the same base AB and between A
the same parallel lines AB and MN. Prove that
area of ' AOM = area of 'BON.

M AN

f) In the given figure, 'ABC and parallelogram MBCN are on
the same base BC and between the same parallels MN and
BC. Prove that,

area of ' ABC = area of rectangle APCN. B P C
X C
D

g) In the given parallelogram ABCD, X and Y are B
any points on CD and AD respectively. Prove that Y
the area of ' AXB = the area of ' BYC.

A

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Geometry – Area of Triangles and Quadrilateral

A D
C
h) In the adjoining parallelogram ABCD, A is joined to any
point E on BC. AE and DC produced meet at F. Prove that
the area of 'BEF = the area of ' CDE

B E

F

P AS B

i) In the given figure, rectangle PQRS and
parallelogram AQRB are on the same base QR and
between the same parallels PB and QR. Prove that

(i) ' PQA # ' SRB Q R
(ii) Area of rectangle PQRS = Area of AQRB.

j) In the adjoining parallelogram ABCD, PQ // AB and AP D
RS // BC. Prove that the area of ROPA = the area of
QCSO. R S
O C

BQ

D MC
B
k) In the given figure, ABCD is a parallelogram. If M and N are N
any points on CD and DA respectively,
prove that ' AMB = ' CDN + 'ANB.

A

A PB
Q
l) In the given diagram, ABCD and PQRD are two
C
parallelograms. Prove that ABCD = PQRD. D R

AB

m) In the given figure, if AB // DC // EF, AD // BE and D HC
AF // DE, prove that parm. DEFH = parm . ABCD. G

EF

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4. a) In the given 'ABC, P is any point on the median AD. Prove A
that ' APB = ' APC. P

B D C
Q C
A

b) In the adjoining ' ABC, P and Q are the
mid-points of the sides AB and AC respectively
and R be any point on BC. Prove that, P

area of ' PQR = 1 area of ' ABC. B
4
R

P S
R
c) In the figure, PQRS is a parallelogram. Q and S are M
joined to any point M on the diagonal PR. Prove that,
the area of ' PQM = the area of ' PSM.

Q

A

d) In the given figure, DE // BC. Prove that D E
(i) ' BOD = ' COE C
(ii) ' BAE = ' CAD O
B C

e) In the given figure, P is the mid-point of AB and Q A
is any point on the side BC.CR meets AB at R and
R
CR // PQ. Prove that P

area of ' BQR = 1 area of ' ABC. B Q
2

DC

f) The line drawn through the vertex C of the quadrilateral
ABCD parallel to the diagonal DB meets AB produced at E.
Prove that,
A B E
C
the area of quad. ABCD = the area of ' DAE.

A

g) In the given triangle ABC, two medians BE and CD are D E
intersecting at O. Prove that O

the area of 'BOC = the area of quad. ADOE

B

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Geometry – Area of Triangles and Quadrilateral

A D
E
h) In the adjoining figure, it is given that AD // BC and
BD // CE. Prove that, area of ' ABC = area of ' BDE.

i) In the given figure, BEST is a parallelogram. BC
Diagonal BS is produced to point L. Prove that L
'LST and 'LSE are equal in area.
TS
j) In the given figure, AD // BC. If the area of 'ABE and
'ACF are equal, prove that EF // AC. B E
A ED

F

B C
AD
k) In the given figure, AD // BE // GF and AB // DG // EF.
If the area of parallelograms ABCD and CEFG are equal, B CE
then prove that DE // BG.
G F
l) In the adjoining figure, PQRS and LQMN are P S
two parallelograms of equal in area. Prove that
LR // SN. L N

m) In the given triangle ABC, medians BN and CM are Q RM
intersected at O. Prove that A
the area of 'BOC = the area of quadrilateral AMON.
MN
n) In the given figure, M is the mid-point of AE, O
prove that area of 'ABE is equal to the area
of parallelogram ABCD. BC

E

D MC

AB

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Unit Geometry - Construction

14

14.1 Construction of a triangle whose area is equal to the area
of a given quadrilateral.

Example 1: Construct a quadrilateral ABCD in which AB = 3.5 cm, BC = 4.8 cm,
AC = 5.5 cm, CD = 2.6 cm, and AD = 4 cm. Construct a triangle equal in area to
the quadrilateral.

Steps 1 2.6 cm C
(i) Draw AB = 3.5 cm

(ii) From the point B, draw an arc with radius D 4.8 cm
BC = 4.8 cm and from A draw another arc with 4 cm
radius AC = 5.5 cm. These two arcs intersect each
other at C.

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

(iv) From C, draw an arc with radius CD = 2.6 cm and A 3.5 cm B
from A, draw another arc with radius AD = 4 cm.
These two arcs intersect each other at D. Join C, D
and D, A. Thus, ABCD is the given quadrilateral.

Step 2 E D C
A B
(i) From the point D, draw DE parallel to CA to
meet BA produced at E.

(ii) Join C and E.
Now, ∆ BCE is the required triangle.
Here, ∆ BCE = ∆ ABC + ∆ ACE
= ∆ ABC + ∆ ACD
= Quad. ABCD

[∆ ACE and ∆ ACD are on the same base AC and between the same Parallels AC
and DE]

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Geometry – Construction

14.2 Construction of a parallelogram whose area is equal to the
area of a given triangle.

Example 2: Construct a triangle ABC in which a = 5.4 cm, b = 4.5 cm and c = 3.8 cm.
Construct a parallelogram equal in area to the triangle and having an angle
60°.

Step 1 A
(i) Draw a = BC = 5.4 cm

(ii) From B, draw an arc with radius c = BA = 3.8 cm and c = 3.8 cm b = 4.5 cm
from C, draw another arc with radius b = CA = 4.5 cm.
These two arcs intersect each other at A.

(iii) Join B, A and C, A. Thus, ABC is the given triangle. B a = 5.4 cm C

Step 2 A E FX
(i) Through the point A, draw AX parallel to BC.

(ii) With the help of the perpendicular bisector of
BC, mark the mid–point D on BC.

(iii) Draw ‘ EDC = 60° such that DE meet AX at E.

(iv) From the point E, draw an arc with radius B 60° C
EF = DC to cut AX at F. D

(v) Join F and C.
Now, EDCF is the required parallelogram.

Example 3: Construct a triangle ABC in which AB = 4.7 cm, BC = 5.8 cm and
CA = 3.5 cm. Construct a parallelogram with a side 6.5 cm and equal in
area to the triangle.

Step 1 4.7 cm A
(i) Construct the triangle ABC from the given 3.5 cm

measurements

Step 2 B 5.8 cm C
AE FX
(i) Follow the process (i) and (ii) of step 2 from
example 2.

(ii) From the point D, draw an arc with radius
DE = 6.5 cm to cut AX at E.

(iii) From the point E, draw an arc with radius 6.5 cm
EF = DC to cut AX at F.

(iv) Join F and C. D C
5.8 cm
B

Now, EDCF is the required parallelogram.

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Geometry – Construction

14.3 Construction of a triangle whose area is equal to the area of
a given parallelogram.

Example 4: Construct a parallelogram ABCD in which AB = 3.6 cm, BC = 5.2 cm
and ‘ ABC = 60°. Construct a triangle with one side 6.8 cm and equal
in area to the parallelogram.

Step 1 D X
(i) Draw AB = 3.6 cm C
(ii) At B, draw ‘ ABX = 60°

(iii) From B, draw an arc with radius
BC = 5.2 cm to cut BX at C.

(iv) From C, draw an arc with radius 5.2 cm
CD = AB and from A draw another arc
with radius AD = BC. These two arcs
intersect each other at D.

(v) Join, C, D and A, D. Thus, ABCD is the 60° B
given parallelogram.
A 3.6 cm

Step 2
(i) Produce AB to E such that AB = BE
(ii) From A, draw an arc with radius AF = 6.8 cm to cut DC (or DC produced, if necessary)

at F.
(iii) Join A, F and F, E. Now, AEF is the required triangle.

X
D CF

5.2 cm

60° B

A 3.6 cm E

14.4 Construction of a triangle whose area is equal to the area
of a given triangle.

Example 5: Construct a triangle ABC in which AB = 5.5 cm, ‘BAC = 60°and
AC = 5 cm . Construct another triangle DAB with the side AD = 7.8 cm and
the area equal to the area of ' ABC.

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Geometry – Construction Geometry – Construction

Step 1 D
(i) Construct the triangle ABC CX

from the given measurements. 5 cm

Step 2 A 5.5 cm B
(i) Through the point C, draw CX

parallel to AB.
(ii) From the point A, draw an arc

of radius AD = 7.8 cm to cut
CX at D.
(ii) Joing D, A and D, B.
Now, DAB is the required
triangle.

EXERCISE 14.1

General section
1. Construct triangle ABC by using the following measurements:

a) AB = 4.2 cm, BC = 3.6 cm, and CA = 5.5 cm b) AB = 5.4 cm., BC = 6.3 cm and CA = 4.5 cm

c) AB = 5.7 cm, ‘ A = 45°, and AC = 4.2 cm d) BC = 4.8 cm, ‘ B = 60° and CA = 6.8 cm

e) BC = 3.6 cm, ‘ B = 30°, and ‘ C = 45°.

2. Construct quadrilaterals ABCD by using the following measurements:
a) AB = 3.5 cm, BC = 4.2 cm, AC = 6.3 cm, CD = 2.5 cm, and AD = 5.2 cm

b) AB = 4.6 cm, BC = 3.9 cm, AC = 6.5 cm, CD = 3.7 cm, and AD = 4.8 cm

c) AB = 4.5 cm, BC = 4.2 cm, AC = 5.8 cm, BD = 6.4 cm, and AD = 5.3 cm

d) AB = 3.8 cm, ‘ B = 120°, BC = 4.6 cm, CD = 5.4 cm, and AD = 5 cm
e) AB = BC = 5.5 cm, CD = AD = 6.2 cm, and ‘ BAD = 75°.
3. Construct parallelograms ABCD by using the following measurements:
a) AB = 5.6 cm, BC = 4.5 cm, and ‘ ABC = 60°
b) AB = 4.3 cm, BC = 3.5 cm, and ‘ BAD = 120°
c) AB = 4.5 cm, BC = 5.5 cm, and AC = 6.7 cm
d) AB = 3.8 cm, ‘ ABD = 60°, and AD = 5.2 cm
e) AB = 5.5 cm, diagonal AC = 7.6 cm, and diagonal BD = 5.6 cm.

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Geometry – Construction Geometry – Construction

Creative section

4. a) Construct a quadrilateral PQRS in which PQ = QR = 5 cm, RS = SP = 4 cm,
and ‘P = 60°. Then, construct a triangle whose area is equal to the area of the
quadrilateral.

b) Construct a quadrilateral PQRS having PQ = QR = 5.9 cm, RS = PS = 6.1 cm and
‘QPS = 75°. Then, construct ' PST which is equal in area to the given quadrilateral.

c) Construct a quadrilateral MNOP in which MN = 4 cm, NO = 5 cm, OP = 5.5 cm,
PM = 3.5 cm, and ‘PMN = 30°. Construct a triangle MPQ whose area is equal to
the quadrilateral MNOP.

d) Construct a quadrilateral ABCD having AB = 5.2 cm, BC = 5 cm, CD = 4.2 cm,
AD = 4 cm, and ‘ABC = 60°. Construct ' ADE whose area is equal to the
quadrilateral ABCD.

e) Construct a quadrilateral ABCD in which AB = 4 cm. BC = 5 cm, CD = 5.5 cm,
DA = 4.5 cm, and diagonal AC = 6 cm. Also construct a triangle ADE equal in area
to the quadrilateral.

f) Construct a quadrilateral ABCD in which AB = 5.4 cm, BC = 5.1 cm, CD = 4.9 cm,
AD = 6.1 cm, and the diagonal BD = 5.7 cm. Also construct a triangle equal in area
to the quadrilateral ABCD.

g) Construct a quadrilateral PQRS in which QR = 5.3 cm, RS = 5 cm PS = 5.7 cm,
PQ = 6.2 cm, and PR = 5.6 cm. Then, construct a triangle RQT equal in area to the
quadrilateral PQRS.

h) Construct a 'WXB equal in area to the quadrilateral WXYZ having WX = XY = 5.5
cm, YZ = ZW = 4.5 cm, and ³WXY = 75°.

5. a) Construct a triangle in which a = 7 cm, b = 6 cm and c = 5 cm and also construct a
parallelogram whose area is equal to the area of triangle ABC and one of the angles
being 60°.

b) Construct a triangle ABC in which a = 7.8 cm, b = 7.2 cm and C = 6.3 cm. Then
construct a parallelogram DBEF equal in area to ' ABC and ‘ DBC = 75°.

c) Construct a triangle PQR in which PQ = 7.5 cm, QR = 6.8 cm, and RP = 6 cm.
Then construct a parallelogram TQSU equal in area to ' PQR having a side
TQ = 6.4 cm.

d) Construct a triangle ABC from the following data. Also construct a parallelogram
having one side 7.5 cm and equal in area to the triangle.
AB = 7.1 cm, ‘ BAC = 60° and AC = 5.7 cm

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Geometry – Construction

e) Construct a triangle ABC in which AB = 5 cm, ‘ ABC = 75°and ‘ ACB = 60°.
Then, construct a rectangle equal in area to the triangle ABC.

f) Construct a triangle ABC having sides AB = 4 cm, BC = 6.8 cm, and CA = 6.5 cm.
Then, construct a rectangle equal in area of ' ABC.

6. a) Construct a parallelogram ABCD in which AB = 6 cm, BC = 4 cm and
‘BAD = 45°. Construct a triangle APQ having one angle 60° and equal in area of
the parallelogram.

b) Construct a parallelogram ABCD having AB = 5 cm, BC = 4 cm, and ‘B = 120°.
Also construct a triangle with the area equal to the parallelogram.

c) Construct a parallelogram ABCD in which AB = 6 cm, BC = 4 cm and
‘ BAD = 60°. Then construct a triangle AEF equal in area to the parallelogram and
the side AE = 7.5 cm.

d) Construct a parallelogram PQRS in which PQ = 5 cm, diagonal PR = 6 cm, and
diagonal QS = 8cm. Construct a triangle PSA whose area is equal to the area of the
parallelogram.

7. a) Construct a triangle ABC having sides a = 6.4 cm, b = 6cm, and c = 5.6 cm. Also
construct another triangle equal in area to the 'ABC and having a side 7 cm.

b) Construct a triangle PQR in which PQ = 5.2 cm, QR = 6 cm and ‘Q = 60°.
Construct another triangle SPQ which is equal in area to the 'PQR and side
PS = 8.5 cm.

c) Construct a triangle XYZ in which XY = 6.3 cm, ‘X = 30°, and ‘Y = 45°. Construct
another triangle WXY equal in area to ' XYZ and a side WY = 7.5 cm.

8. a) Construct a parallelogram ABCD in which AB = 3.5 cm, AC = 5.9 cm, and
BC = 4.2 cm. Construct another parallelogram ABEF where BE = 6.5 cm.

b) Construct a parallelogram PQRS in which PQ = 4 cm, ‘P = 45°, and PS = 4.5 cm.
Construct another parallelogram PQXY where ‘PQX = 30°.

9. a) Construct a rectangle with length 7.1 cm and breadth 6.1 cm. Also, construct a
triangle having one angle 60° and equal to the area of the rectangle.

b) Construct a rectangle ABCD with length 5.5 cm and breadth 4.5 cm. Also, construct
a triangle EBF having one angle 45° and equal to the area of the rectangle.

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Unit Geometry - Circle

15

15.1 Definition of terms related to circle P2 P1
O P4
A circle is the locus traced out by a moving point which is
equidistant from a fixed point called the centre of the circle. In P3
the figure, O is the centre of a circle.
O
(i) Circumference

The perimeter of a circle is called its circumference. It is
the total length of the curved line of the circle.

(ii) Radius B A
C
It is the line segment that joins the centre of a circle and
any point on its circumference. In the figure OA is the O
radius. The plural form of radius is radii. All the radii of a D
circle are equal,
i.e. OA = OB = OC = OD = ... R

(iii) Diameter PO Q
S
A line segment that passes through the centre of a circle
and joins any two points on its circumference is called the X
diameter of the circle. In the figure PQ, RS, etc. are the
diameters. The length of a diameter of a circle is two times P OQ
its radius. So, PQ = 2OQ or 2OP and RS = 2OR or 2OS. A Y
diameter divides a circle into two halves. Q

(iv) Semi-circle PR
O
A diameter divides a circle into two halves and each half is S
called the semi-circle. In the figure, PXQ and PYQ are the
semi-circles.

(v) Arc

A part of the circumference of a circle is called an arc. It is
denoted by a symbol ‘ ‘. In the figure PQR is the minor
and PSR is major arcs of the circle. A minor arc is less than
half of the circumference and a major arc is greater than
half of the circumference.

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(vi) Chord Geometry – Circle

The line segment that joins any two points on the circumference Q
of a circle is called the chord of the circle. In the figure PQ, and P
RS are any two chords of the circle. Diameter is the longest chord
of a circle. O
R
(vii) Segment
S
The region enclosed by a chord and corresponding arc is called
the segment of a circle. In the figure, the shaded region is the PQ
minor segment and none-shaded region is the major segment of O
the circle. A minor segment is less than half of a circle whereas
the major segment is greater than half of the circle. The minor and
major segments of a circle are known as the alternate segments
to each other.

(viii) Sector O

The region enclosed between any two radii and the corresponding Q
arc of a circle is called a sector of the circle. In the figure, OPQ is P
a sector of a circle.
Q CZ R
(ix) Concentric circles Y
B
Two or more circles are said to be concentric circles if they have
the same centre but different radii. In the figure, circles PQR, P XAO
XYZ, and ABC are three concentric circles with the centre at O.
P
(x) Intersecting circles
O O'
If two circles intersect each other at two points, they are said to Q
be intersecting circles. In the figure, two circles are intersecting
each other at P and Q. PQ is the common chord of these two
intersecting circles.

(xi)Tangent to a circle O
TQ
A line that intersects the circle at exactly one point is called a
tangent to the circle. The point at which the tangent intersects P BQ
the circle is called the point of contact. In the figure, PQ is the
tangent and T is the point of contact.

(xii) Secant of a circle A O
P
A line that intersects a circle in two distinct points is called a
secant of the circle. In the figure, PQ is the secant of the circle.

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Geometry – Circle

15.2 Theorems related to arcs and the angles subtended by them

In the figure alongside, O is the centre of a circle, OP and OQ are the radii, and PQ
is the arc of the circle.

Here, PQ subtends an ‘POQ at O. O

So, ‘POQ is the angle at the centre of the circle. Of course,

‘POQ is standing on PQ. So, we may call the opposite arc of P

‘POQ is PQ. Q

Furthermore, if we consider the radius OP starts rotating from the point P to the
point Q, it describes an arc PQ and ‘POQ. Thus, the degree measurements of PQ
and ‘POQ are exactly the same.

? ‘POQ PQ

Thus, the degree measurement of the angle at the centre of a circle is equal to the
degree measurement of its opposite arc.

A

Again, ‘PAQ is subtended by the arc PQ at the circumference

of the circle. ‘PAQ is said to be an angle at the circumference

of the circle. It is also called inscribed angle. Here, the opposite

arc of ‘PAQ is PQ. P Q
A
Now, let's establish the relation between an inscribed angle and
its opposite arc.

In 'OAP, ‘OAP = ‘OPA [? OA = OP]

? 2‘OAP = 180q – ‘AOP [From sum of the O
angles of 'OAP] PQ

In 'OAQ, ‘OAQ = ‘OQA [OA = OQ]

? 2‘OAQ = 180q – ‘AOQ [From sum of the angles of 'OAQ]

Now, 2‘OAP + 2‘OAQ = 180q – ‘AOP + 180q – ‘AOQ

or, 2‘PAQ = 360q – (‘AOP + ‘AOQ)

or, 2‘PAQ = 360q – Ref. ‘POQ

or, 2‘PAQ circumference – PAQ [Circumference 360q and PAQ Ref. ‘POQ]

or, 2‘PAQ PQ

? ‘PAQ 1 PQ.
2

Thus, the degree measurement of the angle at the circumference of a circle is half

of the degree measurement of its opposite arc.

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Theorem 5

If two arcs of a circle subtend equal angles at the centre of the circle, the arcs are equal.

Experimental verification
Step 1: Three circles with centre O and different radii are drawn.
Step 2: Two equal angles AOB and COD are drawn at the centre of each circle.

CD A

O D O OD
A

C AB B
B C

(i) (ii) (iii)
Step 3: The lengths of AB and CD are measured with the help of thread and ruler and the

results are tabulated.

Figure arc AB arc CD Result
(i) AB = CD
AB = CD
(ii) AB = CD

(iii)

Conclusion: If two arcs of a circle subtend equal angles at the centre of the circle, the arcs are
equal.

Converse of Theorem 5

Equal arcs of a circle subtend equal angles at the centre of the circle.

Experimental verification
Step 1: Three circles with centre O and different radii are drawn.
Step 2: Two equal arcs AB and CD are drawn with the help of pencil and compasses.

C

AB

OD B D

O O
(iii)
A

C DC A
B

(i) (ii)

Step 3: ‘AOB and ‘COD are measured and the results are tabulated.

Figure ‘AOB ‘COD Result
(i) ‘AOB = ‘ COD
(ii) ‘AOB = ‘ COD
(iii) ‘AOB = ‘ COD

Conclusion: Equal arcs of a circle subtend equal angles at the centre of the circle.

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Geometry – Circle

Theorem 6

Equal chords of a circle form equal arcs in the circle.

Experimental verification

Step 1: Three circles with centre O and different radii are drawn.

Step 2: Two equal chords AB and CD are drawn in each circle.

B A
A
AC

O CO O B
BD D D
C

(i) (ii) (iii)

Step 3: The length of arcs AB and CD are measured with the help of thread and ruler and

the results are tabulated.

Figure arc AB arc CD Result
(i) AB = CD
AB = CD
(ii) AB = CD

(iii)

Conclusion: Equal chords of a circle form equal arcs in the circle.

Converse of Theorem 6

If the arcs formed by chords of a circle are equal, the chords are equal.

Experimental verification

Step 1: Three circles with centre O and different radii are drawn.

Step 2: Two equal arcs AB and CD are drawn with the help of pencil and compasses in each
circle.

Step 3: A, B and C, D are joined to form chords AB and CD.

AC B A C

A

O D CO O
B D
BD
(i) (ii) (iii)

Step 4: The lengths of chords AB and CD are measured and the results are tabulated.

Figure AB CD Result

(i) AB = CD

(ii) AB = CD

(iii) AB = CD

Conclusion: If the arcs formed by chords of a circle are equal, the chords are equal.

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Theorem 7

The angle at the centre of a circle is twice the angle at its circumference standing on
the same arc.

Experimental verification
Step 1: Three circles with centre O and different radii are drawn.
Step 2: ‘AOB is drawn at the centre and ‘ACB is drawn at the circumference of each

circle. Here, ‘AOB and ‘ACB are standing on the same arc AB.

C AB

O A O
AB O C
(iii)
B
C

(i) (ii)

Step 3: ‘AOB and ‘ACB are measured and the results are tabulated.

Figure ‘AOB ‘ACB Result
(i) ‘AOB = 2 ‘ACB
(ii) ‘AOB = 2 ‘ACB
(iii) ‘AOB = 2 ‘ACB

Conclusion: The angle at the centre of a circle is twice the angle at its circumference standing
on the same arc.

Theoretical proof C
O
Given: O is the centre of a circle. ‘AOB is the angle at the centre
and ‘ACB is the angle at the circumference of the circle. DB
They are standing on the same arc AB.

To prove: ‘AOB = 2 ‘ACB. A

Construction: C, O are joined and the line is produced to D.

Proof

Statements Reasons
1. In 'OAC, ‘OAC = ‘ OCA 1. OA = OC (radii of the same circle),

2. ‘AOD = ‘OAC + ‘OCA so base angles of isosceles 'OAC
2. Exterior and opposite interior
3. ‘AOD = ‘OCA + ‘OCA = 2 ‘OCA
4. Similarly, ‘BOD = 2 ‘OCB angles of 'OAC
5. ‘AOD + ‘BOD = 2 ‘OCA + 2 ‘OCB 3. From statement (1)
6. ‘AOB = 2 (‘OCA + ‘OCB) = 2‘ACB 4. Same as above facts and reasons
5. Adding statements (3) and (4)
6. Whole part axiom

Proved

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Geometry – Circle

Alternative process Reasons C
1. Relation between angle at the centre O
Statements AB
1. ‘AOB AB of a circle and its opposite arc
2. Relation between inscribed angle
2. ‘ACB 1 AB
2 and its opposite arc
i.e. AB 2‘ACB
3. From the statements (1) and (2)
3. ‘AOB = 2 ‘ACB Proved

Theorem 8 C

The angle in the circumference of a semi-circle is one right angle.

A O B

Given: O is the centre of a circle and AB is a diameter of the circle.

To prove: ‘ACB = 90q

Proof

Statements Reasons

1. ‘AOB = 180q 1. ‘AOB is a straight angle.

2. ‘ACB = 1 ‘AOB 2. Inscribed angle is half of the angle at the
2 centre of a circle standing on the same arc.

3. ‘ACB = 1 u 180q = 90q 3. From the statements (1) and (2)
2

Proved

Note: A diameter divides a circle into two halves and each half is called a semi-circle.
Angle in the same-circle is always a right angle (90q).

Theorem 9

Angles on the same segment of a circle are equal.
Or

The angles in the circumference of a circle (inscribed angle) standing on the same arc
are equal.

Experimental verification

Step 1: Three circles with centre O and different radii are drawn.

Step 2: Inscribed angles ABC and ADC are drawn in the same segment ABDC and on the

same arc AC. A C
A
BD B
O O
O C
A CD B
(ii) D
(i) (iii)

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Geometry – Circle
Step 3: ‘ABC and ‘ADC are measured and the results are tabulated.

Figure ‘ABC ‘ADC Result
(i) ‘ABC = ‘ADC
(ii) ‘ABC = ‘ADC
(iii) ‘ABC = ‘ADC

Conclusion: Angles in the same segment of a circle are equal. BD
Theoretical proof

Given: O is the centre of a circle. ‘ABC and ‘ADC are the O
C
inscribed angles standing on the same arc AC and in the A
same segment ABDC.

To prove: ‘ABC = ‘ADC

Construction: O, A and O, C are joined. So, ‘AOC is the angle at the centre of the circle.
Proof

Statements Reasons

1. ‘ABC = 1 ‘AOC 1. Inscribed angle is half of the angle at the centre
2 of a circle standing on the same arc.

2. ‘ADC = 1 ‘AOC 2. Same reason as above
2

3. ‘ABC = ‘ADC 3. From statements (1) and (2)

Theorem 10 Proved

The opposite angles of a cyclic quadrilateral are supplementary.

Experimental verification

Step 1: Three circles with centre O and different radii are drawn. A
Step 2: Cyclic quadrilaterals ABCD are drawn in each circle.
B
CC

DB O D O
O B D
C
A A
(i) (ii) (iii)

Vedanta Excel in Mathematics - Book 10 216 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Geometry – Circle

Step 3: ‘A, ‘B, ‘C and ‘D of each cyclic quadrilateral are measured and the results are
tabulated.

Figure ‘A ‘C ‘A + ‘C ‘B ‘D ‘B + ‘D Result

(i) ‘A + ‘C = 180q,

‘B + ‘D = 180q

(ii) ‘A + ‘C = 180q,

‘B + ‘D = 180q

(iii) ‘A + ‘C = 180q,

‘B + ‘D = 180q

Conclusion: The opposite angles of a cyclic quadrilateral are supplementary.

Theoretical proof D

Given: O is the centre of a circle. ABCD is the cyclic quadrilateral.

To prove: ‘ABC + ‘ADC = 180q and ‘BAD + ‘BCD = 180q OC
Construction: O, A and O, C are joined.
A
Proof B

Statements Reasons
1. Inscribed angle is half of
1. ‘ABC = 1 Ref ‘AOC
2 the angle at the centre of

a circle standing on the

same arc.

2. ‘ADC = 1 obt. ‘AOC 2. Same reasons as above
2

3. ‘ABC + ‘ADC = 1 (Ref. ‘AOC + obt. ‘AOC) 3. Adding the statements (1)
2 and (2)

4. ‘ABC + ‘ADC = 1 u 360q 4. The sum of reflex and
2 obtuse angles AOC forms a
i.e. ‘ABC + ‘ADC = 180q complete turn.

5. Similarly, ‘BAD + ‘BCD = 180q 5. By joining O, B and O, D
and following the same

facts and reasons.

Worked-out examples Proved

Example 1: Find the sizes of unknown angles in the following figures:

a) b) B c)

O x z° 110°
yR O
P 110° 30° O 25°
AC
Q

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Geometry – Circle

Solution:

a) Ref. ‘POR = 2 ‘PQR [Angle at the centre of a circle is twice the inscribed

= 2 u 110q = 220q angle standing on the same arc]

Now, y = 360q – 220q = 140q. [Sum of yq and Ref. ‘POR is an angle of a complete turn]

B

b) Let’s join B, O.

(i) ‘ABO = ‘OAB = 30q [OA = OB and base angle of 'OAB] 30° O 25°
(ii) ‘CBO = ‘OCB = 25q [OC = OB and base angle of 'OBC] AC
(iii) ‘ABO + ‘CBO = 30q + 25q = 55q

or, ‘ABC = 55q

(iv) Obt. ‘AOC = 2 ‘ABC [Angle at the centre of a circle is twice the inscribed

= 2 u 55q = 110q angle, standing on the same arc]

(v) Ref. ‘AOC = x = 360q – 110q = 250q

c) x + 110q = 180q [sum of a linear pair] z°
O
or, x = 180q – 110q = 70q x 110°

Now, z + 70q = 180q [Opposite angles of cyclic

quadrilateral are supplementary.]

or, z = 180q – 70q = 110q

Example 2: Find the sizes of unknown angles in the following figures:

a) A D b) B c)

40° y A O D
O x y
O
55° x x20° C
B AC
C 30°
BE
Solution:

a) ‘BDC = ‘BAC = 40q [Inscribed angles standing on the same arc]
[Sum of the interior angles of 'BCD]
Now,‘BDC + ‘BCD + ‘CBD = 180q
[OA = OC and base angles of isosceles 'OAC]
or, 40q + x + 55q = 180q [Sum of the interior angles of 'OAC]

or, x = 180q – 95q = 85q [Inscribed angle is half of the angle at the
centre of a circle standing on the same arc]
b) ‘OAC = ‘OCA = 20q

Now, x + 20q + 20q = 180q

or, x = 180q – 40q = 140q
1
Again, y = 2 u 140q= 70q

c) y = ‘CBE = 30q [Exterior angle of a cyclic quadrilateral ABCD
is equal to the opposite interior angle]
Now, x = 2y [Inscribed angle and angle at the centre of a
= 2 u 30q = 60q. circle standing on the same base]

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Geometry – Circle

Example 3: In the given figure, two chords PQ and RS R
XQ
intersect at right angle at the point X. Prove that P
arc (PS – QS) = arc (PR – QR) S
R
Solution: XQ

Given: Chords PQ and RS intersect at right angle at the point X.

To prove: arc (PS – QS) = arc (PR – QR) P

Construction: P, S are joined.

Proof S

Statements Reasons

1. ‘SPQ 1 QS 1. Relation between inscribed angle and its
2 opposite arc.

2. ‘PSR 1 PR 2. Same reason as above
2
3. 3. Adding statements (1) and (2)
‘SPQ + ‘PSR 1 (QS + PR )
2 4. 'PXS is a right angled triangle and ‘X is a
4. right angle and the sum of two acute angles
90q 1 (QS + PR ) of a right angled '
2

i.e. QS + PR 180q

5. PS + QR 180q 5. PS + SQ + QR + RP 360q and from
statement (4)

6. PS + QR = QS + PR 6. From statements (4) and (5)

i.e. arc (PS – QS) = arc (PR – QR)

Proved

Example 4: In the adjoining circle, AD // BC. Prove that ‘AYC = ‘BXD. Y X
Solution:

Given: In the given circle, AD // BC. AD
BC
To prove: ‘AYC = ‘BXD

Proof

Statements Reasons
1. AB = CD 1. AD // BC and they subtend equal arcs.
2. AB + BC = BC + CD
3. ABC = BCD 2. Adding BC to both sides of statement 1.
4. ‘AYC= ‘BXD 3. Whole part axiom

4. Circumference angles of the circle
standing on the equal arcs.

Proved

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Example 5: In the given figure, two circles intersect at A and X B
B. Through A two straight lines PAQ and XAY
are drawn terminated by the circumference. Q
Prove that ‘PBX = ‘QBY AY

Solution: P
Given:
A and B are the points of intersection of two circles.
Two straight lines PAQ and XAY are passing through A.

To prove: ‘PBX = ‘QBY

Proof

Statements Reasons
1. ‘PBX = ‘PAX 1. Circumference angles standing on the same XP
2. Vertically opposite angles
2. ‘PAX = ‘QAY 3. Circumference angles standing on the same QY
3. ‘QAY = ‘QBY 4. From statements (1), (2) and (3.)

4. ‘PBX= ‘QBY

Proved

Example 6: In the given figure, if MN = IU, prove that U

(i) MU = IN N
(ii) MI // UN

Solution: M
Given: I

MN and IU are two chords of a circle and MN = IU.

To prove: (i) MU = IN and (ii) MI // UN

Proof

Statements Reasons
1 UMI = NIM 1. Arcs made by two equal chords MN and IU.

2. UM + MI = NI + MI 2. Whole parts axiom

3. UM = NI 3. MI is cancelled from both sides of statement (2).
4. From statement (3), corresponding arcs are equal.
4. MU = IN

5. ‘MNU = ‘IMN 5. The inscribed angled standing on the equal arcs
6. MI // UN (UM = NI )

6. From statement (5), being the alternate angles equal.

Proved

Example 7: In the given figure, O is the centre of the circle. If two O
chords AB and CD intersect at a point P,
AP D
prove that 2‘APC = ‘AOC + ‘BOD C B

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Solution:

Given: O is the centre of a circle. Chords AB and CD intersect at the point P.

To prove: 2‘APC = ‘AOC + ‘BOD

Construction: B and C are joined.

Proof

Statements Reasons
1. Relation between inscribed angle and
1 ‘ABC = 1 ‘AOC
2 central angle standing on the same arc.
2. Same reason as in (1)
2. ‘BCD = 1 ‘BOD
2 3. Adding statements (1) and (2)

3. ‘ABC + ‘BCD =12 (‘AOC + ‘BOD) 4. In 'BCP, ‘APC is the exterior angle.

4. ‘APC = 1 (‘AOC + ‘BOD)
2
or, 2 ‘APC = ‘AOC + ‘BOD

Proved

Example 8: In the given figure, PQRS is a cyclic quadrilateral P Q T
in which PQ and SR are produced to meet at T. If S R
QT = RT, prove that (i) PS // QR (ii) PR = QS.

Solution: PQRS is a cyclic quadrilateral. PQ and SR are
Given:
produced to meet at T and QT = RT.
To prove:
(i) PS // QR (ii) PR = QS

Proof

Statements 1. Reasons
1. ‘TQR = ‘TRQ 2. QT = RT and base angles of isosceles
'TQR
2. ‘TQR = ‘PSR The exterior angle of a cyclic
quadrilateral is equal to the opposite
3. ‘TRQ = ‘SPQ 3. interior angle.
Same reason as above
4. ‘PSR = ‘TRQ and ‘SPQ = ‘TQR 4. From statements (1), (2) and (3)
From statement (4), corresponding
5. PS // QR 5. angles are equal

6. In 's PRT and SQT 6. Inscribed angles standing on the same
(i) ‘TPR = ‘TSQ (i) arc
(ii) ‘PTR = ‘STQ Common angle
(iii) RT = QT (ii) Given
(iv) ? 'PRT # 'SQT (iii) A.S.A. axiom
(iv)
7. PR = QS 7. Corresponding sides of congruent
triangles

Proved

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Geometry – Circle

Example 9: WXYZ is a cycle quadrilateral. If the bisectors of ‘XWZ and ‘XYZ meet

the circle at the points P and Q respectively, prove that PQ is the diameter

of the circle. Q

Solution: WZ

Given: WXYZ is a cyclic quadrilateral.

WP is the bisector of ‘XWZ and YQ is the bisector of ‘XYZ. X PY
To prove: PQ is the diameter of is the diameter of the circle.
Construction: W, Q and Z, Q are joined.
Proof:

Statements Reasons

1. ‘XWZ + ‘XYZ = 180° 1. Sum of opposite angles of cyclic
quadrilateral
or, 1 ‘XWZ + 1 ‘XYZ = 90°
2 2
‘PWZ + ‘QYZ = 90°

2. ‘PWZ + ‘QWZ = 90° 2. ‘QYZ = ‘QWZ , angles on the same
segment

3. ‘PWQ = 90° 3. From statement 2, whole part axiom

4. ‘PWQ is in a semicircle 4. From statement 3

5. PQ is a diameter 5. From statement 4

Proved

EXERCISE 15.1

General section

1. Find the size of unknown angles in the following figures:

a) D b) C c) d)
110°
B y A B A O QO
x O x 140° x
O 28°
150° C x B R
P

AC

e) f) P g) P h) A

B OA x 40° Ox40°30°
35° O O BC
B R
x 30° x
C A Q

i) A j) D k) l) A

x A D O x°
B O 60° y°
x O 120° 39° O C B
DC CB x° 38°
y°

AB C

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Geometry – Circle

m) P n) A o) R p) A D

x 48° E C S Ox y Ox
35°
O 60y° x y 65°
x 80° B Q CB
QR D P

2. Find the sizes of unknown angles in the following figures:

a) Q R b) B c) d)
x
115° x C A O O
O 32° C
P S D 40° O A C
x x
AB
B

3. Find the sizes of unknown angles in the following figures:

a) B b) A c) B d) A D

x

A 79° yC 40°O 30° 72°
x xC BE

x B A D C
D 120° P
D g)
f) P C h) D
e) A S

96° x O
xO
D B 140°
E
S G
Rx T xF
C QR Q FD H

i) aR j) A k) l) M

S O A x 60° L

44° 98° x x y
b QT BD K 85°N
C 72° E J
P C 55°
E
B

R

Creative section

4. S 65° T
O
a) In the figure alongside, O is the centre of the circle and 40° 35°

PQRS is the cyclic quadrilateral. Find the sizes of ‘PRQ Q
and ‘PRS.

P

A O

b) In the given figure, O is the centre of the circle. AB and DC 40° B
C
are two parallel chords. If ‘BAD = 40°, find the measures of: D

(i) ‘BCD (ii) ‘BOD (iii) ‘ADC

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Geometry – Circle B

c) In the adjoining figure, AB is the diameter of the circle CO
with centre O. If ‘ COD = 50°, find the size of ‘ CED. 50°

EDA
Q
T

d) In the figure given alongside, O is the centre of the circle R 70° O
and PQ is a diameter. Chord ST is parallel to the diameter. If
‘QRS = 70°, find the measure of ‘QST. S
C
P
D

e) In the given figure, AB = AC, ‘ADE = 35°, 35° A
and ‘BED = 70°, Prove that ADBE is a cyclic
quadrilateral. 70°
E
B

5. a) In the adjoining figure, O is the centre of the circle and PQ is the R Q
diameter. Show that ‘PRQ is a right angle. O

P

b) In the given figure, O is the centre of the circle. Prove that Q
‘POR = 2 ( ‘PRQ + ‘QPR). R

P

O

A

C B
O
c) In the given figure, O is the centre of the circle. If AC = AB and BC
is an angle bisector of ‘ABO, prove that ' ABO is an equilateral
triangle.

DC

d) In the given figure, O is the centre of the circle, AB the diameter OB

A

and BC = CD. Prove that AD // OC.

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Geometry – Circle

DC

e) In the adjoining figure, AB is the diameter. OC and OD are the A OB
radii where ‘ BOC = ‘ COD. Prove that AD // OC.

M

f) In the figure alongside, chords MN and RS of a circle N X
intersect externally at X. S G
D
Prove that ‘MXR 1 (MR – NS ) R
2

g) In the given circle, chords DE and FG intersect within the circle F X
1 E
at X. Prove that ‘DXF 2 (DF + GE ).

M N
B
h) In the given figure, O is the centre of a circle. C and D are two A O
points on the chord AB. If AC = BD and AM = BN , prove that CD
‘ACM = ‘BDN.

A

i) In the given figure, two chords AB and CD intersect at right

angle at X. Prove that, AD – CA = BD – BC . CX D
E
B

D

j) In the adjoining figure, O is the centre of a circle in which C
1 OB
' OCE is an isosceles triangle. Prove that BC = 3 AD A

YX

k) In the figure alongside, AD // BC. Prove that ‘AYC = ‘BXD.

A D
B C
D
C
EB
l) In the figure, X is the centre of a circle. AB is a diameter of the
circle. If DE A AB and AC = BD , prove that ‘ABC = ‘BDE. A

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Geometry – Circle

m) In the given figure, O is the centre of the circle, AB the diameter CD
O
and DO A AB. Prove that ‘ AEC = ‘ ODA.

A B

E

n) In the adjoining figure, O is the centre of the circle. If X M A
AMO A XOZ, prove that ‘OAZ = ‘XYO. O Y
Z

P

A Q
C
o) In the given figure, ‘ APC = ‘ BQD. Prove that AB // CD.

B
D

p) In the adjoining figure, POQ and ROT are two diameters of a O T
circle with centre at O. If Q is the mid-point of the arc TQS P Q
and ‘ QOR is obtuse, prove that PQ // RS.
S
R

D C

6. a) In the given figure, ABCD is a cyclic quadrilateral. Prove that BE
A
‘CBE = ‘ADC. A

b) ' ABC is an isosceles triangle and XY // BC. If XY cuts AB X Y
at X and AC at Y, prove that the four points X, B, C and Y
C
are concyclic.
Q
B T

P R
D
c) In the given figure, QR is the bisector of ‘ SQT and PQRS
E
is a cyclic quadrilateral. Prove that ' PSR is an isosceles C

triangle. S

A

d) In the adjoining figure, AC = BC and ABCD is a cyclic

quadrilateral. Prove that DC bisects ‘ BDE. B

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Geometry – Circle

e) In the adjoining figure, TPQ is an isosceles triangle in P R
which TP = TQ. A circle passing through P and Q cuts Q
TP and TQ at R and S respectively. Prove that PQ // RS. T
S

ED C

f) ABCD is a parallelogram. The circle through A, B, and C A B
intersects CD produced at E. Prove that AE = AD.

B

g) In the adjoining figure, ABC is a triangle in which C D
AB = AC. A circle passing through B and C intersects the A A
sides AB and AC at the points D and E respectively. Prove
that AD = AE. E

B

h) In the given figure, ABCD is a quadrilateral in which AB // DC DC
and ‘ ADC = ‘ BCD. Prove that the points A, B, C, and D are R
concyclic.

i) In the adjoining figure, PQRS is a cyclic quadrilateral. If the Q
diagonals PR and QS are equal, prove that, QR = PS and S P
MQ
PQ // SR.
NR
P

j) In the figure, PQRS is a parallelogram. The inscribed circle

cuts PQ at M and Rs at N. Prove that ‘ MNS = ‘ PQR. S

k) In the given triangle ABC, AD A BC and CE A AB. A
Prove that ‘BDE = ‘BAC. E

BD C
P Q

l) If PC is the bisector of ‘ APB, prove that XY // AB. XY

A B
C
P
A D

m) In the given figure, ABCD is a cyclic quadrilateral. C
A side CD is produced to the point P. If AB = AC, B
prove that AD is a bisector of ‘BDP.

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Geometry – Circle

n) The bisectors of opposite angles ³A and ³C of a cyclic quadrilateral ABCD
intersect the corresponding circle at the points X and Y respectively. Prove that
XY is the diameter of the circle.
S

7. a) In the given figure, PQRS is a parallelogram. PR
Prove that UTRS is a cyclic-quadrilateral. T

b) In the given figure, AB and EF are parallel to each other. Prove A U Q
that CDEF is a cyclic quadrilateral. B D E

C F
S

c) In the given figure, NPS, MAN and RMS are straight lines. M
Prove that PQRS is a cyclic quadrilateral.
PA

N R
Q

C
Q

8. a) In the given figure, AP is the radius of circle ABC and diameter A P
of circle APQ. Prove that AQ = QC.

B

B

A

b) In the adjoining figure, X and Y are the centres of the circles X Y
which intersect at A and C. XA and XC are produced to meet C D
the other circle at B and D. Prove that AB = CD.

P AQ

c) In the adjoining figure, two circles intersect at A and B.

The straight line PAQ meets the circles at P and Q and the

straight line RBS meets the circles at R and S. Prove that R BS
S
PR // QS. M Q
N
P

d) In the given figure, two circles are intersecting at M R
and N. PQ and RS pass through M. If R, P, S, and Q

are joined to N, prove that ‘ PNR = ‘ SNQ

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Geometry – Circle

9. a) prove that the angle at the centre of a circle is twice the angle at the circumference
standing on the same arc.

b) The points S, T and U lie on the circumference of a circle with centre M. Prove that,
1
‘ TSU = 2 ‘ TMU.

c) Prove that the angles at the circumference of a circle standing on the same arc are
equal.

d) In a cyclic quadrilateral PQRS, prove that ‘P + ‘R = ‘Q + ‘S = 180°.

e) If ABCD is a cyclic parallelogram, prove that ABCD is a rectangle.

f) If one side of a cyclic quadrilateral is produced, prove that the exterior angle is
equal to the opposite interior angle of the quadrilateral.

g) Prove that a cyclic trapezium is always isosceles and its diagonals are equal.

h) Prove that the quadrilateral formed by angle bisectors of a cyclic quadrilateral is
also cyclic.

i) Prove that the sum of the angles in the four segments exterior to a cyclic quadrilateral
is equal to six right angles.

15.3 Tangent and Secant RSecant

A tangent of a circle is the line that meets the circle only at Q
one point on the circumference of the circle. In the given T Tangent
figure, PQ is the tangent of the circle at the point T. Here, the
point T at which the tangent meets the circle is called the
point of contact.

On the other hand, a secant of a circle is the line that intersect P
the circle at any two points. In the figure PR is the secant.

15.4 Angles in alternate segments A

In the figure, PQ is a tangent to a circle at the point of contact T. C OB
A chord TA is drawn from the point of contact T. Here, the P TQ
chord TA divides the circle into two segments, TBA and TCA.
These two segments are called alternative segments.

Here, for ‘ATP, the alternate segment is TBA.
For ‘ATQ, the alternate segment is TCA.

So, ‘ATP and ‘TBA are said to be the alternate segment angles, and, ‘ATQ and
‘TCA are also said to be the alternate segment angles.

The alternate segment angles are always equal.

i.e., ‘ATP = ‘TBA and ‘ATQ = ‘TCA.

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Geometry – Circle

Theorem – 11

A tangent to a circle is perpendicular to the radius of the circle drawn at the point of
contact.

Given: O is the centre of a circle PQ is a tangent to the circle at T.

OT is the radius of the circle passes through T. O

To prove: OT A PQ. S
T RQ
Construction: Any point R is taken on PQ and O, R are joined. P
Proof

Statements Reasons

1. OT = OS 1. Radii of the same circle

2. OS < OR 2. OS is a part of OR.

3. OT < OR 3. From statements (1) and (2)

4. OT is the shortest length of all the 4. It is true for every line joining O to PQ.
lines drawn from O to PQ

5. OT A PQ 5. Perpendicular is the shortest line
segment drawn from a point to a line.

Proved

Theorem – 12

The length of two tangents to a circle at the point of contact from the same external point

are equal. T2 R

Given: O is the centre of a circle PQ and PR are two

tangents from the same point P to the circle at T1 P
and T2.

To prove: PT1 = PT2 T1 Q

Construction: O, P; O, T1 and O, T2 are joined.

Proof

Statements Reasons

1. In rt. ‘ed 's OPT1 and OPT2 1.

(i) ‘OT1P = ‘OT2P (R) (i) Both of them are right angle

(ii) OP = OP (H) (ii) Common side

(iii) OT1 = OT2 (S) (iii) Radii of the same circle

(iv) ? 'OPT1 # 'OPT2 (iv) R.H.S. axiom
2. PT1 = PT2 2. Corresponding sides of the congruent

triangles

Proved

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Geometry – Circle

Theorem – 13

The angles formed by a tangent to a circle and a chord drawn from the point of contact are

equal to respective angles in the alternate segments. R
A
Given: O is the centre of a circle. PQ is a tangent to the circle

at T. TA is a chord. ‘ATP and ‘TBA, ‘ATQ and C O
‘TCA are two pairs of alternate segment angles. B

To prove: ‘ATP = ‘TBA and ‘ATQ = ‘TCA PT Q

Construction: O, T are joined and TO is produced to R, a point on the circumference. Hence,
TR is the diameter of the circle. R, A are joined.

Proof

Statements Reasons

1. ‘RAT = 90q 1. Angle in the semi-circle

2. ‘RTA + ‘TRA = 90q 2. Sum of two acute angles of rt. ‘ed. 'RAT
3. ‘RTA + ‘ATQ = 90q 3. RT A PQ

4. ‘RTA + ‘TRA = ‘RTA + ‘ATQ 4. From statements (2) and (3)

i.e. ‘TRA = ‘ATQ 5. ‘TCA and ‘TRA are standing on the same
5. ‘TCA = ‘ATQ

6. Similarly, ‘ATP = ‘TBA arc ABT.
6. Same facts and reasons as above

Worked-out examples Proved

M B
O
Example 1: In the given figure, O is the centre of the circle. MN is A
Solution: a tangent where A is the point of contact. If AB is the N
bisector of ‘OAM, find the size of ‘OBA.

Here, ‘OAM = 90q [OA A MN]

? ‘OAB = 1 ‘OAM = 1 u 90q = 45q
2 2

'OBA is an isosceles triangle [OA = OB, radii of the same circle]

? ‘OBA = ‘OAB = 45q [Base angles of isosceles triangle]

Example 2: In the given figure, PT is a tangent to a circle whose O
Solution: centre is O. If OP = 17 cm and OT = 8 cm, find the length T
of the tangent segment PT.

P

Here, OPT is a right angled triangle. [OT A PT]

? PT = OP2 – OT2 = 172 – 82 = 289 – 64 = 225 = 15 cm

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Geometry – Circle

Example 3: Find the sizes of unknown angles in the following figures:
Solution:
a) b) R c) A
O 65° S
O Q y
Px Bx

50° 60° yQ 110°
Px T
PT Q

a) In 'OPT, ‘OPT = x = 90q – ‘POT [One of the acute angles of rt. ‘ed 'OPT]
= 90q – 50q = 40q [One of the acute angles of rt. ‘ed 'OQT]

In 'OQT, ‘OQT = 90q – ‘QOT

= 90q – 60q = 30q

b) ‘QOR = 2 ‘QSR [Angle at the centre of a circle is twice the inscribed

= 2 u 65q = 130q angle standing on the same arc]

In quad. PQOR, ‘PQO + ‘QOR + ‘ORP + ‘RPQ = 360q

or, 90q + 130q + 90q + x = 360q

or, x = 360q – 310q = 50q

c) ‘ATQ = 180q – ‘ATP [The sum of ‘ATQ and ‘ATP is a straight angle]

= 180q – 110q = 70q

‘TBA = ‘ATQ = 70q [Angles in the alternate segments]

Also, ‘BTA = ‘TBA = 70q [Base angles of the isosceles 'TBA]

Again, ‘TBA + ‘BTA + ‘TAB = 180q [Sum of the angles of 'TAB]

or, 70q + 70q + y = 180q

or, y = 180q – 140q = 40q D
E

Example 4: In the given figure, O is the centre of a

circle. If AD, AC and EB are tangents, OF A

prove that AD + AC = AB + BE + AE. B
C
Solution:

Given: O is the centre of a circle.

AD, AC and EB are the tangents to the circle at D, C, and F respectively.

To prove: AD + AC = AB + BE + AE

Proof

Statements Reasons

1. BC = BF 1. Two tangents drawn from a common external

point to the points of contact are equal.

2. ED = EF 2. Same reason as above

3. BC + ED = BF + EF 3. Adding statements (1) and (2)

4. BC + ED = BE 4. BF and EF are the parts of the same straight

line BE.

5. AD + AC = AB + BC + AE + ED 5. AB and BC are the parts of AC, AE and ED

are the parts of AD

6. AD + AC = AB + BE + AE 6. From statements (4) and (5)

Proved

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Geometry – Circle

EXERCISE 15.2

General section

1. Find the lengths asked in each of the following figures:

a) b) c) d)

O O O B 3cm O
M8cm P 4cm A
3 cm R 5 cm
P 4 cm A Q 12cm P B 16cm A Find PB.

Find OP. Find QR. Find the radius.

e) In the given figure, O is the centre of the circle. PQ is R P
a tangent and P the point of contact. RS is produced to 15 cm
the point Q. If PQ = 15 cm and SQ = 9 cm, find the
value of OS. O S 9cm Q

f) In the given figure, O is the centre of the circle and PQ the R O
P
tangent where P is the point of contact. If OQ = 25 cm and 18cm
QR = 18 cm, find the length of PQ. Q

2. Find the sizes of unknown angles in the following figures:

a) b) T 40° P 35° A c)

O x O xB
O
Tx 50° 60° y 55°
T A
A N N

d) A B Q f) E
C
P 30° O e)
x x 60°
x 120° A x
BQ O RB B D

AP

g) T h) i) S

x D z C T O 160° x
y
O 110° Q R
x P
2 30° 50° x
AB C AT
B

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Geometry – Circle

j) k) A l) A

O 30° O xE
xE O y 32°
40°
E F B CD
ABC yx
B C FD

m) n) P C o) T

O Q x
T
S 57° A 32° B Q xS
x O P 30°

PR QR

p) C q) B r)
E x
C x
D 30°
23° x
A B y 84°
T AN

s) A O B t) S R
26° N B
D P 110°
T x x

C AQ

3. Find the sizes of known angles in the following figures:

a) b) Z Q c) D B

O

Cx yB O 150° x X A 35° x O

50° P E
Y C
A
e) DB f) A
d) O 60° F
Ax N
P

40°

O 30° xT

S T EC M B
O R
C
Q x° 30°
M
OD
Creative section

4. a) In the given figure BD is the diameter of the circle B A
with centre O. AC is a tangent to the circle at C.
If ‘ACD = 30°, show that ' ACD is an isosceles triangle.

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Geometry – Circle Geometry – Circle
b) In the given figure, BD is a secant and CD is a tangent.
Prove that ' ACD and ' BCD are similar. CD
A

B
A

c) In the given figure, A and P are contact points of T O
tangents. Prove that TA = TP. P

d) In the adjoining figure, PQ is tangent to the circle at A. C is ON B
the middle point of arc AB. CM A PQ and CN A AB. Prove C
that CM = CN.
A MQ
P

P

e) In the given figure, O is the centre of the circle, TS tangent O R
and Q is the point of contact. If PR // OS, prove that Q S

' PQR a ' OSQ.

T

Objective Questions

Tick the correct alternatives.

1. Which of the following statement is true?

(A) The median of a triangle bisects it. (B) The diagonal of parallelogram bisects it.

(C) The diameter of circle bisects it. (D) All of the above

2. Which of the following statement is NOT true?

(A) Parallelograms standing on the same base and between the same parallels are equal

in area.
(B) Triangles standing on the same base and between the same parallels are equal in

area.
(C) The area of triangle and the parallelogram standing on the same base and between

the same parallels are equal.

(D) The rectangle and rhombus standing on the same base and between the same

parallels are equal in area.

3. A farmer has a field ABCD in the form of parallelogram. If he joins the corners A and B to any

point P on the side CD by thread, what is the ratio of areas of ∆ABP to that of his field?

(A) 1:2 (B) 2:1 (C) 1:4 (D) 4:1 DC

4. In the given figure, ABCD is a parallelogram. Which one of the E

triangle is equal in area to ∆BCD?

(A) ∆ABD (B) ∆BED (C) ∆ABE (D) both (A) and (C) AB

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5. In the given diagram, PQRS is a square. If PQ//ST and SQ = 3 2 cm S RT
then the area of ∆PQT is P Q

(A 9 cm2 (B) 12 cm2 (C) 18 cm2 (D) 36 cm2

6. In the given diagram, GOAT is a parallelogram in which AE = ET. What G T

percent of area of parallelogram GOAT is occupied by the trapezium E

GOAE? O A
C
(A) 20% (B) 25% (C) 75% (D) 80%
O
7. In the given figure, O is the centre of the circle. If OA = AB, what is the

measure of ‘ACB?

(A) 600 (B) 900 (C) 300 (D) 450 AB

8. Smarika wishes to draw two angles at the circumference having equal measurements.

Which of the following mathematical facts should she know?

(A) The angles at the circumference standing on the same arc of a circle are equal.

(B) The angles at the circumference subtended by the equal arcs of a circle are equal.

(C) The angles in the circumference standing on the same segment of a circle are equal.

(D) All of the above. R
9. In the figure, O is the centre of the circle. If PR = QR, what is the measure

of ‘QSR? P OQ
S
(A) 300 (B) 450 (C) 600 (D) 900

10. The opposite angles of a cyclic quadrilateral are

(A) Supplementary (B) Complementary (C) Equal (D)Right angles

11. Under which condition, any quadrilateral can NOT be a cyclic quadrilateral? A
(A) If the opposite angles are supplementary.
(B) If the opposite angles are complementary.
(C) If the exterior angle is equal to the opposite interior angle.
(D) If a pair of angles subtended by any side at opposite vertices are equal.

12. In the semi-circle given alongside, O is the centre. If OC = 5 cm and C O

BC = 6 cm, what is the length of AC?

(A) 5 cm (B) 6 cm (C) 10 cm (D) 8 cm B

C

13. In the given figure, arc AD = arc AC and ‘CAD = 1400, what is the

measure of ‘ABC? A 140° B
T
(A)1400 (B) 700 (C) 400 (D) 200 D
S
14. In the given figure, NT // MS then which pair of angles are equal? N
M P
(A) ‘AMS and ‘ANT (B) ‘ASM and ‘ATN

(C) ‘MAN and ‘SAT (D) ‘ANT and ‘ATN A

15. In the given figure; O is the centre of circle, PQ the tangent and

A the point of contract. If OR = 3cm and QR = 2cm, the O
perimeter of ∆POQ is: R
(A) 5 cm (B) 4 cm (C) 6 cm (D) 12 cm Q

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Unit Trigonometry

16

16.1 Trigonometric ratios of an acute angle of a right-angled

triangle - review C

In the given right-angled triangle ABC, right angled at hypotenuse (h) perpendicular (p)
B. ‘BAC = T is an acute angled and taken as angle of
reference.
Here, side AC is opposite to the right angled and it is
hypotenuse (h).

Side BC is opposite to the angle of reference (T) and it is T B

perpendicular (p). A base (b)

Side BC is adjacent to the angle of reference (T) and it is base (b).

Now, we can make six different ratios of the various pairs of the sides of a right-
angled triangle. These ratios are given below:

(i) perpendicular = p which is called sine of T or sinT .
hypotenuse h

(ii) base = b which is called cosine of T or cosT.
hypotenuse h

(iii) perpendicular = p Which is called tangent of T or tanT.
base b

(iv) hypotenuse = h Which is called cosecant of T or cosecT
perpendicular p

(v) hypotenuse = h which is called secant of T or secT.
base b

(vi) base = b which is called cotangent of T or cotT
perpendicular p

Here, sinT, cosT, and tanT are the main ratios and cosecT, secT, and cotT are inverse
ratios of the main ratios respectively.

16.2 Values of trigonometric ratios of some standard angles

The angles like 0°, 30°, 45°, 60°, and 90° are commonly known as the standard angles.
The values of trigonometric ratios of these standard angles can also be obtained
geometrically without using the table. The table given below shows the value of
trigonometric ratios of these standard angles

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Trigonometry

Trigonometric ratios Angles
sin
cos 0° 30° 45° 60° 90°
tan 0 1
1 1 13 0
0 2 22 f

3 1 1
2 2 2

1 1 3
3

16.3 Height and distance A
B
One of the important applications of trigonometry

is to find the heights of objects like, tree, building, Line of sight Height
tower, etc. and distance between objects. In such
cases, we have to solve right angled triangles.

For example, in the figure, AB is the height of a

tree, BC is the distance between the tree and the C T

observer at C. Distance

Here, if T and BC are known, we can calculate the height AB or if T and AB are
known, we can calculate the distance between the tree and the observer by using a
trigonometric ratio.

16.4 Angle of elevation and angle of depression Angle of eleLivnateioofnsight

In the adjoining figure, an observer is looking at the top Horizontal
of a tower. Here, OA is called the line of sight and OB
is the horizontal line. ‘BOA is the angle between the
upward line of sight and the horizontal line. ‘BOA is
called the angle of elevation.

Thus, the angle made by upward line of sight with
horizontal line is known as angle of elevation.

In this figure, ‘BOA is the angle between the OB
downward line of sight and the horizontal line. Angle of depression
‘BOA is called the angle of depression.

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Trigonometry

Thus, the angle made by downward line of sight with horizontal line is known as

angle of depression. C A
In the adjoining figure, the horizontal lines AC and BO are Angle of depression

parallel to each other. The angle of depression CAO and the

angle of elevation BOA are alternate angles between parallel Angle of elevation B

lines. O

? ‘ CAO = ‘ BOA

i.e., Angle of depression= Angle of elevation.

The instrument which is used to measure the angle of elevation or the angle of
depression is called Theodolite.

Worked-out examples

Example 1: A man observes the top of a tree 30 3 m high and finds the elevation of
30°. Find the distance between the man and the foot of the tree.

Solutions:

Let AB be the height of the tree and BC be the distance A
between the foot of the tree and the man. 30 3 m
Here, AB = 30 3 m B

Angle of elevation, ‘ BCA = 30°. C 30°

In rt. ‘ ed ' ABC, tan 30° = p = AB = 30 3
b BC BC
1 30 3
or, 3 = BC

or, BC = 30 3 × 3 = 30 × 3 = 90 m

Hence, the required distance is 90 m.

Example 2: An electric pole 15 m high is supported by a wire fixing its one end on the

ground at some distance from the pole. If the wire joining the top of the pole

is inclined to the ground at an angle of 60q, find the length of the wire.

Solution:

Let AB be the height of the pole and AC be the length of wire. A

Here, AB = 15 m.

Angle of elevation BCA = 60q 15 m

In rt. ‘ed ' ABC, B 60° C

sin60q = p = AB
b AC

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Trigonometry

or, 3 = 15
2 AC

or, AC 3 = 30 m

AC = 30 × 3 = 10 3= 10 × 1.732 = 17.32 m
3 3

Hence, the required length of the wire is 17.32 m.

Example 3: The diameter of a circular pond is 90 m and a pole is fixed at the centre of
Solutions: the pond. The height of the pole is 48 m and the pond is 3 m deep. Find the
angle of elevation of the top of the pole from a point in the circumference
of the pond.

Let BC be the height of the pole above the surface of water. C
AB be the radius of the circular pond.
Here, diameter of the pond = 90 m

Then, radius of the pond = 90 m = 45 m
2

Height of the pond above the surface of water, BC = 48 m – 3 m 45 m

= 45 m A T B
45 m 3m

In right-angled triangle ABC,

tan T = p = BC = 45 =1
b AB 45

or, tan T = tan45°

? T = 45°
Hence, the required angle of elevation is 45°.

Example 4: A man 1.5 m tall standing at a distance of 50 m from a tree observes the
angle of elevation of the top of the tree to be 45°. Find the height of the tree.

Solution:

Let CE be the height of the tree and AB be the height of the man. BC be the distance between

the foot of the tree and the man. E
Here, AB = 1.5 m

BC = 50 m A 45° D
Angle of elevation DAE = 45q 1.5 m 50 m C
In rectangle ABCD,
B

DC = AB = 1.5 m and AD = BC = 50 m

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Trigonometry

In rt. ‘ed ' ADE,

tan45q = p = ED
b AD

or, 1 = ED
50

or, ED = 50 m

Now, CE = ED + DC = 50 m + 1.5 m = 51.5 m

So, the required height of the tree is 51.5 m.

Example 5: A girl, 1.2 m tall, is flying a kite. When the length of the string of the kite is

180 m, it makes an angle of 30° with the horizontal line. At what height is

the kite from the ground?

Solution:

Let CK be the height of the kite from the ground, AB be the height of the girl and AK be the

length of the string. K

Here, AB = 1.2 m 180 m
AK = 180 m

Angle of elevation DAK = 30q. A 30° D
1.2 m C
In rectangle ABCD, AB = DC = 1.2 m.
B
In rt. ‘ ed ' ADK,

sin30q = p = KD
h AK

or, 1 = KD
2 180

or, KD = 90 m

Now, CK = KD + DC = 90 m + 1.2 m = 91.2 m

So, the kite is at a height of 91.2 m from the ground.

Example 6: A woman is 1.6 m tall and the length of her shadow in the sun is 1.6 3 m.
Find the altitude of the sun.

Solution:
Let AB be the height of the woman and BC be the length of her shadow. Let T be the altitude of
the sun.

Here, AB = 1.6 m A

BC = 1.6 3 m

In rt. ‘ ed ' ABC,
p AB
tanT = b = BC 1.6 m
B
or, tanT = 1.6 T =?
1.6 3 C

1.6 3 m

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Trigonometry

or, tanT = 1
3

or, tanT = tan30q
? T = 30q
So, the required altitude of the sum is 30q.

Example 7: A boy 3 m tall is 72 m away from a tower 25 3 m high. Find the angle of
elevation of the top of the tower from his eyes.

Solutions:

Let AB be the height of the boy and CD be the height of the tower. Let BD be the distance
between the boy and the tower.

Here, BD = AE = 72 m C

AB = 3 m and CD = 25 3 25 3 m
E
? CE = CD – ED = 25 3 – AB = 25 3 – 3 = 24 3 m D

Now, in rt. ‘ ed ' AEC,

tan A = p = CE A
b AE 3m

24 3 B 72 m
72
or, tan A =

tan A = 3 = 3 3 = 1
3 3× 3

A = tan–1 1 = 30°
3

So, the required angle of elevation is 30°.

Example 8: The top of a tree which is broken, by the wind makes an angle of 60° with the
Solutions: ground at a distance 3 3 m from the foot of the tree. Find the height of the tree
before it was broken.

A

Let AC be the height of the tree before it was broken, BD be the broken part
of the tree and CD be the distance between the foot of the tree and the point
on the ground at which the top of the tree touched.

Here, CD = 3 3 m B

The angle of elevation, ‘ CDB = 60°

Now, in rt. ‘ed ' BCD, 60°
3m
tan 60° = p = BC = BC C 3 D
b CD 33

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Trigonometry

or, 3 = BC
or, 33
Also,
BC = 9 m
or,
cos 60° = b = 33
h BD

1 = 33
2 BD

or, BD = 6 3 = 6 × 1.732 = 10.39 m
Then, the height of the tree before broken = AC

= AB + BC
= BD + 9 m
= 10.39 m + 9 m
= 19.39 m
So, the required height of the tree was 19.39 m.

Example 9: Two vertical poles are fixed 60 m apart. The angle of depression of the top of
the first pole from the top of the second pole which is 150 m high is 30°. Find
the height of the first pole.

Solutions:

Let AB be the height of the first (shorter) pole, CD be the height of the second (taller) pole
and BD be the distance between them.

Here, BD = AE = 60 m C
CD = 150 m 30°

The angle of depression = angle of elevation, ‘ CAE = 30°.

Now, in rt. ‘ ed ' CEA, A 30° 150 m
E
p CE CE
tan 30° = b = AE = 60

or, 1 = CE
3 60
D
B 60 m
60
or, CE = 3

= 60 × 3 = 20 3 = 20 × 1.732 = 34.64 m
3 3

Again, in rectangle ABDE,

AB = DE = CD – CE = 150 m – 34.64 m = 115.36 m
So, the required height of the the first pole is 115.36 m

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Trigonometry

Example 10: Two men are on the opposite side of a tower of 30 m high. They observed

the angle of elevation of the top of the tower and found to be 30q and 60q

respectively. Find the distance between them.

Solution:

Let AB be the height of the tower and CD be the distance between the two men. A

Here, AB = 30 m

Angles of elevation BCA = 30q and ‘ BDA = 60q. 30 m

In rt. ‘ed ' ABC, C 30° 60°

tan30q = p = AB B D
b BC

or, 1 = 30
3 BC

or, BC = 30 3 = 30 u 1.732 = 52.05 m

Again, in rt. ‘ ed ' ABD,

tan60q = p = AB
b BD

or, 3 = 30
BD

or, BD = 30 = 30 × 3 = 30 3 = 10 3 = 17.32 m
3 3 3 3

Now, CD = BC + BD = 52.05 m + 17.32 m = 69.37 m

So, the required distance between the men is 69.37 m.

EXERCISE 16.1

General section

1. Find the unknown lengths of the sides of the following figures:

a) b) c) d)

AC P 30° E
ym

G 80 3 m F
ym

40 3 m

xm
xm
xm 150 m
45° ym
C 20 m B
A 60° B Q 30° R
xm ym

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Trigonometry

e) C f) A 30° g) R h) A

60° 121.5 m P 45° T B
xm xm

2m
xm
200 m
xm
A E E 60° C
1.5m D Q 80 m S 60°
B 60 3 m D C 10 3 m
B
d) C
D

2. Find the unknown sizes of angles in the following figures:

a) A b A c) P x S

45 m 50 m Ax
40 m
E
20 m
2m

32 m
x x B Q 25 2 m R
C
B 15 3 m C
Creative section B 10 3 m D

3. a) The angle of elevation of the top of a tree observed from a point 60 m away from its
foot is 45°. Find the height of the tree.

b) The angle of depression at a point on the ground from the top of a building is 60°. If
the point is 36 m far from the foot of the building, find the height of the building.

c) From the top of a tower of height 120 m the angle of depression of an object lying
in front of the tower is observed and found to be 60°. Find the distance between the
object and the foot of the tower.

d) A tower on the bank of a river is of 20 m high and the angle of elevation of the top
of the tower from the opposite bank is 30°. Find the breadth of the river.

e) When the sun's altitude is 45°, the length of the shadow of a vertical pole is 12.5 m.
Find the height of the pole.

f) Find the length of the shadow cost by a tree of height 15 m at a time of the sun's
attitude to be 30°.

g) A man was flying a kite. The string was 300 m long and an angle of 45° was formed
with horizon. What was the height of the kite above the horizon?

h) A ladder 7.5 m long is resting against a wall. If the ladder makes an angle of 60°
with the ground, find

(i) the height where the ladder is resting on the wall.

(ii) The distance between the feet of the ladder and the wall.

4. a) A circular pond has a pole standing vertically at its centre. The top of the pole is
30 m above the water surface and the angle of elevation of it from a point on the
circumference is 60°. Find the length of radius of the pond.

b) The diameter of a circular pond is 120 m. The angle of elevation of the top of the
pillar situated in the middle of the pond observed from the edge of the pond is
found to be 60°. Find the height of the pillar above the surface of water.

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Trigonometry

c) The circumference of a circular meadow is 220 m and a pillar is situated at the
centre of the meadow. If the angle of elevation of the top of the pillar from a point
at the circumference of the meadow is found to be 30°, find the height of the pillar.

5. a) An observer finds the angle of elevation of the top of a pole to be 60°. If the height
of the pole and the observer are 25.3 m and 1.3 m respectively, find the distance
between the observer and the pole.

b) A woman 1.6 m tall, observes the angle of elevation of the top of a house and it is
found to be 60°. If the distance of the woman from the foot of the house is 100 m,
find the height of the house.

c) A man observes the top of a pole of 52 m height situated in front of him and finds
the angle of elevation to be 30°. If the distance between the man and the pole is
86m, find the height of the man.

d) An observer observed from the top of a house of height 8 m, the angled of elevation
of the top of the temple is 60°. If the distance between the temple and the house is
15 3 m, find the height of the temple.

e) From the top of a hill 51.6 m high, the angle of depression of the top of a tree of
height 6.6 m situated in front of the hill, was observed and found to be 30°. Find
the distance between the hill and the tree.

f) The horizontal distance between two trees of different heights is 30 m. The angle
of depression of the top of the first tree when seen from the top of the second tree
is 30°. If the height of the second tree is 40 m, find the height of the first tree.

6. a) On a windy day, a girl of height 1.1 m was flying her kite. When the length of the
string of the kite was 33 metres, it makes an angle of 30° with horizon. At what
height was the kite above the ground?

b) On the roof of a house 9 m high, a 1.5 m tall man was flying a kite and the kite was
at a height of 58 m above the ground. If the string of the kite makes an angle of 30°
with the horizon, calculate the length of the string of the kite.

7. a) From the top of a building 20 m high, a 1.7 m tall man observes the elevation of the
top of a tower and finds it 45°. If the distance between the building and the tower
is 50 m, find the height of the tower.

b) A 2 m tall man standing on the top of a building 18 m high observes the elevation
of the top of a tower 60 m high and finds it 60°. Find the distance between the
building and the tower.

c) A man of height 1.8 m observes the angle of elevation of the top of a building of
height 107 m from the roof of a house and it is found to be 30°. If the distance
between the house and the building is 25 3 m, find the height of the house.

d) The heights of a house and a temple are 13 m and 25 m respectively. If a man
observes the roof of the house from the roof of the temple he finds the angle of
depression 45°. Find the distance between the house and the temple.

8. a) A man is 2m tall and the length of his shadow in the sun is 2 3 m. Find the altitude
of the sun.

b) A woman observes the top of a tower of 80 3 m height from a point on the ground
240 m far from the foot of the tower. Find the angle of elevation.

c) A man 1.75 m tall is 50 m away from a tower 51.75 m high. Find the angle of
elevation of the top of the tower from his eyes.

9. a) The upper part of a straight tree broken by the wind makes an angle of 45° with
the plane surface at a point 9 m from the foot of the tree. Find the height of the tree
before it was broken.

Vedanta Excel in Mathematics - Book 10 246 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Trigonometry

b) A tree of 14 m height is broken by the wind so that its top touches the ground and
makes an angle of 60° with the ground. Find the length of the broken part of the
tree.

c) The top of a tree broken by the wind makes an angle of 30° with the ground at a
distance of 4 3 m from the foot of the tree. Determine the height of the tree before
it was broken.

d) A tree of 15 m height is broken by the wind so that its top touches the ground and
makes an angle of 30° with the ground. Find the length of the broken part of the
tree.

10. a) Two men are on the opposite side of a tower 40 m high on the same horizontal line.
They observed the angle of elevation of the top of the tower and found to be 45° and
30°. Find the distance between them.

b) From a lighthouse 100 3 m high the angles of depression of two ships on its opposite
sides are observed to be 30° and 60° respectively. Find the distance between the
ships if the bases of the ships and the lighthouse are on the same horizontal line.

Project work
11. a) Make the groups of your friends. Find the heights of the basketball poles, roof of

your school buildings, height of trees, etc. inside your school premises.
b) An inclinometer or clinometer is an instrument used for measuring angles of slopes,

elevation and depression. Use this instrument from your math lab to measure the
height and distance of poles, trees, and so on.

16.5 Area of triangle A

(i) When the base and height of a triangle are given, its area is DC
A
calculated as 1 base u height.
2 b
1 aC
In the figure, area of 'ABC = 2 base u height

= 1 BC u AD B
2

(ii) When the lengths of three sides of a triangle are given, its

area is calculated as s (s – a) (s – b) (s – c), where s is the c
B
semi-perimeter i.e., s = a + b + c.
2

16.6 Area of a triangle when its two sides and angle contained
by them are given

In the given figure, ABC is a triangle. AD A BC is drawn. A
Here, BC = a, CA = b and AB = c

(i) In right-angled 'ABD,

sinB = AD = AD cb
AB c B DC

? AD = c sinB a

Now, the area of 'ABC = 1 BC u AD = 1 ac sinB
2 2

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 247 Vedanta Excel in Mathematics - Book 10

Trigonometry

(ii) Also. In right angled 'ACD,

sinC = AD = AD
AC b

? AD = b sinC

Now, area of 'ABC = 1 BC u AD = 1 ab sinC
2 2

(iii) Similarly, when BE A AC is drawn, A

Area of 'ABC = 1 AC u BE = 1 bc sinA
2 2

Thus, when ‘A is between b and c cE
b
1
Area of 'ABC = 2 bc sinA Ba C

When ‘B is between a and c

Area of 'ABC = 1 ac sinB
2
When ‘C is between a and b

Area of 'ABC = 1 ab sinC
2

Here, 1 bc sinA = 1 ac sinB = 1 ab sinC
2 2 2

i.e., a = b = c
sinA sinB sinC

This relation is known as sine law.

From the above relations, following conclusions can be drawn.

(i) The area of any triangle is calculated as half of the product of any two sides of the
triangle and the sine of the angle between them.

(ii) The formula is valid even if the angle between two sides is obtuse.

(iii) The sides of any triangle are proportion to the sines of the opposite angles.

Worked-out examples

Example 1: Find the area of the given triangle ABC. A
7 cm
Solution: B 30°

Here, area of 'ABC = 1 BC u AB sin30q 8 cm C
2
1 1 A
= 2 u 8 cm u 7 cm u 2 = 14 cm2

Example 2: In the given figure, BD is a median. If BD = 6 cm, 30° D
BC = 10 cm and ‘DBC = 30q, find the area of C
'ABC.

B

Vedanta Excel in Mathematics - Book 10 248 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur