Vedanta Excel in Mathematics Book 9 Final Re_CTP

Unit 13 Geometry - Triangle

13. 1 Triangle –Looking back

1. Let’s fill in the blanks with correct answers as quickly as possible.

a) A triangle in which all three sides are equal is called an ………..

b) Every triangle has at least …………. acute angles.

c) The sum of interior angles of a triangle is always ………. right angles.

d) An obtuse angled triangle has exactly one …………. angle.

e) The base angels of an isosceles triangle are ………

2. Let’s match the following as quickly as possible.

a) Each angle measures 60o (i) Isosceles triangle

b) One right angle and other angles are equal (ii) Acute angled triangle

c) Sum of angles is not 180o (iii) Isosceles right-angled
triangle

d) Base angles are equal (iv) Equilateral triangle

e) No sides are equal (v) No triangle exists

f) Each angle is less than a right angle (vi) Scalene triangle

3. Let’s write ‘True’ for true statements and ‘False’ for false statements.

a) Every equilateral triangle is an isosceles triangle.

b) Every isosceles triangle is an equilateral triangle.

c) Every equilateral angle is an acute angled triangle.

d) Every acute angled triangle is an equilateral triangle.

e) A right-angled triangle may be an isosceles triangle.

13.2 Types of triangle - review

Types of triangles by angles

On the basis of the sizes of angles, there are three types of triangles:

(i) Acute angled triangle (ii) Obtuse angled triangle (iii) Right angled triangle

AA A

B C B CB C

It has three acute angles. It has one obtuse ang le. It has one right angle.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 197 Vedanta Excel in Mathematics - Book 9

Geometry - Triangle

Types of triangles by sides
On the basis of the length of the sides of triangles, there are three types of triangles.

(i) Scalene triangle (ii) Isosceles triangle (iii) Equilateral triangle

AA A

BC BC BC

It has none of the It has two It has all three

equal sides. equal sides. equal sides.

13.3 Median and altitude of a triangle

A median of a triangle is a straight line that joins a vertex of a triangle to the

mid-point of its opposite side. A

AD, BE, and CF are the medians of ∆ ABC. All three F E
medians of a triangle are concurrent and the meeting

point of medians is called the centroid of the triangle. In G
DC
the figure, G is the centroid of ∆ ABC. The centroid of a B
triangle divides each median in the ratio of 2:1.

A perpendicular drawn from a vertex to the opposite side of a triangle is known as the

altitude (or height) of the triangle. A

AD, BE, and CF are the altitudes of ∆ABC. The altitudes F E
of a triangle are also concurrent and the meeting point of O
altitudes of a triangle is called Orthocentre of the triangle.
In the figure, O is the orthocentre of the triangle ABC.

Following are the important facts about the medians and B D C

altitudes of different types of triangles.
(i) In an equilateral triangle, its medians are also the altitudes of the triangle or

vice versa.
(ii) In an isosceles triangle, the median drawn from its vertex to the base is also its

altitude.
(iii) In a right-angled triangle, the median drawn to its hypotenuse is half of the

hypotenuse in length.
(iv) Each median of a triangle divides the triangle in two triangles having equal

area.

13.4 Properties of triangles

Following are a few important properties of triangles. We can verify these properties

experimentally as well as theoretically.
(i) The sum of the angles of a triangle is equal to two right angles (180°).
(ii) The exterior angle of a triangle is equal to the sum of two opposite interior

angles.

Vedanta Excel in Mathematics - Book 9 198 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Geometry - Triangle

(iii) The sum of any two sides of a triangle is greater than the third side.

(iv) In any triangle, the angle opposite to longer side is greater than the angle opposite
to shorter side.

(v) The base angles of an isosceles triangle are equal.

(vi) In an isosceles triangle, perpendicular drawn from the vertical angle to the base
bisects the base.

(vii) A straight line that joins the mid-points of any two sides of a triangle is parallel
to the third side of the triangle, ... and so on.

Theorem 1

The sum of the angles of any triangle is equal to two right angles.

Experimental verification

Step 1: Three triangles with different shapes and Vedanta ICT Corner
sizes are drawn. Suppose the name of each
triangle is ABC. Please! Scan this QR code or
browse the link given below:

https://www.geogebra.org/m/v3vngznk


Step 2: All three angles of each triangle are measured and the measurements are tabulated
in the table.

Figure ∠A ∠B ∠C ∠A + ∠B + ∠C Result
(i) ∠A + ∠B + ∠C = 180°

(ii) ∠A + ∠B + ∠C = 180°

(iii) ∠A + ∠B + ∠C = 180°

Conclusion: The sum of the angles of any triangle is equal to two right angles.

Theoretical proof

Given: ∠ABC, ∠BCA and ∠BAC are the angles of ∆ABC.

To prove: ∠ABC + ∠BCA + ∠BAC = 2 rt. angles
Construction: Through the vertex A, a straight line XY parallel to BC is

drawn.
Proof:

Statements Reasons

1. ∠XAB = ∠ABC 1. XY // BC and alternate angles

2. ∠YAC = ∠BCA 2. XY // BC and alternate angles

3. ∠XAB + ∠BAC + ∠YAC = ∠XAY 3. Whole part axiom

4. ∠XAB + ∠BAC + ∠YAC = 2 rt. angles 4. ∠XAY is a straight angle

5. ∠ABC + ∠BAC + ∠BCA = 2 rt. angles 5. From statements (1), (2) and (4)

Proved

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 199 Vedanta Excel in Mathematics - Book 9

Geometry - Triangle

Theorem 2

The exterior angle of a triangle is equal to the sum of the two opposite interior angles.

Experimental verification Vedanta ICT Corner
Please! Scan this QR code or
Step 1: Three triangles ABC with different shapes browse the link given below:
and sizes are drawn. A side BC is produced
to D in each triangle. https://www.geogebra.org/m/ethbh2t9

Step 2: The exterior angle ACD and the opposite interior angles ABC and BAC are measured
in each triangle. The measurements are tabulated in the table.

Figure ∠ABC ∠BAC ∠ABC + ∠BAC ∠ACD Result
(i) ∠ABC + ∠BAC = ∠ACD
(ii) ∠ABC + ∠BAC = ∠ACD
(iii) ∠ABC + ∠ BAC = ∠ACD

Conclusion: The exterior angle of a triangle is equal to the sum of the two opposite interior
angles.

Theoretical proof B A
CD
Given: In ∆ ABC, the side BC is produced to D. ∠ACD is the
exterior angle so formed. ∠ABC and ∠BAC are the
opposite interior angles.

To prove: ∠ACD = ∠ABC + ∠BAC
Proof

Statements Reasons

1. ∠BCA + ∠ACD = ∠BCD 1. Whole part axiom
2. ∠BCA + ∠ACD = 180° 2. ∠BCD is a straight angle
3. ∠ABC + ∠BCA + ∠BAC = 180° 3. Sum of the angles of a triangle
4. ∠BCA + ∠ACD = ∠ABC + ∠BCA + ∠BAC 4. From statements (2) and (3)
5. ∠ACD = ∠ABC + ∠BAC 5. Cancelling ∠BCA from both

sides of statement (4).

Proved

Worked-out examples A

Example 1: From the figure given alongside, find the size of ∠ABC. (2x–5°)

Solution: (3x+10°) (7x–35°)

Here, ∠BAC = (2x – 5o), ∠ABC = (3x + 10o) and ∠ACD = (7x –35o) B CD

Now,∠BAC + ∠ABC = ∠ACD [The exterior angle of triangle is equal to the sum of two
opposite interior angles.]

Vedanta Excel in Mathematics - Book 9 200 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Geometry - Triangle

or, (2x – 5o) + (3x + 10o) = 7x – 35o
or, 5x + 5o = 7x – 35o
or, -2x = - 40o
or, x = 20o
Hence, ∠ABC = 3x + 10o = 3×20o + 10o = 70o

Example 2: In a triangle, the exterior angle formed by extending a side is 100o. If the
difference between two interior angles opposite to the exterior angle is
40o, find the interior angles of the triangle.

Solution:

Let, x and y be two interior angles opposite to the given exterior angle of a triangle.

Then, x + y = 100o …equation (i) [The exterior angle of triangle is equal to
Also, x – y = 40o the sum of two opposite interior angles.]

or, x = y + 40o … equation (ii)

Now, putting the value of x from equation (ii) in equation (i), we get

(y + 40o) + y = 100o

or, 2y + 40o = 100o

or, 2y = 60o

or, y = 30o

Again, putting the value of y in equation (ii), we get

x = 30 o + 40o = 70o

Hence, the first interior angle (x) = 70o,

the second interior angle (y) = 30o

and the third interior angle = 180o – (x + y) = 180o – (70o + 30o) = 80o

Example 3: In the figure BD and CD are the bisectors of A D
∠ABC and ∠ACE respectively. If ∠BAC = 56°, 56° x
find the value of x.

Solution:

B C E
A
(i) H ere, suppose ∠ABD = ∠DBC = a and ∠ACD = ∠DCE = b. 56° D
In DABC, ∠BAC + ∠ABC = ∠ACE x

a b E
Ba b

C

or, 56° + (a + a) = (b + b) [Being exterior angles of triangle equal
to the sum of opposite interior angles]
or, 56° + 2a = 2b

or, 2(28° + a) = 2b

∴ b = 28° + a .......... (i)

(ii) In DBCD, ∠DBC + ∠BDC = ∠DCE

a + x = b

or, a + x = 28° + a [From (i), b = 28° + a]

∴ x = 28°

Hence, x = 28°

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 201 Vedanta Excel in Mathematics - Book 9

Geometry - Triangle

Example 4: The size of an exterior angle of a triangle is the complement of smaller of
two opposite interior angles. If the greater of the opposite interior angles
exceeds twice the smaller by 10o, find the measure of the interior angles of
the triangle.

Solution:
Let, the smaller interior angle opposite to the exterior angle of a triangle be x.
Then, the exterior angle = (90o – x) and the greater of opposite interior angles = (2x + 10o)
Now, x + (2x + 10o) = 90o – x
or, 4x = 80o
or, x = 20o
Hence, the first interior angle = x = 20o,
the second interior angle = (2x + 10o) = (2 × 20o + 10o) = 50o
and the third interior angle = 180o – (20o + 50o) = 110o

Example 5: In ∆ABC, BN is perpendicular to AC and CM is A

perpendicular to AB. BN and CM intersect at O. Prove MO N
that ∠ BOC = 180°– ∠A
C
Solution: B

Given: In ∆ABC, BN ⊥ AC and CM ⊥AB. BN and CM intersect at O.

To prove: ∠BOC = 180°–∠A

Proof:

S.N. Statements S.N. Reasons

1. ∠BAC + ∠ABN = ∠BNC 1. The exterior angle of triangle is equal to the
i.e., ∠A + ∠ABN = 90o sum of two opposite interior angles.

2. ABN+ ∠BMO = ∠BOC 2. Same as (1)
i.e., ∠ABN+ 90o = ∠BOC

3. ∠A + ∠ABN – (∠ABN + 90o) 3. Subtracting statement (2) from statement (1)
= 90o– ∠BOC

i.e.,∠BOC = 180°–∠A

Proved

Example 6: In the adjoining ∆ABC, AY is the bisector of ∠BAC A
XY
and AX ⊥ BC. Prove that: ∠XAY = 1 (∠B – ∠C).
2
Solution: B C

Given: In ∆ABC, AY is the bisector of ∠BAC and AX ⊥ BC.

To prove: ∠XAY = 1 (∠B – ∠C).
2

Vedanta Excel in Mathematics - Book 9 202 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Geometry - Triangle

Proof:

S.N. Statements S.N. Reasons

1. ∠BAC = 2∠BAY 1. AY is the bisector of ∠BAC.
2. The sum of angles of triangle is 180o.
2. ∠BAC+ ∠ABC +∠ACB = 180o 3. From statements (1) and (2).

3. 2∠BAY + ∠B +∠C= 180o 4. The exterior angle of triangle is equal
to the sum of two opposite interior
i.e., ∠BAY = 1 (180o– ∠B –∠C) angles.
2

4. ∠BAX+ ∠ABX = ∠AXC

i.e., ∠BAX = 90o – ∠B

5. ∠BAY - ∠BAX = 1 (180o – ∠B –∠C) 5. Subtracting statement (4)
2
from statement (3) and using
– (90o – ∠B) ∠BAY – ∠BAX = ∠XAY

i.e., ∠XAY = 1 (∠B – ∠C)
2

Proved

EXERCISE 13.1

General section B A C
1. a) In DABC, M is the mid-point of side BC and AP ⊥ BC. PM
Then, name the altitude and median of the triangle.

X

b) In DXYZ, A, B and C are the mid-points of sides A GC
XY, YZ and XZ respectively. Then, name the medians
of the triangle. What is the point G called? Y B Z
P
Z R
Q Y D
c) In DPQR. PX ⊥ QR, QY ⊥ PR and RZ ⊥ PQ.
O
Then, name the altitudes of the triangle.
X
What is the point O called? A

2. a) From the figure given alongside, write down the relation of
∠BAC, ∠ABC and ∠ACD.
B
C
S

b) In ∆PQR, the side QR is extended to the point S. 58° R
If ∠PRS = 58o, what is the value of ∠P + ∠Q? Q

P

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 203 Vedanta Excel in Mathematics - Book 9

Geometry - Triangle

3. Answer the following questions
a) In a triangle, if an exterior angle is 69o and its opposite interior angles are xo and

2xo, find the sizes of each interior angle of the triangle.
b) When a side of a triangle is extended to a point, the exterior angle so formed is

100o. If the two opposite interior angles are in the ratio 2:3, find the sizes of all
three interior angles of triangle.
c) An exterior angle of a triangle is (4x – 15)° and two opposite interior angles are
(x + 10)o and (2x – 5)o. Find the measure of each interior angle of the triangle.
d) In a triangle, an exterior angle is (153 – 3x)° and two opposite interior angles are
(5x – 3)o and (2x + 6)o. Find the measure of all interior angles of the triangle.

4. Find the unknown sizes of angles: S

a) A b) A c) R
b 140°
B 2x
120° a 50­° 3x+5° p 2p P
C CD 2
D 70° B
Q

d) Q e) A f) P
C
3x C D y
120° T
O 80°

A 2x P B 30° xQ
R 4x B SR

g) h) i)

F A E A
E T
By 35° xD
30° 25°
PS

x 110° B 45° 50° D
AD Qx R
145°
CG C B CE

Creative section

5. a) In a triangle, the exterior angle formed by extending a side is 80o. If the difference
between two interior angles opposite to the exterior angle is 20o, find the interior
angles of the triangle.

b) In a triangle, an exterior angle formed by producing a side is half of a right angle.
One of the two opposite interior angles is twice the other; find the sizes of each
interior angle of the triangle.

c) The size of an exterior angle of a triangle is the supplement of smaller of two
opposite interior angles. If the greater of the opposite interior angles exceeds the
smaller by 30o, find the measure of the interior angles of the triangle.

6. a) In ∆PQR, RX ⊥ PQ and QY ⊥ PR, RX and QY intersect P
at O. Prove that ∠QOR = 180° – ∠P. Y

X
O

QR

Vedanta Excel in Mathematics - Book 9 204 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Geometry - Triangle

D

b) In the given figure ABCD, prove that C
∠BCD = ∠BAD + ∠ABC + ∠ADC.
[Hint: Join A and C then extended AC to the point E.] AB

c) In the adjoining star shaped figure, prove that A E
∠A + ∠B + ∠C + ∠D + ∠E = 180° T

PS

7. a) In the figure alongside, BE and CE are the angular B QR D
C
bisectors of ∠ABC and ∠ACD respectively.
AE
Prove that ∠BAC = 2∠BEC. B
CD

b) In the given figure, the bisector of ∠ACU meets AU at O. T
U
Prove that ∠COT = 1 (∠CAT + ∠CUT). O
2
CA

c) In the adjoining DABC, AY is the bisector of ∠BAC and A
XY
AX ⊥ BC. Prove that ∠XAY = 1 (∠B – ∠C).
2
B C

13.5 Triangle inequality property

Theorem 3

Let’s study the following activities.
The mathematics teacher asked his/her three students Archana, Rahul and Pemba to form
triangles with the sticks of given measurements.

Student’s name Lengths of sticks

Archana 5 cm, 7 cm and 10 cm Archana's work
Rahul 4 cm, 12 cm and 6 cm

Pemba 11 cm, 6 cm and 5 cm Rahul's work Pemba's work

All of them tried to form a triangle. Archana Vedanta ICT Corner
formed the triangle successfully. When Rahul and Please! Scan this QR code or
Pemba tried to join the ends of the two smaller browse the link given below:
sticks, they were unable to form triangles, why?
https://www.geogebra.org/m/jyyx5kkd

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 205 Vedanta Excel in Mathematics - Book 9

Geometry - Triangle

The sum of any two sides of a triangle is greater than the third side.
Experimental verification
Step 1: Three triangles ABC with different shapes and sizes are drawn.

Step 2: All three sides of each of ∆ABC are measured and the lengths are tabulated in the
table.

Figure AB BC CA AB + BC BC + CA AB + CA Result

(i) AB + BC > CA,
BC + CA > AB, AB + CA > BC

(ii) AB + BC > CA,
BC + CA > AB, AB + CA > BC

(iii) AB + BC > CA,
BC + CA > AB, AB + CA > BC

Conclusion: The sum of any two sides of a triangle is greater than the third side.

Theorem 4

In any triangle, the angle opposite to the longer side is greater than the angle opposite to
the shorter side.

Experimental verification
Step 1: Three triangles ABC with different shapes and sizes are drawn in such a way that

BC is the longest and CA is shortest sides in each triangle.

Step 2: The angle opposite to the greatest side BC and the angle opposite to the shortest side
CA are measured and tabulated in the table.

Figure Angle opposite to BC: ∠A Angle opposite to CA: ∠B Result
(i) ∠A > ∠B
(ii) ∠A > ∠B
(iii) ∠A > ∠B

Conclusion: The angle opposite to the longer side of any triangle is greater than the angle
opposite to the shorter side.

Vedanta Excel in Mathematics - Book 9 206 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Geometry - Triangle

Converse of Theorem 4

In any triangle, the side opposite to the greater angle is longer than the side opposite to the
smaller angle.

Experimental verification
Step 1: Three triangles ABC with different shapes and sizes are drawn in such a way that

∠A is the greatest and ∠B is the smallest angles in each triangle.

Step 2: The side opposite to the greatest angle A and the side opposite to the smallest angle
B are measured and tabulated in the table.

Figure Side opposite to ∠A: BC Side opposite to ∠B: CA Result.
(i) BC > CA
(ii) BC > CA
(iii) BC > CA

Conclusion: The side opposite to the greater angle of any triangle is longer than the side
opposite to the smaller angle.

Theorem 5

Of all straight line segments drawn to a given line from a point outside it, the perpendicular
is the shortest one.

Experimental verification

Step 1: Three straight line segments XY with different lengths and position are drawn.
A point P is marked outside the each line segment. Three line segments PA, PB, PC
and a perpendicular PQ are drawn from P to XY.

(i) (ii) (iii)

Step 2: The lengths of the line segments PA, PB, PC, and PQ are measured and tabulated in
the table.

Figure Length of the line segments Result

(i) PA PB PC PQ
(ii)
(iii) Perpendicular PQ is the shortest one.

Perpendicular PQ is the shortest one.

Perpendicular PQ is the shortest one.

Conclusion: Of all line segments drawn to a given line from a point outside it, the
perpendicular is the shortest one.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 207 Vedanta Excel in Mathematics - Book 9

Geometry - Triangle

Worked-out examples

Example 1: Can you construct a triangle with sides 3.5 cm, 9 cm and 4 cm? Why?
Solution:
Here, 3.5 cm + 9 cm = 12.5 cm > 4 cm,
9 cm + 4 cm = 13 cm > 3.5 cm
4 cm + 3.5 cm = 7 cm which is not greater than 9 cm
So, we cannot construct the triangle with sides 3.5 cm, 9 cm and 4 cm.

Example 2: In ∆ABC, if AB = 4 cm and BC = 7 cm, between what two measures should
the third side AC fall?

Solution:
Here, in ∆ABC, AB = 4 cm and BC = 7 cm
Since, for any triangle; the sum of any two sides should be greater than the third side and
their difference should be smaller than the third side.
So, 4 cm + 7 cm = 11 cm and 7 cm – 4 cm = 3 cm
Hence, the length of the third side AC lies between 3 cm and 11 cm i.e., 3 cm < AC < 11 cm.

Example 3: In the given quadrilateral ABCD, D C
A B
prove that AB + BC + CD > AD.

Solution:

Given: ABCD is a quadrilateral. D C
To prove: AB + BC + CD > AD B

Construction: B and C are joined.

Proof: A

S.N. Statements S.N. Reasons
1. AB + BC > AC 1. In ∆ABC, the sum of two sides is greater
than the third side.
2. AC + CD > AD 2. In ∆ACD, the sum of two sides is greater
than the third side.
3. AB + BC + CD > AD 3. From statements (1) and (2).

13.6 Congruent triangles

Two or more triangles are said to congruent if they have the same shape and size. In this
case, when one triangle is placed over another triangle, their corresponding parts exactly
coincide to each other. The congruent triangles are always similar and they have equal area.

Vedanta Excel in Mathematics - Book 9 208 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Geometry - Triangle

13.7 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 as per the
following conditions. We use these conditions 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 ∆s ABC and PQR,

a) AB = PQ (S)
b) BC = QR (S)
c) CA = RP (S)
d) ∴ ∆ ABC ≅ ∆ PQR [S.S.S. axiom]

Corresponding parts of congruent triangles are also equal.

∴ ∠A = ∠P, ∠B = ∠Q and ∠C = ∠R.

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

(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 ∆s ABC and PQR

a) AB = PQ (S)

b) ∠B = ∠Q (A)

c) BC = QR (S)

d) ∴∆ABC ≅ ∆PQR [S.A.S. axiom]

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

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 ∆s ABC and PQR,
a) ∠B = ∠Q (A)
b) BC = QR (S)
c) ∠C = ∠R (A)
d) ∴ ∆ABC ≅ ∆PQR [A.S.A. axiom]

Now, AC = PR and AB = PQ [Corresponding sides of congruent triangles]
∠A = ∠P [Corresponding angles of congruent triangles]

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 209 Vedanta Excel in Mathematics - Book 9

Geometry - Triangle

(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 congruent triangles.

In right angled ∆s ABC and PQR,

a) ∠B = ∠Q (R)

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]

AB = PQ [Corresponding sides of congruent triangles]

(v) A.A.S. axiom

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

This axiom can be verified by using A.S.A. axiom.

Here,
a) ∠A = ∠P [Given]
b) ∠B = ∠Q [Given]

c) ∠A + ∠B + ∠C = ∠P + ∠Q + ∠R [Sum of the angles of any triangle is 180°]

d) ∴ ∠C = ∠R [From (a), (b) and (c)]

Now, in ∆s ABC and PQR,

e) ∠B = ∠Q (A) [Given]

f) BC = QR (S) [Given]

g) ∠C = ∠R (A) [from (d)]

h) ∴ ∆ABC ≅ ∆PQR [A.S.A axiom]

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

EXERCISE 13.2

General section
1. a) Identify and name the longest and the shortest sides of the following triangles.

FR

60° DE PQ
35° 85°

Vedanta Excel in Mathematics - Book 9 210 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Geometry - Triangle

b) Identify and name the greatest and the smallest angles of the following triangles.

B

X E
G

Y Z CA F



c) In ∆ABC, if ∠A = 85°, ∠B = 65° and ∠C = 30°, name the longest and the shortest
sides of the triangle.

d) In ∆PQR, if ∠Q = 75°, and ∠R = 25°, name the longest and the shortest sides of the
triangle.

e) In ∆XYZ, if XY = 5.6 cm, YZ = 4.7 cm and ZX = 6.5, name the greatest and the
smallest angles of the triangle.

2. a) Can you construct a triangle with sides 5 cm, 7 cm and 10 cm? Why?

b) Can you construct a triangle with sides 6 cm, 6 cm and 6 cm? Give reason.

c) Can you construct a triangle with sides 4 cm, 7 cm and 11 cm? Why?

d) Can a triangle be constructed with the sides 5.2 cm, 3.7 cm and 8.5 cm? Why?

Creative section - A

3. a) In ∆ABC, if AB = 5 cm and BC = 8 cm, between what two measures should the third
side AC fall?

b) If the measures of two sides of a triangle are 6 cm and 4 cm, between what two
measures should the third side lie?

4. Use the necessary axioms to verify that the following pairs of triangles are congruent.
Also, write the equal corresponding parts of the triangles.

Y

a) A P b) D X

B CQ R E FZ S
BQ R R
d)
P FT
G

c) C E

A

5. a) In the adjoining figure, show that 70°
∆ABC ≅ ∆PQR. Also find the unknown Vedanta Excel in Mathematics - Book 9
sizes of angles of ∆ABC and ∆PQR.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 211

Geometry - Triangle

b) In the figure alongside, show that
∆RST ≅ ∆XYZ. Also find the
unknown sizes of x cm, y cm and
a°.

Creative section - B P

6. a) In ∆PQR, O is the interior point. Prove that O

OP + OQ + OR > 1 (PQ + QR + RP) . Q R
2 A C
O 8 cm
b) In the figure alongside, O is the interior point of DABC. 7 cm
If AB = 7 cm, BC = 11 cm and CA = 8 cm, show that B 11 cm
OA + OB + OC > 13 cm.

7. a) In the given figure, AB // PQ and AB = PQ.
Prove that ∆ABO ≅ ∆POQ. Also show that AO = OQ and
BO = OP

b) In the given figure, PQ // SR and PS // QR.
Prove that ∆PQR ≅ ∆PRS.
Also show that PQ = RS and PS = QR.

Project work and activity section

8. a) Take three sticks of different lengths such that the total length of two shorter sticks
is greater than longer stick. Now, make a triangle by joining these sticks.

b) Take three sticks of different lengths such that the total length of two shorter
sticks is less than the longer stick. Can you make a triangle by joining these sticks?
Discuss in your class.

9. a) Fold a rectangular sheet of paper diagonally and cut into two halves.

(i) What type of triangles did you get?

(ii) Are these triangles congruent? Discuss in the class.

(iii) Which are the corresponding sides and angles in these two triangles? Discuss
in the class.

Vedanta Excel in Mathematics - Book 9 212 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Geometry - Triangle

Theorem 6

If any two sides of a triangle are equal, the angles opposite to them are equal.
Or

Base angles of an isosceles triangle are equal.

Experimental verification
Step 1: Three isosceles triangles ABC with different shapes and sizes are drawn.

Step 2: The angles opposite to the equal sides of each triangle are measured and tabulated
in the table.

Figure Angles opposite to the equal sides (Base angles) Result

(i) ∠ABC ∠ACB ∠ABC = ∠ACB
(ii) ∠ABC = ∠ACB
(iii) ∠ABC = ∠ACB

Conclusion: Base angles of an isosceles triangles are equal.

Theoretical proof

Given: In isosceles ∆ABC, AB = AC

To prove: ∠ABC = ∠ACB

Construction: From the vertex A, AD ⊥ BC is drawn.

Proof

Statements Reasons

1. In ∆s ABD and ACD 1.
(i) By construction, AD ⊥ BC
(i) ∠ADB = ∠ADC (R) (ii) Given
(iii) Common side
(ii) AB = AC (H) (iv) R.H.S. axiom

(iii) AD = AD (S) 2. Corresponding angles of congruent triangle.

(iv) ∴ ∆ABD ≅ ∆ACD

2. ∠ABD = ∠ACD
i.e. ∠ABC = ∠ACB

Proved

Converse of Theorem 6

If two angles of a triangle are equal, the sides opposite to them are also equal.
Or

It two angles of a triangle are equal, it is an isosceles triangle.

Experimental verification
Step 1: Three line segments AB of different lengths are drawn.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 213 Vedanta Excel in Mathematics - Book 9

Geometry - Triangle
Step 2: At the points B and C of each line segment, angles of equal sizes are drawn. Let A be

the point at which the arms of the angles meet. Now, Ds ABC are formed.

Step 3: The lengths of the sides opposite to the equal angles of each triangle are measured
and tabulated in the table.

Figure Lengths of sides opposite to equal angles Result
AB AC
(i) AB = AC
(ii) AB = AC
(iii) AB = AC

Conclusion: The sides opposite to the equal angles of a triangle are equal, i.e. each of triangle
ABC are isosceles triangles.

Theoretical proof

Given: In ∆ABC, ∠B = ∠C

To prove: AB = AC.
Hence ∆ABC is an isosceles triangle.

Construction: From the vertex A, AD ⊥ BC is drawn.

Proof

Statements Reasons
1. In ∆s ABD and ACD
1.

(i) ∠ABD = ∠ACD (A) (i) Given
(ii) By construction, AD ⊥ BC
(ii) ∠ADB = ∠ADC (A) (iii) Common side
(iv) A.A.S. axiom
(iii) AD = AD (S)
2. Corresponding sides of congruent triangles
(iv) ∴ ∆ BD ≅ ∆ACD

2. AB = AC

i.e. ∆ABC is an isosceles triangle.

Proved

Theorem 7

The bisector of the vertical angle of an isosceles triangle is the perpendicular bisector of
the base.

Experimental verification
Step 1: Three isosceles triangles ABC with different shapes and sizes are drawn in such a

way that AB = AC in each triangle.

Vedanta Excel in Mathematics - Book 9 214 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Geometry - Triangle
Step 2: The bisector of ∠A is drawn in each triangle. Suppose it is AD and it meets the

opposite side BC at D.

Step 3: The lengths of BD and DC, and the angles ADB and ADC are measured and tabulated
in the table.

Figure BD DC Result ∠ADB ∠ADC Result
(i) ∠ADB = ∠ADC, AD ⊥ BC
(ii) BD = DC ∠ADB = ∠ADC, AD ⊥ BC
(iii) ∠ADB = ∠ADC, AD ⊥ BC
BD = DC

BD = DC

Conclusion: The angular bisector of the vertical angle of an isosceles triangle is the
perpendicular bisector of its base.

Theoretical proof

Given: ∆ABC is an isosceles triangle in which AB = AC.
AD is the bisector of ∠BAC.

To prove: AD ⊥ BC and BD = DC.

Proof

Statements Reasons

1. In ∆ABD and ∆ACD 1.
(i) Given
(i) AB = AC (S) (ii) Given
(iii) Common side
(ii) ∠BAD = ∠CAD (A) (iv) S.A.S. axiom
(iii) AD = AD (S)

(iv) ∴ ∆ABD ≅ ∆ACD

2. ∠ADB = ∠ADC 2. Corresponding angles of congruent
triangles

3. AD ⊥ BC 3. Adjacent angles in linear pair are equal.

4. BD = DC 4. Corresponding sides of congruent
triangles.

5. AD is the perpendicular bisector 5. From statements 3 and 4
of BC.

Proved

Converse of Theorem 7

The straight line joining the vertex and the mid–point of the base of an isosceles triangle is
perpendicular to the base and bisects the vertical angle.

Experimental verification
Step 1: Three isosceles triangles ABC with different shapes and sizes are drawn in such a

way that AB = AC in each triangle.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 215 Vedanta Excel in Mathematics - Book 9

Geometry - Triangle
Step 2: The mid–point of BC is marked as D. A and D are joined in each triangle.

Step 3: ∠ADB, ∠ADC, ∠BAD, and ∠DAC are measured and tabulated in the table.

Figure ∠ADB ∠ADC Result ∠BAD ∠DAC Result

(i) ∠ADB = ∠ADC, AD ⊥ BC ∠BAD = ∠DAC

(ii) ∠ADB = ∠ADC, AD ⊥ BC ∠BAD = ∠DAC

(iii) ∠ADB = ∠ADC, AD ⊥ BC ∠BAD = ∠DAC

Conclusion: The straight line joining the vertex and the mid-point of the base of an
isosceles triangle is perpendicular to the base and bisects the vertical angle.

Theoretical proof

Given: ∆ABC is an isosceles triangle in which AB = AC.
AD joins the vertex A and mid-point D of the base BC.

To prove: AD ⊥ BC and ∠BAD = ∠CAD

Proof

Statements Reasons

1. In ∆ABD and ∆ACD 1.
(i) Given
(i) AB = AC (S) (ii) Common side

(ii) AD = AD (S) Given
S.S.S. axiom
(iii) BD = DC (S) (iii)
(iv)
(iv) ∴ ∆ABD ≅ ∆ACD

2. (i) ∠ADB = ∠ADC 2. (i) Corresponding angles of congruent
(ii) AD ⊥ BC
triangles

(ii) Adjacent angles in linear pair are equal.

3. ∠BAD = ∠CAD 3. Corresponding angles of congruent
triangles

Proved

Worked-out examples

Example 1: Find the values of x and y from the figure given A
alongside. y

Solution: Bx C D

Here, ∠BAC = 90° and AB = AC = CD.
(i) ∠ABC = ∠ACB = x [Base angle of isosceles ∆ ABC]

Vedanta Excel in Mathematics - Book 9 216 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Geometry - Triangle

(ii) ∠ABC + ∠ACB + ∠BAC = 180° [ Sum of angles of DABC]

or, x + x + 90° = 180°

or, 2x = 90°

∴ x = 45°

(iii) ∠CAD + ∠ADC = y [Base angles of isosceles ∆ ACD]

(iv) ∠CAD + ∠ADC = ∠ACB [Being exterior angle of ∆ACD equal to the
sum of two non-adjacent interior angles]

or, y + y = 45° T
or, 2y = 45° S
y = 22.5°
∴

Example 2: In the adjoining figure, find the values of a and b. ab

Solution: Here, PQ R
(i) ∠SPQ = ∠PSQ = a [ PQ = QS]

(ii) ∠SQR = ∠SPQ + ∠PSQ = a + a = 2a [Exterior angle of DPQS]

(iii) ∠SQR = ∠SRQ = 2a [ SQ = SR]
(iv) ∠SPR + ∠PRS = ∠TSR

or, a + 2a = 90°

∴ a = 30°
(v) ∠PSQ + ∠QSR + ∠RST = 180° [Being parts of straight angle]

or, a + b + 90° = 180°

or, 30° + b + 90° = 180°

∴ b = 60°
Hence, a = 30° and b = 60°.

Example 3: In DABC, AB = AC and AP ⊥ BC. If AB = (2x + 3) cm, A (3y – 1)cm
AC = (3y – 1) cm, BP = (y + 1) cm and PC = (x + 2) cm, (2x + 3)cm
find the values of x and y.

Solution: B (y+1)cm P (x+2)cm C
(i) AB = AC or, 2x + 3 = 3y – 1
2x – 3y = –4 ... (i)

(ii) BP = PC [ AB = AC and AP ⊥ BC]

or, y + 1 = x + 2 y = x + 1 ... (i)

Now,

Putting the value of y in equation (i) from equation (ii), we get,

2x – 3(x + 1) = –4

or, 2x – 3x – 3 = –4

∴ x =1

Again, putting the value of x in equation (ii), we get,

y =1+1=2

Hence, x = 1 and y = 2.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 217 Vedanta Excel in Mathematics - Book 9

Geometry - Triangle

Example 4: In the given figure, AB = AC and BD = EC. Prove A
that ∆ ADE is an isosceles triangle.

Solution:
Given: AB = AC and BD = EC
To prove: (i) ∆ADE is an isosceles triangle.

Proof BD EC

Statements Reasons

1. In ∆ABD and ∆ACE 1.
(i) Given
(i) AB = AC (S) (ii) Base angles of an isosceles ∆ABC.
(iii) Given
(ii) ∠ABD = ∠ACE (A)

(iii) BD = EC (S)

2. ∴ ∆ABD ≅ ∆ACE 2. By S.A.S. axiom

3. AD = AE 3. Corresponding sides of congruent triangles

4. DADE is an isosceles triangle 4. From statement (3)

Proved

Example 5: If the perpendiculars drawn from any two vertices to
their opposite sides of a triangle are equal, prove that
the triangle is an isosceles triangle.

Solution:
Given: In ∆ABC, CP ⊥ AB, BQ ⊥ AC and CP = BQ
To prove: (i) ∆ABC is an isosceles triangle.
Proof

Statements Reasons
1.
1. In ∆PBC and ∆QBC (i) Both are right angles
(ii) Common sides
(i) ∠BPC = ∠BQC (R) (iii) Given
(ii) BC = BC (A) 2. By R.H.S. axiom
3. Corresponding angles of congruent triangles
(iii) CP = BQ (S)
4. Base angles are equal.
2. ∴ ∆PBC ≅ ∆BQC

3. ∠PBC = ∠QCB
i.e. ∠ABC = ∠ACB

4. DABC is an isosceles triangle

Proved

EXERCISE 13.3 A
General section

1. a) In the given figure, AB = AC, write down the relation between
∠ABC and ∠ACB.

B P C
X R
b) In the figure alongside, ∠PQR = ∠PRQ, what is the relation
between PQ and PR?

Q

c) In the adjoining figure, if XY = XZ and YA = AZ, write down

the relation between XA and YZ. YAZ

Vedanta Excel in Mathematics - Book 9 218 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Geometry - Triangle

L

d) In the given figure, if LM = LN and LP ⊥ MN, write down the
relation between MP and PN, and ∠LMP and ∠LNP.

MP N

O

e) In the figure alongside, ON = OE and ∠NOT = ∠TOE, write
down the relation between, NT and TE, OT and NE.

NT E

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

P A E
W

Q RS X YZ B CD F G H
D K A
C
A P

B CL M E BD
D
i) A j) T N l) X
xy
xx Py k)
69° y B
BD 40° x V
330°0° U y
x 30° RA CY Z
CQ S

Creative section - A

3. a) In the given figure, find the sizes of ∠x, ∠y, and ∠z.


b) In the adjoining figure, calculate the sizes of 50
∠p, ∠q, ∠r, and ∠s.

c) In the given figure, AB = AC and
∠ACD = 110°, what will be the measure of
∠BAC + ∠ACB?

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 219 Vedanta Excel in Mathematics - Book 9

Geometry - Triangle

A

4. a) In the given figure, AB = AC, ∠BAC = 44° and ∠ACD = 24°, D 44°
show that BC = CD. 24°

B C
P R

b) In the figure alongside, SR = QR, ∠QPR = 48° and S 48°
∠PRS = 18°, show that PQ = PR.
18° (3y+1)cm (5y–2)cm
Q
P

5. a) In the adjoining figure, find the values of x and y. (x+8)cm

Q(x+4)cm T (y+3)cmR

b) In DABC, AD bisects ∠A and it is perpendicular to base BC. (3x+1)cm A
If AB = (3x + 1) cm, AC = (5y – 2) cm, BD = (x + 1) cm and B(x+1)cm D (y+2)cmC
DC = (y + 2) cm, find the values of x and y.

Creative section - B
6. a) In the given figure, AB = AC and BD = EC. Prove that AD = AE.

b) In the given figure, AB = AC, BD = EC and ∠DAE = 30°. 30°
Prove that ∆ADE is an isosceles triangle. Also calculate the P
size of ∠ADE. O

7. a) In the given figure, PQ = PR, QO bisects ∠PQR and RO bisects A
D
∠PRQ. Prove that OQR is an isosceles triangle. Q R
C
b) In the figure alongside, BD and CD are bisectors of ∠ABC

and ∠ACB respectively. If BD = CD, prove that DABC is an B
isosceles triangle.

8. a) In the given figure, X is the mid-point of QR, XA ⊥ PQ,
XB ⊥ PR and XA = XB. Prove that DPQR is an isosceles triangle.

Vedanta Excel in Mathematics - Book 9 220 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Geometry - Triangle

A

b) In the given figure, BN ⊥ AC, CM ⊥ AB and BN = CM. Prove that M N
DABC is an isosceles triangle. C

B

c) In the given triangle ABC, AB = AC, BP ⊥ AC and CQ ⊥ AB.

Prove that (i) BP = CQ (ii) OP = OQ.

9. a) In the figure alongside, APB and AQC are equilateral
triangles. Prove that PC = BQ.

(Hint: ∆ APC ≅ ∆ AQB, then PC = BQ)

b) In the figure alongside PABQ and ACYX are squares. P X Y
Prove that PC = BX. A

QB D
A
c) In the adjoining figure ABC is an equilateral triangle and BCDE is a E
square. Prove that AE = AD.

BC

d) In the figure alongside PQRS is a square in which PA and S R
SB intersect at O. If PA = SB, prove that PA and SB are A
perpendicular to each other at O.
O
(Hint: ∆ PBS ≅ ∆ PQA by RHS, then show that ∠SOA = ∠SPO + ∠OPB = 90°)

P BQ

Project work and activity section

10. a) Take a square sheet of paper and fold it through its one diagonal. Cut out it through
the diagonal and get two right-angled triangles. Now, fold each right-angled triangle
bisecting the vertical angle.

(i) Is the folding perpendicular to the base in each triangle?

(ii) Does the folding bisect the base in each triangle?

(iii) What conclusion can you make from these activities? Discuss in your class.

b) Draw a triangle of your own. Produce its all sides as shown in the
diagram and get three exterior angle of triangle. Show that sum of
three exterior angles of a triangle is 360°.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 221 Vedanta Excel in Mathematics - Book 9

Geometry - Triangle

13.8 Similar triangles - review
In the figure given alongside, each C

angle of DABC is respectively 30° R

equal to the corresponding angle of 30°

DPQR. Therefore, DABC and DPQR 115°
P
have the same shape and they are 115° 35° B 35° Q
said to be the similar triangle. We A

write it as DABC ~ DPQR. However,

DABC and DPQR do not have the same size. FZ

On the other hand, DDEF and DXYZ are not 85° 40°

the similar triangles. Because, none of the 110°
X
angles of these triangles are equal. D 50° 45° 30° Y
E

Following are the required conditions for the similarity between triangles. These are
also called the properties of similar triangles.

(i) When all angles of one triangle are respectively equal to the corresponding angles of
another triangle, the triangles are said to be similar.

For example A P
B
In ∆s ABC and PQR, CQ R

∠A = ∠P
∠B = ∠Q
∠C = ∠R
∴ ∆ BC ∼ ∆PQR

‘∼' is the symbol of similarity between the similar triangles.

Note: Of course, if two angles of a triangle are respectively equal to two corresponding
angles of another triangle, the remaining angles must be equal. So the triangles are said
to be similar.

(ii) When the ratios of the corresponding sides of triangles are equal, i.e. the corresponding
sides of triangles are proportional, the triangles are said to be similar.

For example K

In ∆s 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.

Vedanta Excel in Mathematics - Book 9 222 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Geometry - Triangle

Theorem 9

Two equiangular triangles are similar.

Experimental verification
Step 1: Three pairs of triangles ABC and PQR with two equal corresponding angles are

drawn.
Here, in ∆ABC and ∆PQR, ∠A = ∠P and ∠B = ∠Q.

Step 2: The remaining angles and all the corresponding sides of each pair of triangles are
measured. The ratios of corresponding sides are found.

Fig. ∠C ∠R Result ∆ABC ∆PQR Ratio of corres- Result
AB BC CA PQ QR RP ponding sides

AB BC CA

PQ QR RP

(i) ∠C = ∠R AB BC CA
PQ = QR = RP

(ii) ∠C = ∠R AB BC CA
PQ = QR = RP

(iii) ∠C = ∠R AB BC CA
PQ = QR = RP

Conclusion: As the corresponding sides of equiangular triangles are proportional, each pair
of triangles are similar.

Theorem 10

If any two corresponding sides of triangles are proportional and the angles included by
them are equal, the triangles are similar.

Experimental verification

Step 1: Three pairs of triangles ABC and PQR are drawn in such a way that in each pair

AB = BC and ∠B = ∠Q.
PQ QR

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 223 Vedanta Excel in Mathematics - Book 9

Geometry - Triangle

Step 2: The remaining angles and the remaining sides of each pair of triangles are measured.
Then the ratios of the remaining corresponding sides are found.

Fig Corresponding angles Corresponding sides AC Results
∠A ∠P Result ∠C ∠R AC PR PR
∠A = ∠P Result
(i) ∠C = ∠R AB = BC = AC
PQ QR PR

(ii) ∠A = ∠P ∠C = ∠R AB = BC = AC
PQ QR PR

(iii) ∠A = ∠P ∠C = ∠R AB = BC = AC
PQ QR PR

Conclusion: If two corresponding sides of triangles are proportional and the angle included
by them are equal, the triangles are similar.

Theorem 11

If two triangles have their corresponding sides proportional, the triangles are similar.

Experimental verification

Step 1: Three pairs of triangles ABC and PQR are drawn in such a way that in each pair

AB = BC = AC .
PQ QR PR

Step 2: The corresponding angles of each pair of triangles are measured.

Fig. ∠A ∠P Result ∠B ∠Q Result ∠C ∠R Result

(i) ∠A = ∠P ∠B = ∠Q ∠C = ∠R

(ii) ∠A = ∠P ∠B = ∠Q ∠C = ∠R

(iii) ∠A = ∠P ∠B = ∠Q ∠C = ∠R

Conclusion: If two triangles have their corresponding sides proportional, the triangles are
similar.

Worked-out examples

Example 1: In the adjoining figure, show that A y cm P 9 cm C
∆ABC ∼ ∆PQC. Then, find the lengths of 10 cm
sides marked with x and y. 6 cm Q x cm

Solution: 2 cm
B
1. In ∆ ABC and ∆ PQC,
(i) ∠ACB = ∠PCQ [Common angle]

Vedanta Excel in Mathematics - Book 9 224 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Geometry - Triangle

(ii) ∠ABC = ∠PQC [AB // PQ and corresponding angles]
(iii) ∠BAC = ∠QPC [Remaining angles of triangles]
(iv) ∆ABC ∼ ∆PQC [A.A.A. axiom]

2. AB = BC = AC [Corresponding sides of similar triangles]
PQ QC PC

or, 10 = 2 + x = 9+y
6 x 9
5 2 + x 5 9+y
or, 3 = x Also, 3 = 9
27 + 3y = 45
or, 5x = 6 + 3x or,

or, 2x = 6 or, 3y = 18

∴ x = 3 ∴ y = 6

Hence, x = 3 cm and y = 6 cm.

Example 2: In the adjoining figure, AC // DE. Prove that
AB.DE = AC.BE. Also, find the value of x.

Solution:
1. In ∆ ABC and ∆BDE,

(i) ∠ACB = ∠BDE [AC // DE and alternate angles]
(ii) ∠ABC = ∠DBE [Vertically opposite angles]
(iii) ∠BAC = ∠BED [Remaining angles of the triangles]

(iv) ∴ ∆ABC ∼ ∆BDE [A.A.A. axiom]

2. AB = AC = BC [Corresponding sides of similar triangle are proportional]
BE DE BD

or, AB.DE = AC.BE proved.

Again, AB = BC
BE BD

or, 5.6 = x
3.6 4.5

or, x = 7 cm.

Example 3: In the given figure, show that DPQR ~ DXQR. If 4 cm P
PX = 5 cm and QX = 4 cm, find the length of QR. 5 cm R

Solution: X

1. In ∆ PQR and ∆ XQR,
(i) ∠PQR = ∠XQR [Common angle]
(ii) ∠QPR = ∠XRQ [Given]
(iii) ∠PRQ = ∠QXR [Remaining angles of the triangles]
Q
(iv) ∴ ∆PQR ∼ ∆XQR [A.A.A. axiom]

2. PQ = QR [Corresponding sides of similar triangles]
QR XQ

or, 9 cm = QR Vedanta ICT Corner
QR 4 cm Please! Scan this QR code or
browse the link given below:
or, QR2 = 36 cm2 ∴ QR = 6 cm
https://www.geogebra.org/m/jkh6f87x
Hence, the length of QR is 6 cm.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 225 Vedanta Excel in Mathematics - Book 9

Geometry - Triangle

A M D
O
Example 4: In the figure alongside, ABCD is a
parallelogram, in which M is the mid-point B C
of AD, prove that: OB = 2OD.

Solution:

Given: In parallelogram ABCD, M is the mid-point of
AD.

To prove: OB = 2OD

Proof:

Statements Reasons
1.
1. In DMOD and ∆ BOC (i) AD // BC and alternate angles
(ii) Same as (i)
(i) ∠OMD = ∠OCB (A) (iii) Vertically opposite angles.
(iv) By A.A.A axiom
(ii) ∠ODM = ∠OBC (A)

(iii) ∠MOD = ∠BOC (A)

(iv) ∆MOD ~ ∆BOC

2. OM = OD = MD 2. Corresponding sides of similar triangles
OC OB BC

3. BC = AD = 2MD 3. Opposite sides of parallelogram and
given.

4. OD = MD 4. From statements (3) and (4)
OB 2MD

∴ OB = 2OD

Proved

EXERCISE 13.4

General section P A
B Q
1. a) In the given figure, ∆ABC ~ ∆APQ. If AB = 15 cm, C
AP = 10 cm and BC = 12 cm, find the length of PQ.

b) In the adjoining figure, PQ // XZ, PX = 6 cm, XY = 15 cm X
P
and YQ = 3 cm, find the length of QZ. Y
QZ

c) In the adjoining figure, ∆PQR ~ ∆ABR, QR = 20 cm, P
A
QB = 8 cm, AB = 6 cm and PR = 15 cm. Find the
B
length of PQ and AR. Q R

Vedanta Excel in Mathematics - Book 9 226 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Geometry - Triangle

10 cm P
R
d) In the adjoining figure, ∆PEM ~ ∆REM, ∠EPM = ∠RME. 8 cm
Find the length of EM.

EM

A

e) In the given figure, ∆ABC ~ ∆ABD, ∠BAD = ∠ACB, 4 cm
find the length of BD.
D8 cm C
B

Creative section - A DA
x cm
2. a) In the figure alongside AB // DE. Show that
∆ABC ~ ∆DCE and find the value of x. C
9 cm B
E
10 cm
5 cm

A6 cm 10 cm
8Dcm
b) In the adjoining figure, ∠BAD = ∠ACB. Prove that C
∆ABC ~ ∆ABD and find the lengths of AD and BD.

B

c) In the figure alongside, ∠ABC = ∠ACP. Prove that P
∆ABC ~ ∆APC and find the lengths of AB and PB.

A

d) In the given figure, AX ⊥ BC. Prove that ∆ABC ~ ∆AXC 3 cm 4 cm
and find the lengths of CX and AX. C

BX RQ
A 3 cm

e) In the adjoining figure, AB // QC, PR = 2RQ and P C
QC = 3 cm. Find the length of AP with the suitable

reasons. B

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 227 Vedanta Excel in Mathematics - Book 9

Geometry - Triangle

3. a) A boy 1.8 m tall casts the shadow of length 3 m at
2:30 p.m., what is the height of the tree which casts the
shadow of length 30 m at the same time?

1.8 m

30 m 3 m

b) The width of a LED TV of screen size 32 inch

is 27.9 inch. What is the screen size of the TV 32 in
which is 37.5 inch wide?


27.9 in 37.5 in

D

c) A man standing in front of a tower observes the top of

the tower in a mirror on the ground. The mirror is placed

at a distance of 36 ft. from the foot of the tower. If the A
man is 2.5 ft. away from the mirror, and the distance of

his eye level from the ground is 5 ft, what is the height 5 ft.

of the tower? B2.5ft.C 36 ft. E
Creative section - B

5. a) In the given figure, AB // CD. Prove that
(i) ∆AOB ~ ∆COD
(ii) AB.DO = CD.AO

b) In the adjoining figure, ∠P = ∠S. P S
Prove that R
(i) ∆POR ~ ∆QOS Q
(ii) PR.OQ = QS.OR A
(iii) PO.OQ = OR.OS
M
c) In the given right angled triangle ABC, ∠ABC = 90° and C N
NM ⊥ AC. Prove that B

(i) ∆ABC ~ ∆AMN

(ii) BC.AM = AB.MN

d) In the adjoining figure, ∠PQS = ∠PRQ. Prove that
(i) ∆PQS ~ ∆PQR
(ii) PQ2 = PS.PR

Vedanta Excel in Mathematics - Book 9 228 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Geometry - Triangle

e) In the given right-angled triangle ABC right-angled at B,
BD ⊥ AC and ∠CBD = ∠BAC. Prove that

(i) ∆ABC ~ ∆BCD (ii) ∆BCD ~ ∆ABD

(iii) BC2 = AC.CD (iv) BD2 = AD.CD

(v) BC2 = AC
BD2 AD

f) In the trapezium ABCD, AB // DC and the diagonals AC
and BD intersect each other at O.
Prove that AB.DO = CD.BO

g) In the adjoining figure, ∠ADE = ∠ ACB, and A B
∠DAC = ∠BAE. Prove that AD.BC = AC.DE DE

6. a) In the given figure, AB = DC, AB // DC, M is the C
mid-point of AB. Prove that

(i) DAOM ~ DCOD (ii) CO = 2AO

b) In the adjoining diagram, AD // BC and AD = BC. AM E
ND
Prove that
C
(i) DEBC ~ DEMN (ii) BE.MN = AD.ME

B

c) In the given figure, ABCD is a parallelogram. Prove A D
that
F
(i) DADF ~ DCEF CE
(ii) BC.EF = AF.CE

B

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 229 Vedanta Excel in Mathematics - Book 9

Geometry - Triangle

OBJECTIVE QUESTIONS

Let’s tick (√) the correct alternative.

1. The sum of interior angles of any triangle is

(A) 1 right angle. (B) 2 right angles (C) 3 right angles (D) 4 right angles

2. Which of the following measurements cannot be the side lengths of a triangle?

(A) 1 cm, 2 cm, 3 cm (B) 3 cm, 4 cm, 5 cm

(C) 7 cm, 4.5 cm, 3.8 cm (D) 6.2 cm, 10.5 cm, 8.4 cm

3. When a side of a triangle is produced to a point, then

(A) The exterior angle so formed is equal to the sum of interior angles.

(B) The exterior angle so formed is equal to the sum of any two interior angles.

(C) The exterior angle so formed is equal to the sum of two opposite interior angles.

(D) The exterior angle so formed is equal to adjacent interior angle.

4. An exterior angle of a triangle is 100° and two opposite interior angles are equal.
Then measure of each of these angle will be

(A) 100° (B) 50° (C) 80° (D) 70°

5. In a ∆ABC, if AB = 3.6 cm, BC = 5.2 cm and AC = 4.5 cm, then the smallest angle is

(A) ∠A (B) ∠B (C) ∠C (D) None

6. In ABC, if ∠A = 60o and ∠B = 70o then which of the following statements is correct?

(A) AB <BC < AC (B) BC <AC < AB (C) AC < BC < AB (D) AB <AC < AB

7. For any ∆ABC, which of the following is not the triangle inequality property?

(A) AB + BC < AC (B) BC + AC < AB (C) AC – AB > BC (D) AB – AC < AB

8. In an isosceles triangle ABC, AB = AC and AM is the median drawn from vertex A
upon base BC then which of the following statement is true?

(A) AM ⊥ BC (B) ∆ABM ≅∆ACM

(C) ∠BAM = ∠CAM (D) All of the above

9. Which of the following is NOT the criterion for congruency of triangles?

(A) S.S.S. (B) S.A.S. (C)R.H.S. (D) A.A.A.

10. In ∆PQR and ∆XYZ, PQ = QR then they will be similar if
XY YZ

(A) ∠P = ∠X (B) ∠Q = ∠ X (C) ∠R = ∠Z (D) ∠Q = ∠Y

Vedanta Excel in Mathematics - Book 9 230 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Unit 14 Geometry: Parallelogram

14.1 Special types of Quadrilaterals

The quadrilateral ABCD given alongside is a closed plane figure Q
P
bounded by four line segments called the sides of the quadrilateral.

AC is a diagonal of the quadrilateral.

DP ⊥ AC and BQ ⊥ AC are drawn.

Therefore, DP = h1, and BQ = h2 are the heights of DACD and DABC respectively.
Here, area of quadrilateral = Area of (DACD + DABC)

= 1 AC × h1 + 1 AC × h2 = 1 AC (h1 + h2)
2 2 2

Thus, area of quadrilateral = 1 diagonal × sum of perpendiculars drawn from
2

opposite vertices to the diagonal

Trapezium, parallelogram, rectangle, square, rhombus, and kite are the special types

of quadrilaterals.

Trapezium

The quadrilateral in which any two opposite sides are
parallel is called a trapezium. In the adjoining figure, PQRS
is a trapezium in which PQ // SR.

Properties of trapezium

(i) The parallel sides of a trapezium are called its bases. In the figure, PQ and SR

are the bases.

(ii) The non-parallel sides of a trapezium are called its legs. In the figure, PS and

QR are the legs.

(iii) The straight line segment that joins the mid-points of legs is called the median

of the trapezium. XY is the median.

(iv) The perpendicular distance between the parallel sides (bases) of a trapezium is

called its altitude (height). h is the height of the trapezium.

(v) Area of trapezium = 1 Height × Sum of the bases = 1 h (PQ + SR)
2 2

Parallelogram

The quadrilateral in which opposite sides are parallel
is called a parallelogram. In the figure, ABCD is a
parallelogram.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 231 Vedanta Excel in Mathematics - Book 9

Geometry - Parallelogram

(i) The opposite sides of a parallelogram are equal.

(ii) The opposite angles of a parallelogram are equal. h
b
(iii) The diagonals of a parallelogram bisect each other.

(iv) The area of a parallelogram = base × height

= b × h.

Rectangle, square, and rhombus are some special types of parallelograms. Let's
discuss the important properties of these special types of parallelograms.

Rectangle

A rectangle is a parallelogram in which each angle is 90°.

(i) The opposite sides of a rectangle are equal.

(ii) The diagonals of a rectangle are equal.

(iii) The diagonals of a rectangle bisect each other.

(iv) The area of a rectangle = length × breadth

= l × b.

Square

A parallelogram in which all sides are equal and each angle is 90° is called a square.

(i) The diagonals of a square are equal. l

(ii) Each diagonal bisects the vertical angles.

(iii) Diagonal bisect each other perpendicularly.

(iv) The triangles formed by the diagonals are congruent.
1
(v) The area of a square = (side)2 = l2 or = 2 (diagonal)2

Rhombus

A rhombus is a parallelogram in which all sides are equal.

(i) The opposite angles of a rhombus are equal. d1
d2
(ii) Each diagonal bisects the vertical angles.

(iii) The diagonals are not equal and bisect each other

perpendicularly.

(iv) The triangles formed by the diagonals are congruent.

(v) The area of a rhombus = 1 × product of diagonals
2

= 1 × d1 × d2
2

Vedanta Excel in Mathematics - Book 9 232 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Geometry - Parallelogram

Theorem 13

The straight line segments that join the ends of two equal and parallel line segments
towards the same sides are also equal and parallel.

Given: AB = CD and AB // CD. The ends A, C and B, D
are joined.

To prove: AC = BD
AC // BD

Construction: B and C are joined.

Proof

Statements Reasons
1. In ∆s ABC and BCD
1.

(i) AB = CD (S) (i) Given
(ii) AB // CD and alternate angles
(ii) ∠ABC = ∠BCD (A) (iii) Common side
(iv) S. A. S. axiom
(iii) BC = BC (S)

(iv) ∴ ∆ABC ≅ ∆BCD

2. AC = BD 2. Corresponding sides of congruent triangles

3. ∠ACB = ∠CBD 3. Corresponding angles of congruent triangles

4. AC // BD 4. Alternate angles being equal

Proved

Theorem 14

The straight line segments that join the ends of two equal and parallel line segments
towards the opposite sides bisect each other.

Given: AB = CD and AB // CD. The opposite ends A, D and
B, C are joined. Let, AD and BC intersect at O.

To prove: AD and BC bisect each other at O.
i.e. AO = OD and BO = OC

Proof

Statements Reasons

1. In ∆s AOB and COD 1.
(i) AB // CD and alternate angles
(i) ∠ABO = ∠OCD (A) (ii) Given
(iii) AB // CD and alternate angles
(ii) AB = CD (S) (iv) A. S. A. axiom

(iii) ∠BAO = ∠ODC (A) 2. Corresponding sides of congruent triangles

(iv) ∴ ∆AOB ≅ ∆COD

2. AO = OD and BO = OC

3. AD and BC bisect each other at O. 3. From statement (2)

Proved

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 233 Vedanta Excel in Mathematics - Book 9

Geometry - Parallelogram

Theorem 15

The opposite angles and sides of a parallelogram are equal.

Given: ABCD is a parallelogram in which AB // DC and
AD // BC.

To prove: (i) ∠ABC = ∠ADC, ∠BAD = ∠BCD
(ii) AB = DC, AD = BC

Construction: Diagonal AC is drawn.

Proof

Statements Reasons
1.
1. In ∆s ABC and ACD (i) AB // DC and alternate angles

(i) ∠BAC = ∠ACD (A) (ii) Common side

(ii) AC = AC (S) (iii) AD // BC and alternate angles

(iii) ∠ACB = ∠CAD (A) (iv) A. S. A. axiom
2. Corresponding angles of congruent triangles
(iv) ∴ ∆ABC ≅ ∆ACD 3. Drawing the diagonal BD and same as above

2. ∠ABC = ∠ADC in ∆s ABD and BCD
4. Corresponding sides of congruent triangles
3. ∠BAD = ∠BCD
Proved.
4. AB = DC and AD = BC

Converse (I) Theorem 15

If the opposite sides of a quadrilateral are equal, the quadrilateral is a parallelogram.
Given: ABCD is a quadrilateral in which AB = DC and

AD =BC.
To prove: ABCD is a parallelogram, i.e. AB // DC and AD // BC.
Construction: Diagonal AC is drawn.
Proof

Statements Reasons
1.
1. In ∆s ABC and ACD (i) Given
(ii) Given
(i) AB = DC (S) (iii) Common side
(iv) S. S. S. axiom
(ii) BC = AD (S) 2. Corresponding angles of congruent triangles

(iii) AC = AC (S) 3. From statement (2), alternate angles being
equal
(iv) ∴ ∆ABC ≅ ∆ACD
4. Opposite sides are parallel
2. ∠BAC = ∠ACD and Proved

∠ACB = ∠CAD

3. AB // DC and AD // BC

4. ABCD is a parallelogram

Vedanta Excel in Mathematics - Book 9 234 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Geometry - Parallelogram

Converse (II) of Theorem 15

If the opposite angles of a quadrilateral are equal, the quadrilateral is a parallelogram.

Given: ABCD is a quadrilateral in which ∠ABC = ∠ADC and
∠BCD = ∠BAD.

To prove: ABCD is a parallelogram, i.e., AB // DC and AD // BC.
Proof

Statements Reasons

1. ∠A + ∠B + ∠C + ∠D = 360° 1. Sum of the angles of a quadrilateral
2. ∠A + ∠B + ∠A + ∠B = 360° 2. Given, ∠A = ∠C and ∠B = ∠D.

or, 2 (∠A + ∠B) = 360°

or, ∠A + ∠B = 180° 3. Being the sum of co-interior angles 180°
3. ∴ AD // BC 4. Same as above
4. Similarly, ∠A + ∠D = 180° 5. Same as reason (3)
5. ∴ AB // DC 6. Opposite sides are parallel
6. ABCD is a parallelogram

Theorem 16 Proved

The diagonals of a parallelogram bisect each other.

Given: ABCD is a parallelogram. Diagonals AC and BD intersect
at O.

To prove: AC and BD bisect each other at O, i.e., AO = OC and
BO = OD.

Proof

Statements Reasons
1.
1. In ∆s AOB and COD (i) AB // DC and alternate angles
(ii) Opposite sides of a parallelogram
(i) ∠OAB = ∠OCD (A) (iii) AB // DC and alternate angles
(iv) A. S. A. axiom
(ii) AB = DC (S) 2. Corresponding sides of congruent triangle.
3. From statement (2)
(iii) ∠OBA = ∠ODC (A)
Proved
(iv) ∴ ∆AOB ≅ ∆COD
2. AO = OC and BO = OD
3. AC and BD bisect each other at O.

Converse of Theorem 16

If the diagonals of a quadrilateral bisect each other, the quadrilateral is a parallelogram.

Given: ABCD is a quadrilateral in which diagonals AC and
BD bisect each other at O.
∴ AO = OC and BO = OD

To prove: ABCD is a parallelogram, i.e., AB // DC, AD // BC,
AB = DC, AD = BC.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 235 Vedanta Excel in Mathematics - Book 9

Geometry - Parallelogram

Proof

Statements Reasons
1. In ∆s AOB and COD
1.
(i) AO = OC (S) (i) Given
(ii) Vertically opposite angles
(ii) ∠AOB = ∠COD (A) (iii) Given
(iv) S. A. S. axiom
(iii) BO = OD (S) 2. Corresponding sides of congruent triangle
3. Corresponding angles of congruent triangles
(iv) ∴ ∆ AOB ≅ ∆ COD 4. From statements (3), alternate angles are equal
2. AB = DC 5. AD and BC join the ends of two equal and parallel
3. ∠OAB = ∠OCD
4. AB // DC lines towards the same side.
5. AD = BC and AD // BC 6. Opposite sides are equal and parallel.

6. ABCD is a parallelogram Proved

Worked-out examples

Example 1: In the given figure, PQRS is a parallelogram. P T S
If PT = SR, ∠ PSR = 4x° and ∠ QPT = x°, find the x° 4x°
value of ∠ QRS.
Q R
Solution:

(i) ∠PQR = ∠PSR = 4x° [Opposite angles of parallelograms are equal]

(ii) PQ = PT [ PQ = SR and PT = SR]

(iii) ∠PQT + ∠PTQ = 4x° [ PQ = PT]

(iv) ∠PQT + ∠PTQ + ∠QPT = 180° [Sum of angles of DPQT]

or, 4x° + 4x° + x° = 180°

or, 9x° = 180°

∴ x = 20°

(v) ∠QRS = 180° – ∠PQR [ PQ // SR and co-interior angles]

= 180° – 4 × 20° = 100° A D
35° C
Example 2: In the figure alongside, ABCD is a rhombus.
If ∠DAC = 35°, find the measure of ∠ABC. B

Solution:
(i) ∠DAC = ∠ACD = 35° [AD = CD, sides of rhombus]
(ii) ∠ADC + ∠DAC + ∠ACD = 180° [Sum of angles of DADC]
or, ∠ADC + 35° + 35° = 180°

∴ ∠ADC = 110°

(iii) ∠ABC = ∠ADC [Opposite angles of rhombus]

∴ ∠ABC = 110°

Vedanta Excel in Mathematics - Book 9 236 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Geometry - Parallelogram

Example 3: In the figure alongside, WXYZ is a square. ZY
If ∠ZNX = 115°, find the sizes of ∠WZN and ∠ZMY.
M
Solution: 115°

(i) ∠WZN + ∠ZWN = ∠ZNX [∠ZNX is the exterior angle in DZWN] W NX

or, ∠WZN + 90° = 115°

∴ ∠WZN = 25°

(ii) ∠ZWY = ∠XWY=45° [Diagonal of square bisects its vertical angle]

(iii) ∠ZMY = ∠WZM+∠ZWM [In DWZM, ∠ZMY is the exterior angle]

∴ ∠ZMY = 25° + 45° = 70°

Example 4: In the figure, ABCD is a parallelogram. The A Q
diagonal BD is produced in either sides to D

the points P and Q such that BP = DQ. Prove

that: B C
(i) ∠BCP = ∠DAQ P
(ii) AQ // PC.
Solution:

Given: In parallelogram ABCD, the diagonal BD is produced to the points P and Q

such that BP = DQ.

To prove: (i) ∠BCP = ∠DAQ (ii) AQ // PC.

Proof

Statements Reasons
1. AD // BC, alternate angles
1. ∠DBC = ∠ADB 2. Supplements of equal angles, ∠DBC = ∠ADB
3.
2. ∠PBC = ∠ADQ (i) Given
3. In DBPC and DADQ (ii) From statement(2)
(i) BP = DQ (S) (iii) Opposite sides of parallelogram ABCD
(ii) ∠PBC = ∠ADQ (A) (iv) By S.A.S, axiom
(iii) BC = AD (S) 4. Corresponding angles of congruent triangles
(iv) DBPC ≅ DADQ
4. ∠BCP = ∠DAQ and 6. From statements (4), alternate angles ∠BPC
and ∠AQD are equal.
∠BPC = ∠AQD Proved
5. AQ // PC

Example 5: In the figure alongside, ABCD is a parallelogram.
If 2MO = OD, prove that M is the mid-point of BC.

Solution:

Given: ABCD is a parallelogram in which BC // AD and
To prove: AB // DC, 2 MO = OD
M is the mid-point of BC, i.e. BC = 2 MC

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 237 Vedanta Excel in Mathematics - Book 9

Geometry - Parallelogram

Proof: Statements Reasons

1. In ∆ AOD and ∆ MOC 1.
(i) (i) AD // BC and alternate angles
(ii) ∠ADO = ∠CMO (ii) Vertically opposite angles
(iii) ∠AOD = ∠COM (iii) AD // BC and alternate angles
(iv) ∠DAO = ∠MCO (iv) A.A.A. axiom
2. (i) ∆AOD ~ ∆MOC
2. (i) Corresponding sides of similar triangles
AD = OD = AO
MC MO CO

(ii) BC 2MO (ii) AD = BC (opposite sides of parallelogram)
MC = MO and OD = 2MO (given)

or, BC = 2 MC (iii) From statement (ii)
(iii) M is the mid-point of BC

Proved

EXERCISE 14.1 A B
C
General section D
P Q
1. a) In the figure alongside, AB // DC and AB = DC. Write down
the relation between AD and BC.

b) In the given figure, PQ // RS and PQ = RS. What is the relation

between PS and QR? R S
T
B
S
c) In the figure alongside, BEST is a parallelogram. Write down Y
the relation between BT and ES, BE, and TS.

E
Z

d) In the parallelogram WXYZ, diagonals, ZX and WY intersect O X
at O. Write the relation between OW, OY and OZ, OX. W D

2. a) In the given figure, ABCD is a rectangle. AC and BD are A O C
diagonals. If the length of diagonal AC is 10 cm, what is the B P S
length of diagonal BD?

b) In the square PQRS given alongside, what is the value of ∠QPR?

Q R
R D

c) In the given figure, READ is a rhombus. If ∠RED = 40°, what is the A
measure of ∠RDE?

E

Vedanta Excel in Mathematics - Book 9 238 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Geometry - Parallelogram

d) In the given rectangle WXYZ, O is the point of intersection W Z

of its diagonal WY and XZ. If ∠XOY = 120°, find the size of O
∠OWZ.
Y
X
3. Calculate the size of unknown angles:

a) A D b) P Q c) L E d) A 60° D
x
3x y a 5y p+q 2p–50°

2x zw E b 4y p+10° r O 2x
B C S R I K B CE

e) L D f) F g) P T S h) A F
b x y
x A DE 100°
50° E
y 115° 55° x 110° D
E TA
aa R 60° C
BCQ B

4. a) In the given figure, ABCD is a parallelogram.
If ∠BCD = 125° and AB = AE, find the size of
∠BAE.

b) In the figure alongside, DBCE is a parallelogram. 110°
If ∠AFE =110° and ∠FCE = 40°, find the value
of ∠DBC.

c) In the adjoining figure, PQRS is a parallelogram. H ND25°
Find the value of ∠SMR. A
M
d) In the given figure, HEAD is a parallelogram. If E
AM ⊥ HE, AN ⊥ HD and ∠NAD = 25°, find the
measure of ∠MAE and ∠MAN.

5. a) In the given figure, ABCD is a rhombus. If ∠ADB = 40°,
find the size of ∠CDE.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 239 Vedanta Excel in Mathematics - Book 9

Geometry - Parallelogram

b) In the given figure, ABCD is a rhombus. If
∠BCD = 114°, find the measure of ∠ABD.

c) In the adjoining figure, ABCD is a rhombus. If CD
∠OAB = 30°, find the measures of ∠ODC and
∠ABC. O
30°
B A
S
P 124°
d) In the given figure, PQRS is a rectangle. Q O
If ∠POS = 124°, find the sizes of ∠PQO and
∠OQR. R
SR
e) In the figure alongside, PQRS is a square. If ∠SMR = 70°,
find the measures of ∠PST and ∠STQ. 70°
M

f) In the given figure, FIVE is a square. If PQ // FV and PTQ
∠QPR = 75°, find the measure of ∠FIR and ∠FPI. F PE

75°

RQ
M

I V
S
P
55° M
g) In the given figure, PQRS is a rhombus and SRM RN
is an equilateral triangle. If SN ⊥ RM and
∠PRS = 55°, find the size of ∠QSN.

Q

6. a) In the given figure, ABCD is a parallelogram. If AB = 3x cm, A 19 cm D3x cm12 cm
BC = (5y – 1) cm, CD = 12 cm, and AD = 19 cm, find the B (5y – 1) C
values of x and y.

b) In the figure alongside PQRS is a parallelogram. If (x – y) cmP (x + y) cm S 10 cm
PQ = (x – y) cm, QR = 20 cm, RS = 10 cm, and Q 20 cm R
PS = (x + y) cm, find the value of x and y.

Vedanta Excel in Mathematics - Book 9 240 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Geometry - Parallelogram

L P
5cm
c) In the adjoining figure, LMNP is a rhombus. If
MN = (2a – 3) cm, OM = (3b + 2) cm, OP = 5 cm, and (3b+2)Ocm
perimeter = 28 cm, find the values of a and b.
M (2a – 3)cm N

d) In the adjoining figure, PQRS is a rectangle. If S R
OP = (2x – 3) cm and OR = (x + 1) cm, find the length (x+1)cm Q
of diagonal QS.
(2x–3)cm O
P

e) RACE is a rectangle in which diagonal RC = 18 cm, R E
OA = (p + q) cm and OE = 3p cm, find the values of p 3p cm C
and q.
(p+q)cmO
A

DC

7. a) In the adjoining figure ABCD is a square and ABE is an E
equilateral triangle. Find the measure of ∠ADE and ∠DCE.

A A B
B E
b) In the given figure, ABCD is a square and BEC is an
equilateral triangle. Find ∠AEB and ∠DAE.

Creative section -A D C
P
8. a) In the adjoining figure, PQ = AB and PQ // AB. Q

Prove that (i) AP = BQ (ii) AP // BQ.

AB

b) In the given quadrilateral, AB = DC and BC = AD. A D

Prove that the quadrilateral ABCD is a parallelogram.

BC

c) In the figure alongside, diagonals AC and BD of the D C
quadrilateral bisect each other at O. Prove that ABCD is a O B
parallelogram.

A

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 241 Vedanta Excel in Mathematics - Book 9

Geometry - Parallelogram

9. a) In the given figure, ABCD is a parallelogram. AP bisects
∠A. Prove that DP = BC.

E P M

b) In the given figure, EXAM is a parallelogram, the A
bisector of ∠A meets the mid-point of EM at P. Prove that B
AX = 2AM.
C
X
A

c) In the adjoining figure, ABCD is a rhombus in which CD is

produced to E such that CD = DE. Prove that ∠EAC = 90°.

E D

Creative section -B

10. a) ABCD is a parallelogram. P and Q are two points on the
diagonal BD such that DP = QB. Prove that APCQ is a
parallelogram.

b) ABCD is a parallelogram. DE ⊥ AC and BF ⊥ AC. Prove
that BEDF is a parallelogram.

P S
M
c) In parallelogram PQRS, the bisectors of ∠PQR and
∠PSR meet the diagonal at M and N respectively. Prove N
d) that MQNS is a parallelogram.

11. a) QR
b)
c) AD
d)
In the given figure, ABCD is a parallelogram. If P and Q

Q are the points of trisection of diagonal BD, prove that P
PAQC is a parallelogram.

BC

Prove that the diagonals of a rectangle are equal.

If the diagonals of a rhombus are equal, prove that it is a square.

Prove that the diagonals of a rhombus bisect each other perpendicularly.

Prove that the diagonal of a parallelogram divides it into two congruent triangles.

Vedanta Excel in Mathematics - Book 9 242 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Geometry - Parallelogram

Project work and activity section

12. a) Take two rectangular sheets of paper of the same size. Fold one sheet through one
diagonal and another sheet through other diagonal. Then, cut out each sheet of
papers through diagonals.

(i) Are two diagonals equal? (ii) Do these diagonals bisect each other?

b) Take a square sheet of paper and fold it through both diagonals.

(i) Are two diagonals equal?

(ii) Do these diagonals bisect each other perpendicularly?

OBJECTIVE QUESTIONS

Let’s tick (√) the correct alternative.
1. The properties of the quadrilaterals are given below.

(i) The opposite sides are equal. (ii) The opposite angles are equal

(iii) The diagonal bisect each other (iv) The diagonals are equal

(v) The diagonal bisect each other at a right angle

Which of the above are the properties of a parallelogram?

(A) (i), (ii) and (iv) (B) (i), (iv) and (v)

(C) (i), (ii) and (iii) (D) (i), (ii) and (v)

2. Which of the following statements is correct?

(A) Every square is a rectangle. (B) Every parallelogram is a rhombus.

(C) Every parallelogram is a rhombus. (D) Every rectangle is a square.

3. The line segments that join the ends of two equal and parallel line segments towards

the same sides are

(A) parallel (B) equal (C) parallel and equal (D) bisect each other

4. The line segments that join the ends of two equal and parallel line segments towards

the opposite sides bisect each other.

(A) equal (B) parallel (C) parallel and equal (D) equal and perpendicular

5. The opposite angles of a parallelogram are

(A) equal (B) unequal (C) complementary (D) supplementary

6. The opposite sides of a parallelogram are

(A) equal (B) parallel (C) Both (A) and (B) (D) None

7. The quadrilateral with equal opposite sides is always a

(A) parallelogram (B) square (C) rectangle (D) rhombus

8. The quadrilateral with equal opposite angles is a

(A) parallelogram (B) square (C) rectangle (D) rhombus

9. The diagonals of a parallelogram are

(A) equal (B) unequal (C) bisect each other (D) perpendicular to each other

10. The angles made by the diagonals with a side of rhombus are

(A) equal (B) complementary (C) supplementary (D) unequal

Vedanta ICT Corner
Please! Scan this QR code or
browse the link given below:

https://www.geogebra.org/m/z5w6dnby

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 243 Vedanta Excel in Mathematics - Book 9

Unit 15 Geometry: Construction

15.1 Construction of quadrilaterals C

A quadrilateral has ten parts in all: four sides, four angles, and two D B
diagonals. To construct a quadrilateral, we shall usually need data about A
five specified parts of it. However, in the case of regular quadrilaterals
(square, rectangle, parallelogram, rhombus, etc.), the measurements of a
few required number of parts may be sufficient.

Of course, to construct any type of quadrilateral, we should know its properties and we
apply these properties in its construction.

Let’s study the steps of construction of different quadrilaterals under the following conditions.

A. Construction of rhombus

Let’s construct the rhombus under the following conditions.

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

Example: Construct a rhombus ABCD in which AB = 4.5 cm and ∠ ABC = 45°.

Steps of construction D 4.5 cm X
(i) Draw AB = 4.5 cm
C
(ii) At B, construct ∠ABX = 45°. 4.5 cm 4.5 cm

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

(iv) With centres at A and C and A 45°
radius 4.5 cm, draw two arcs B
intersecting each other at D.

(v) Join A, D and C, D. Thus, ABCD is the required rhombus.

2. When two diagonals are given D
O
Example: Construct a rhombus ABCD in which diagonals
AC = 5.4 cm and BD = 4.2 cm.

Steps of construction

(i) Draw AC = 5.4 cm and draw its A C
perpendicular bisector XY. Mark
the mid-point of AC as O.

(ii) With centre at O and radius 2.1 cm B

(12 of BD), draw two arcs to cut OX at D and OY at B.
(iii) Join A, B; B, C; C, D and D, A.

Thus, ABCD is the required rhombus.

Y

Vedanta Excel in Mathematics - Book 9 244 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Geometry - Construction

B. Construction of scalene quadrilateral
Let’s observe the shape and size of the following quadrilateral having side lengths
3 cm, 5 cm, 7 cm and 4 cm. Do the quadrilaterals with same side lengths have exactly
the same shape and size?
A D
B 3 cm C
7 cm

D

5 cm 4 cm B 7 cm

4 cm 4 cm 5 cm

3 cm 5 cm

C 7 cm D A 3 cm B A C

Each interior angle of convex quadrilateral is less than 180o. However, at least one of
the interior angles of concave quadrilateral is more than 180o. Thus, the shape of the
quadrilaterals having same measurements may have different shapes.

XX

3 cm C C
D
2.8 cm 3.4 cm3 cm
A 60°3.4 cm
4.5 cm 2.8 cm D
B A 4.5 cm 60° B

Concave quadrilateral Convex quadrilateral

1. When all four adjacent sides and one of the diagonals are given

Example: Construct a quadrilateral ABCD in which AB = 5.4 cm, BC = 5.1 cm,
CD = 4.9 cm, AD = 6.2 cm and the diagonal BD = 5.8 cm.

Solution:

Here, AB = 5.4 cm, BC = 5.1 cm, CD = 4.9 cm, AD = 6.2 cm and the diagonal
BD = 5.8 cm

Steps of construction D
(i) Draw AB = 5.4 cm. 4.9 cm
(ii) From A, draw an
C
arc with radius
AD = 6.2 cm and 6.2 cm
from B draw another 5.8 cm
arc with radius 5.1 cm
BD = 5.8 cm. These
two arcs intersect
each other at D.

(iii) From B, draw an
arc with radius BC
= 5.1 cm and from
D, draw another arc A 5.4 cm B

with radius DC = 4.9 cm. These two arcs intersect to each other at C. Join A, D;
B, C and D, A.

Thus, ABCD is the required quadrilateral.
2. When all four sides and one of the angles between adjacent sides are given

Example: Construct a quadrilateral ABCD in which AB = 5.2 cm, BC = 5 cm,
CD = 4.2 cm, AD = 4 cm and ABC = 60°.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 245 Vedanta Excel in Mathematics - Book 9

Geometry - Construction

Solution:

Here, AB = 5.2 cm, BC = 5 cm, CD = 4.2 cm, AD = 4 cm and ∠ABC = 60°.

Steps of construction C
(i) Draw AB = 5.2 cm.
(ii) At B construct ∠ABX = 60°. D 4.2 cm

(iii) With the centre at B and radius 5 cm 4 cm
draw an arc to cut BX at C. 5 cm

(iv) From A, draw an arc with radius 4 cm
and from C, draw another arc with
radius 4.2 cm. These two arcs intersect
to each other at D. Join C, D and A, D.

Thus, ABCD is a required quadrilateral. A 5.2 cm B

3. When any three adjacent sides and two diagonals are given

Example: Construct a quadrilateral ABCD in which AB = 4 cm, BC = 5.5 cm,

Solution: DA = 3.4 cm, diagonal AC = 7.2 cm and diagonal BD = 5.8 cm.

Here, AB = 4 cm, BC = 5.5 cm, DA = 3.4 cm, diagonal AC = 7.2 cm and diagonal
BD = 5.8 cm.
Steps of construction
(i) Draw AB = 4 cm. C

(ii) From A, draw an arc with radius D 7.2 cm 5.5 cm
7.2 cm and from B, draw another arc
with radius 5.5 cm. These two arcs 3.4 cm 5.8 cm
intersect to each other at C. Join C to
A and B A 4 cm B

(iii) Also, from A, draw an arc with radius
3.4 cm and from B, draw another arc
with radius 5.8 cm. These two arcs
intersect to each other at D. Join A, D;
B, D and C, D.


Thus, ABCD is a required quadrilateral.

4. When any three adjacent sides and two angles are given

Example: Construct a quadrilateral ABCD in which AB = 5.2 cm, BC = 4.8 cm,
DA = 4.1 cm, ABC = 75o and ∠BAD = 60o

Solution:
Here, AB = 5.2 cm, BC = 4.8 cm, DA = 4.1 cm, ∠ABC = 75o and ∠BAD = 60o.
Steps of construction
(i) Draw AB = 5.2 cm.
(ii) At A construct ∠BAX = 60°.

Vedanta Excel in Mathematics - Book 9 246 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur