Lesson: Triangles

Exercise 7.1 (8)

Question: 1

In quadrilateral ACBD,

AC AD

and AB bisects

A

(See Fig. 7.16). Show that

ABC ABD

. What

can you say about BC and BD?

Solution:

Given:

In quadrilateralACBD,

AC AD

and AB bisects

A

We need to prove that

ABC ABD

Now, in

ABC

and

ABD

,

AB AB

(Common)

CAB DAB

(AB is the angle bisector)

AC AD

(Given)

Therefore,

ABC ABD

by SAS congruence

condition.

BC BD

(By CPCT)

Question: 2

ABCD is a quadrilateral in which AD=BC and

DAB CBA

(see Fig7.17). Prove that

(i)

ABD BAC

(ii)

BD AC

(iii)

.ABD BAC

Solution:

Given:

AD BC

and

DAB CBA

(i)In

ABD

and

,BAC

AB BA

(Common)

DAB CBA

(Given)

AD BC

(Given)

Therefore,

ABD BAC

by SAS congruence

condition.

(ii) Since,

ABD BAC

Therefore, BD=AC(By CPCT)

(iii) Since,

ABD BAC

Therefore,

ABD BAC

(By CPCT)

Question: 3

AD and BC are equal perpendiculars to a line segment

AB (see Fig. 7.18). Show that CD bisects AB.

Solution:

Given:

AD AB

and

.BC AB

To Prove:

OA OB

Proof:

In and ,AOD BOC

A B

(As

AD AB

and

BC AB

)

AOD BOC

(Vertically opposite angles)

AD=BC (Given)

Therefore,

AOD BOC

(By AAS congruence

condition)

This implies,

AO OB

(By CPCT)

or CD bisects AB.

Question: 4

l and m are two parallel lines intersected by another

pair of parallel lines p and q (See Fig. 7.19). Show that

.ABC CDA

Solution:

Given:

l m

and

p q

To prove:

ABC CDA

Proof:

In

ABC

and

,CDA

BCA DAC

(Alternate interior angles)

AC CA

(Common)

BAC DCA

(Alternate interior angles)

Therefore,

ABC CDA

(By ASA congruence

condition.)

Question: 5

Line l is the bisector of angle

A

and B is any point

on l. BP and BQ are perpendiculars from B to the arms

of

A

(see Fig. 7.20). Show that:

(i)

APB AQB

(ii)BP=BQ or B is equidistant from the arms of

.A

Solution:

Given:

l is the bisector of angle

A

.

BP and BQ are perpendiculars on arms of

A

.

(i)

In and ,APB AQB

P Q

(Right angles)

BAP BAQ

(l is the bisector)

AB AB

(Common)

Therefore,

APB AQB

by AAS congruence

condition.

(ii)

BP BQ

by CPCT. Therefore, B is equidistant

from the arms of

A

.

Question: 6

In Fig. 7.21,

,AC AE AB AD

and

BAD EAC

.

Show that

.BC DE

Solution:

Given:

,AC AE AB AD

and

BAD EAC

We need to prove that

BC DE

Now,

BAD EAC

(Given)

Adding

DAC

on both the sides, we get

BAD DAC EAC DAC

BAC EAD

In

ABC

and

,ADE

AC AE

(Given)

BAC EAD

(As proved above)

AB AD

(Given)

Therefore,

ABC ADE

by SAS congruence condition.

BC DE

(By CPCT).

Question: 7

AB is a line segment and P is its midpoint. D and E are

points on the same side of AB such that

BAD ABE

and

EPA DPB

(See Fig. 7.22).

Show that:

(i)

DAP EBP

(ii)

AD BE

Solution:

Given:

P is the midpoint of AB. i.e.,

.AP PB

BAD ABE

and

EPA DPB

…(i)

Since,

EPA DPB

Adding

DPE

both the sides

EPA DPE DPB DPE

DPA EPB

Now, In

DAP

and

,EBP DPA EPB

(As proved above)

AP BP

(Given)

BAD ABE

(Given)

Therefore,

DAP EBP

by ASA Congruence condition.

(ii)

AD BE

by CPCT.

Question: 8

In right triangle ABC, right angled at C, M is the

midpoint of hypotenuse AB.C is joined to M and

produced to a point D such that

.DM CM

Point D

is joined to point B. Show that:

(i)

AMC BMD

(ii)

DBC

is a right angle.

(iii)

DBC ACB

(iv)

1

CM = AB

2

Solution:

Given,

90 ,MC

is the midpoint of AB i.e.,

AM MB

and

DM CM

(i)In

AMC

and

,BMD

AM BM

(Given)

CMA DMB

(Vertically opposite angles)

CM DM

(Given)

Therefore,

ΔAMC ΔBMD

by SAS congruence

condition.

i ..AC BD

(By CPCT)

(ii)

ACM BDM

(By CPCT)

Therefore,

||AC BD

as alternate interior angles are

equal.

Now,

180ACB DBC

(Co-interiors angles)

90 180DBC

90 DBC

(iii)In

DBC

and

,ACB

BC=CB (Common)

ACB DBC

(Right angles)

DB AC

(Using (i))

Therefore,

DBC ACB

by SAS congruence condition.

iiDC AB

(By CPCT)

(iv)DC=AB (Using (ii))

DM CM AB

AsCM CM AB DM CM

2

2

CM AB

AB

CM

Exercise 7.2 (8)

Question: 1

In an isosceles triangle ABC, with

,AB AC

the

bisectors of

B

and

C

intersect each other at O.

Join A to O. Show that:

(i)

OB OC

(ii)AO bisects

A

Solution:

Given,

,AB AC

the bisectors of

B

and

C

intersect each

other at O.

(i) Since, ABC is an isosceles with

,AB AC

B C

1 1

2 2

B C

(As OB and OC are angle bisectors)

OBC OCB

OB OC

(Side opposite to the equal angles are

equal.)

(ii)

In and ,AOB AOC

AB AC

(Given)

AO AO

(Common)

OB OC

(As above proved)

Therefore,

AOB AOC

by SSS congruence

condition.

BAO CAO

By CPCT

Thus, AO bisects

A

.

Question: 2

In

,ABC AD

is the perpendicular bisector of BC (see

Fig. 7.30). Show that

ABC

is an isosceles triangle in

which

.AB AC

Solution:

Given,

AD is the perpendicular bisector of BC.

To show,

AB AC

Proof,

In

ADB

and

,ADC

AD AD

(Common)

ADB ADC

BD CD

(AD is the perpendicular bisector)

Therefore,

ADB ADC

by SAS congruence

condition.

AB=AC (By CPCT).

Question: 3

ABC

is an isosceles triangle in which altitudes BE

and CF are drawn to equal sides AC and AB

respectively (see Fig. 7.31). Show that these altitudes

are equal.

Solution:

Given,

in ABC

BE and CF are altitudes.

AC AB

To show,

BE=CF

Proof:

In and ,AEB AFC

A A

(Common)

AEB AFC

(Right angles)

AB AC

(Given)

Therefore,

AEB AFC

by AAS congruence

condition.

Thus,

BE CF

by CPCT.

Question: 4

∆ABC is a triangle in which altitudes BE and CF to

sides AC and AB are equal (See Fig. 7.32). Show that:

(i)

ABE ACF

(ii)

,AB AC

i.e., ABC is an isosceles triangle.

Solution:

Given,

In

, &ABC BE CF

are two altitudes such that,

BE CF

(i)

In and ,ABE ACF

A A

(Common)

AEB AFC

(Right angles)

BE=CF(Given)

Therefore,

ABE ACF

by AAS congruence

condition.

(ii) Thus,

AB AC

by CPCT and therefore ABC is an

isosceles triangle.

Question: 5

andABC DBC

are two isosceles triangles on the

same base BC (See Fig. 7.33). Show that

.ABD ACD

Solution:

Given,

ABC

and

DBC

are two isosceles triangles.

To show:

.ABD ACD

Proof:

Let us join AD

In

ABD

and

,ACD

AD AD

(Common)

AB AC

(

ABC

is an isosceles triangle.)

BD CD

(

BCD

is an isosceles triangle.)

Therefore,

ABD ACD

(By SSS congruence condition).

Thus,

ABD ACD

by CPCT.

Question: 6

ΔABC is an isosceles triangle in which

AB AC

.

Side BA is produced to D such that

AD AB

(See Fig.

7.34). Show that

BCD

is a right angle.

Solution:

Given,

AB AC

and

AD AB

AC AD

To show:

BCD

is a right angle.

Proof:

In ,ABC

AB AC

(Given)

ACB ABC i

(Angles opposite to the equal

sides are equal.)

In ,ACD

AC AD

(Given)

CDA DCA ii

(Angles opposite to the equal

sides are equal.)

Now,

In ,BCD

180

180

DBC BCD CDB

ABC BCD CDA

180ACB BCD DCA

(From (i) and (ii))

180

180

2 18

9

( )

0

0

ACB DCA BCD

BCD BCD

BCD

BCD

Hence, it is proved.

Question: 7

ABC is a right angled triangle in which

90A

and

.AB AC

Find

B

and

C

Solution:

Given,

90A

andAB AC

B C

(Angles opposite to the equal sides are

equal.)

Now,

180A B C

(Sum of the interior angles of a

triangle is 180

o

.)

90 2 180

2 90

45

B

B

B

Thus,

45B C

Question: 8

Show that the angles of an equilateral triangle are

60

each.

Solution:

Let ABC be an equilateral triangle.

BC AC AB

(Length of all sides is same)

A B C

(Angles opposite to the equal sides

are equal.)

Also,

180

3 180

60

A B C

A

A

Therefore,

60A B C

Thus, the angles of an equilateral triangle are

60

each.

Exercise 7.3 (5)

Question: 1

ABC

and DBC

are two isosceles triangles on the

same base BC and vertices A and D are on the same

side of BC (see Fig. 7.39). If AD is extended to BC

such that it intersects BC at P, show that:

(i)

ABD ACD

(ii)

ABP ACP

(iii)AP bisects

A

as well as

.D

(iv)AP is the perpendicular bisector of BC.

Solution:

Given,

andABC DBC

are two isosceles triangles.

(i)

In and ,ABD ACD

AD=AD (Common)

AB=AC (

ABC

is isosceles)

BD=CD (

DBC

is isosceles)

Therefore,

ABD ACD

by SSS congruence

condition.

(ii)

In and ,ABP ACP

AP=AP (Common)

PAB PAC

ABD ACD

So, by CPCT

AB AC

(

ABC

is isosceles)

Therefore,

ABP ACP

by SAS congruence condition.

(iii)

by .PAB PAC CPCT as ABD ACD

Therefore, AP bisects

1.A

Also,

In and ,BPD CPD

PD=PD (Common)

BD=CD (

DBC

is isosceles.)

soby .( )BP CP ABP ACP CPCT

Therefore,

BPD CPD

by SSS congruence condition.

Thus,

by 2.BDP CDP CPCT

By(1) and (2), we can say that AP bisects

A

as well

as

D

.

y asB ( )iv BPD CPD CPCT BPD CPD

BP CP

Proved above

Also,

180BPD CPD

(BC is a straight line.)

2 180

90

BPD

BPD

Therefore,

AP is the perpendicular bisector of BC.

Question: 2

AD is an altitude of an isosceles triangle ABC in

which

AB AC

. Show that:

(i)AD bisects BC

(ii)AD bisects

A

.

Solution:

Given,

AD is an altitude and

AB AC

(i) In

and ,ABD ACD

90ADB ADC

AB AC

(Given)

AD AD

(Common)

Therefore,

ABD ACD

by RHS congruence condition.

Now,

BD CD

(By CPCT)

Thus, AD bisects BC

(ii)

BAD CAD

(By CPCT)

Thus, AD bisects

A

.

Question: 3

Two sides AB and BC and median AM of one triangle

ABC are respectively equal to sides PQ and QR and

median PN of

PQR

(See Fig. 7.40). Show that:

(i)

ABM PQN

(ii)

ABC PQR

Solution:

Given,

,AB PQ BC QR

and

AM PN

(i)

1

2

BC BM

and

1

2

QR QN

(AM and PN are medians)

also,

BC QR

(Given)

1 1

2

2

BC QR

BM QN

In

ABM

and

,PQN

AM PN

(Given)

AB PQ

(Given)

BM QN

(As proved above)

Therefore,

ABM PQN

by SSS congruence

condition.

(ii)In

ABC

and

,PQR

AB PQ

(Given)

ABC PQR

by CPCT,

Since,

ABM PQN

BC QR

(Given)

Therefore,

ABC PQR

by SAS congruence

condition.

Question: 4

BE and CF are two equal altitudes of a triangle ABC.

Using RHS congruence rule, prove that the triangle

ABC is isosceles.

Solution:

Given,

BE CF

and

BE AC

and

.CF AB

To show triangle ABC is an isosceles triangle.

Now,

In

BEC

and

,CFB

90BEC CFB

(As

BE AC

and

CF AB

)

BC CB

(Common)

BE CF

(Given)

Therefore,

BEC CFB

by RHS congruence

condition.

ECB FBC

(By CPCT)

ACB ABC

Thus,

In

,ABC AB AC

(As sides opposite to the equal angles are equal.)

Thus, ΔABC is an isosceles triangle.

Question: 5

ABC is an isosceles triangle with

AB AC

.

Draw

.AP BC

Show that

.B C

Solution:

Given, ABC is an isosceles triangle with

,AB AC

and

.AP BC

In

ABP

and

,ACP

90APB APC

(Given)

AB AC

(Given)

AP AP

(Common)

Therefore,

ABP ACP

by RHS congruence

condition.

Thus,

B C

(By CPCT).

Exercise 7.4 (6)

Question: 1

Show that in a right angled triangle, the hypotenuse is

the longest side.

Solution:

Let us consider

ABC

be a triangle right angled at A.

Now,

180A B C

(The sum of all the angles of a

triangle is 180°)

as is90 90 .B C A

A

is the largest angle of the triangle, then the side

opposite to it must be the longest line segment.

So, BC is the longest side in

ABC

.

Also, BC is the hypotenuse. (Opposite side to the right

angle)

Therefore, BC is the hypotenuse which is the longest

side of the right-angled triangle ABC.

Question: 2

In Fig. 7.48, sides AB and AC of

ABC

are extended

to points P and Q respectively.

Also,

.PBC QCB

Show that

.AC AB

Solution:

Given, sides AB and AC of

ABC

are extended to

points P and Q respectively.

Also,

PBC QCB

Since, AP is a line segment,

180

180 ..

ABC PBC

ABC PBC i

Also,

Since, AQ is also a line segment,

180

180

ACB QCB

ACB QCB ii

Now,

Given

180 180

PBC QCB

PBC QCB

ABC ACB

Byusing &i ii

Thus,

,AC AB

as side opposite to the greater angle is

longer.

Question: 3

InFig. and

Showthat

7.49, .

.

B A C D

AD BC

Solution:

Given,

and .B A C D

In ,OAB

Since,

B A

(Given)

Therefore,

AO BO i

(Side opposite to the

smaller angle is smaller)

Again,

In ,OCD

Since,

C D

(Given)

Therefore,

OD OC ii

(Side opposite to the greater angle is longer)

Adding (i) and (ii),

AO OD BO OC

AD BC

Henceproved.

Question: 4

AB and CD are respectively the smallest and longest

sides of a quadrilateral ABCD (See Fig. 7.50).

Show that

and .A C B D

Solution:

Given AB and CD are respectively the smallest and

longest sides of a quadrilateral ABCD.

In

,ABD

Since, AB is the smallest side of quadrilateral ABCD.

AB AD ADB ABD i

(Angle opposite to the longer side is larger.)

Similarly,

In ,BCD

Since, CD is the largest side ofquadrilateral ABCD.

BC DC

BDC CBD ii

(Angle opposite to the longer side is larger.)

Adding (i) and (ii) we get,

ADB BDC ABD CBD

ADC ABC

D B

B D

Similarly,

In ,ABC

AB BC

(Since, AB is the smallest side)

ACB BAC iii

(Angle opposite to longer side is larger.)

Similarly,

In ,ADC

AD CD

(Since, CD is the largest side)

DCA DAC iv

Adding (iii) and (iv) we get,

ACB DCA BAC DAC

BCD BAD

C A

A C

Hence, it is proved.

Question: 5

InFig and bisects

Provethat

7.51, .

.

PR PQ PS QPR

PSR PSQ

Solution:

Given, PR>PQ and PS bisects

QPR

In ,PQR

PR>PQ (Given)

PQR PRQ

(i) (

Angle opposite to the

longer side is larger.)

Now,

bisects( )QPS RPS ii PS QPR

In

,PQS

PSR PQS QPS iii

(Exterior angle of a

triangle is equal to the sum of opposite interior

angles.)

Similarly, in

,PSR

PSQ PRS RPS iv

(Exterior angle of a

triangle is equal to the sum of opposite interior angles)

Adding (i) and (ii) we get,

( rom andF

,

)

PQR QPS PRQ RPS

Or PQS QPS PRS RPS

PSR PSQ iii iv

Hence, it is proved.

Question: 6

Show that of all line segments drawn from a given

point not on it, the perpendicular line segment is the

shortest.

Solution:

Let l be a line segment and B is a point lying on it.

We draw a line AB perpendicular to l. Let C (Different

from B) be a point on l.

In ,ABC

90B

Now,

( )

180

90 90

A B C

A C B

C B

AB AC

(Side opposite to the larger angle is

longer.)

Hence, it is proved.

Exercise 7.5 (4)

Question: 1

ABC is a triangle. Locate a point in the interior of

ABC

which is equidistant from all the vertices

of

ABC

.

Solution:

We know that circumcentre of a triangle is always

equidistant from all the vertices of that triangle.

So, circumcentre is the point where perpendicular

bisectors of all the sides of the triangle meet together.

Thus, in

ABC

, to find the circumcentre, we can draw

the perpendicular bisectors of sides AB, BC, and CA of

the triangle.

Hence, O is the point where these bisectors are

meeting together.

Therefore, O is the point which is equidistant from all

the vertices of

ABC

.

Question: 2

In a triangle locate a point in its interior which is

equidistant from all the sides of the triangle.

Solution:

We know that the point which is equidistant from all

the sides of a triangle is called the in centre of the

triangle.

So, in centre of a triangle is the intersection point of

the angle bisectors of the interior angles of the

triangle.

Let

ABC

be a triangle.

Then, in

ABC

, the in-centre of the triangle is the

angle bisectors of the interior angles of the triangle.

Let I is the point where these angle bisectors are

intersecting each other.

Therefore, I is the point which is equidistant from all

the sides of

ABC

.

Question: 3

In a huge park, people are concentrated at three points

(see Fig. 7.52):

A: where there are different slides and swings for

children,

B: near which a man-made lake is situated,

C: which is near to a large parking and exit

Where should an ice cream parlour be set up so that

maximum number of persons can approach

it?

(Hint: The parlour should be equidistant from A, B and

C)

Solution:

To find the point where maximum number of persons

can approach the ice-cream parlour, we need to find

the point which is equidistant from A, B and C.

Now, A, B and C form a triangle. Thus circumcentre is

the only point that is equidistant from its vertices A, B

and C.

So, the ice-cream parlour should be set up at the

circumcentre O of

ABC

.

We know that circumcentre O of the triangle is the

intersecting point of perpendicular bisectors of the

sides of the triangle.

Question: 4

Complete the hexagonal and star shaped Rangolies

[see Fig. 7.53 (i) and (ii)] by filling them with as many

equilateral triangles of side 1 cm as you can. Count the

number of triangles in each case. Which has more

triangles?

Solution:

We can observe that hexagonal-shaped rangoli has 6

equilateral triangles in it.

side

2

3 3

5

4 4

Now, area of

OAB

2

3 25 3

25 cm

4 4

Thus, area of hexagonal-shaped rangoli

2

25 3 75 3

6 cm

4 2

Now, area of equilateral triangle having its side as

2

2

3 3

1 cm 1 cm

4 4

So, number of equilateral triangles of 1 cm side that

can be filled in this hexagonal-shaped rangoli

75 3

2

150

3

4

Therefore, star-shaped rangoli has 12 equilateral

triangles of side 5cm in it.

Here, area of star-shaped rangoli

2

3

12 5 75 3

4

So, number of equilateral triangles of 1cm side that

can be filled in star-shaped rangoli

75 3

3

300

4

Therefore, star-shaped rangoli has more equilateral

triangles in it.