 Lesson: Triangles
Exercise 7.1 (8)
Question: 1
and AB bisects
A
(See Fig. 7.16). Show that
ABC ABD
. What
can you say about BC and BD?
Solution:
Given:
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)
(Given)
Therefore,
ABC ABD
by SAS congruence
condition.
BC BD
(By CPCT)
Question: 2
DAB CBA
(see Fig7.17). Prove that
(i)
ABD BAC
(ii)
(iii)
.ABD BAC Solution:
Given:
and
DAB CBA
(i)In
ABD
and
,BAC
AB BA
(Common)
DAB CBA
(Given)
(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:
and
.BC AB
To Prove:
OA OB
Proof:
In and ,AOD BOC
A B
(As
and
BC AB
)
AOD BOC
(Vertically opposite angles)
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,
and
.
Show that Solution:
Given:
and
We need to prove that
BC DE
Now,
(Given)
DAC
on both the sides, we get
In
ABC
and AC AE
(Given)
(As proved above)
(Given)
Therefore,
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
and
EPA DPB
(See Fig. 7.22).
Show that:
(i)
DAP EBP
(ii) Solution:
Given:
P is the midpoint of AB. i.e.,
.AP PB
and
EPA DPB
(i)
Since,
EPA DPB
DPE
both the sides
EPA DPE DPB DPE
DPA EPB
Now, In
DAP
and
,EBP DPA EPB
(As proved above)
AP BP
(Given)
(Given)
Therefore,
DAP EBP
by ASA Congruence condition.
(ii)
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
(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
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,
Proof,
In
and
(Common)
BD CD
Therefore,
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.
To show,
BE=CF
Proof:
In and ,AEB AFC A A
(Common)
AEB AFC
(Right angles)
(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,
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:
In
ABD
and
,ACD
(Common)
(
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
.
Side BA is produced to D such that
(See Fig.
7.34). Show that
BCD
is a right angle. Solution:
Given,
and
To show:
BCD
is a right angle.
Proof:
In ,ABC
(Given)
ACB ABC i
(Angles opposite to the equal
sides are equal.)
In ,ACD (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
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
(
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.)
soby .( )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
. Show that:
A
.
Solution: Given,
(i) In
and ,ABD ACD
(Given)
(Common)
Therefore,
ABD ACD
by RHS congruence condition.
Now,
BD CD
(By CPCT)
(ii)
(By CPCT)
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
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
.
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)
(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
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)
AO OD BO OC
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.
(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,
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,
(Since, CD is the largest side)
DCA DAC iv
Adding (iii) and (iv) we get,
ACB DCA BAC DAC
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.