NCERT Solutions For Class 9 Maths Chapter 7

NCERT Solutions Class 9 Maths Triangles

NCERT solutions for Class 9 Maths Chapter 7 Triangles is one of the most important topic for the students to practice. For solving questions, it is crucial to have good knowledge of the topic. Students can study various concepts of maths by visiting our site BYJU’S. To ease the fear of maths for the students, we at BYJU’S provide NCERT Solution for Class 9 Maths Chapter 7 Triangles. Student can download the NCERT Solution for class 9 Maths Chapter 7 pdf or can view it online by following the link. These solution are provided in detailed manner, where one can find step-by-step solution to all the questions of NCERT Solutions for class 9 maths chapter 7.

NCERT solutions for class 9 maths chapter 7 pdf are prepared by our subject experts under the guidelines of NCERT to assists students in their examination.

NCERT Solutions For Class 9 Maths Chapter 7 Exercises

Exercise 7.1

Question 1:

In quadrilateral PQRS, PR = PQ and PQ bisects P (look at the fig.). Show that the PQR = PQS. What can you say about QR and QS?

1

Solution:

In PQR and PQS , we have

PR = PS

RPQ=SPQ (PQ bisects P)

PQ = PQ (common)

PQR=PQS (By SAS congruence)

Hence Proved.

Therefore, QR = QS (CPCT)

 

Question 2:

ABCD is a quadrilateral in which AD = BC and DAB= CBA (see Fig.). Prove that

(i) ABD=BAC

(ii) BD = AC

(iii) ABD=BAC

2

Solution:

In the given figure, ABCD is a quadrilateral in which AD = BC and DAB= CBA.

In DAB and BAC, we have

AD = BC [Given]

DAB=CBA [Given]

AB = AB [Common]

ABD=BAC [By SAS congruence]

BD = AC [CPCT]

and ABD=BAC [CPCT]

Hence Proved.

 

Question 3:

AD and BC are equal perpendiculars to a line segment AB (see Fig.). Show that CD bisects AB.

3

Solution:

In AOD=BOC, we have,

AOD=BOC [Vertically opposite angles)

CBO=DAO(Each=90o)

AD = BC [Given]

AOD=BOC [By AAS congruence]

Also, AO = BO [CPCT]

Hence, CD bisects AB Proved.

 

Question 4:

L and m are two para;;e; lines intersected by another pair of parallel lines p and q (see fig.). Show that ABC+CDA .

4

Solution:

In the given figure, ABCD is a parallelogram in which AC is a diagonal, i.e., AB || DC and BC || AD.

In ABCandCDA, we have,

BAC=DCA (Alternate angles)

BCA=DAC (Alternate angles)

AC = AC (Common)

ABC=CDA (By SAS congruence)

Hence Proved.

 

Question 5:

Line l is the bisector of an angle A and B is any point on l. BP and BQ are perpendicular from B to the arms of A (See fig.). Show that:

(i)  APB and AQB

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

5

Solution:

In APB and AQB , we have

PAB=AQB (l is the bisector of A)

APB=AQB (Each=90o)

AB = AB

APB=AQB (By AAS congruence)

Also, BP = BQ (By CPCT)

i.e, B is equidistant from the arms of A

Hence proved.

 

Question 6:

In the fig, AC = AE, AB = AD and BAD=EAC . Show that BC = DE.

6

Solution:

BAD=EAC (Given)

BAD+DAC=EAC+DAC(AddingDAC to both sides) BAC=EAC

Now, in ABCandADE , We have

AB = AD

AC = AE

BAC=DAE

ABC=ADE (By SAS congruence)

BC = DE (CPCT)

Hence Proved.

 

Question 7:

AB is a line segment and p is its mid-point. D and E are points on the same side of AB such that BAD=ABE and EPA=DPB (see fig.). Show that

            (i) DAP=EBP

            (ii) AD = BE

7

Solution:

In DAP and EBP, we have

AP = BP (P is the midpoint of the line segment AB)

BAD=ABE (Given)

EPB=DPA (EPA=DPBEPA+DPE=DPB+DPE) DPA=EPB

AD = BE (ASA)

AD = BE

Hence Proved.

 

Question 8:

In right triangle ABC, right angles at C, M is the mid-point 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 (see fig.). Show that :

            (i) AMC=BMD

            (II) DBC is a right angle.

            (iii) DBC=ACB

            (iv) CM=12AB

8

Solution:

In BMBandDMC , we have

(i) DM = CM (given)

BM = AM (M is the midpoint of AB)

DMB=AMC (Vertically opposite angles)

AMC=BMD (By SAS)

Hence Proved.

(ii) AC || BD (DBM and CAM are .alternate angles)

DBC+ACB=180o (Sum of co-interior angles)

DBC=90o

Hence proved.

(iii) In DBC amd ACB, We have

DB = AC (CPCT)

BC = BC (Common)

DBC=ACB(Each=90o)

DBC=ACB (By SAS)

Hence proved.

(iv) AB = CD

12AB=12CD (CPCT)

Hence, 12AB=CM Proved.

 

Exercise 7.2:

Question 1:

In an isosceles triangle XYZ with XY=XZ ,the bisector of B and C

Intersect each other at O .join A at O .Show that:

(1)OY=OZ    (2)XO bisects A

9

Solution:

12XYZ=12XZY ZYO=XZY

[OY and OZ are bisector of

Y and Z  respectively]

OBY=OZ [ sides opposite to equal angles are equal]

Again,     12XYZ=12XZY

XYO=XYO               [ OY and OZ are bisector  of Y and  Z respectively   ]

In ΔXYO=ΔXYO , we have

XY=XZ  [Given]

OY=OZ  [proved XYove]

XYO=XZO         [proved above]

ΔXYO=ΔXZO          [SAS congruence]

ΔXYO= ZXO   [CPCT]

XO bisects X             proved    

 

Question 2:

In ΔXYZ   ,XO is the perpendicular bisector of YZ .Prove that ΔXYZ is an  isosceles triangle in which  XY=XZ

10

 Solution:

10

In ΔXYO=ΔXZO , we have

XOZ=XOZ    [] each =90°]

YO=ZO [ XO bisects YZ]

XO=XO [common]

ΔXYO=ΔXZO   [SAS]

XY=XZ    [CPCT]

Hence, ΔXYZ is an isosceles triangle .Proved

 

Question 3:

ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and

AB respectively (see figure).Show that these altitudes are equal.

11

Solution:

In ΔABC,

AB=AC   [Given]

B=C     [angles opposite to equal sides of a triangle are equal]

11

Now in right triangles BFC and CEB ,

BFC=CEB                  [Each =90° ]

FBC=ECB                 [proved above]

BC=BC

ΔBFC=ΔCEB            [AAS]

Hence, BE=CF [CPCT] proved

 

Question 4:

ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see.fig) show that

11

(1) ΔABEΔACF

11

(2) AB=AC   i,e ABC is an isosceles triangle.

Solution(1) in ΔABE and ACF , we have

BE=CF         [Given]

BAE=CAF  [common]

BEA=CFA    [ Each =90°]

So, ΔABE=ACF        [AAS]proved

(2)  also, AB=AC                 [ CPCT]

i,e., ABC is an isosceles triangle Proved.

 

Question 5:

ABC and DBC are two isosceles triangle on the same base BC .show that

12

ABD=ACD

Solution. In isosceles ΔABC, We have

AB= AC

12

ABC=ACB   ———-(i)

[Angles opposite to equal sides are equal]

Now , in isosceles ΔDCB, We have

BD=CD

DBC=DCB —————–(ii)

[Angle opposite to equal sides are equal]

Adding (i) and (ii)  , we have

ABC+DBC=ACB+DCB

ABD=ACD .proved

 

Question 6:

ΔABC is an isosceles triangle in which AB=AC side BA is produced to D such that AD =AB  .show that  BCD is a right angle.

13

Solution:

AB=AC

ACB=ABC—————-(1)

[Angles opposite to equal sides are equal]

AB=AD          [Given]

13

AD=AC             [AB=AC]

ACD=ADC ——————(ii)

[Angles opposite to equal sides are equal]

Adding (i) and (ii)

ACB+ACD=ABC+ADC BCD=ABC+ADC

Now in ΔBCD,Wehave

BCD+DBC+BDC== 180  [Angle sum property of a triangle]

BCD+BCD=180°

2BCD=180 2BCD=180

 

Question 7:

ABC is a right angled triangle in which A=90 and AB=AC.Find  

14

BandC

Solution.  In ΔABC, We have

A=90

AB =AC

We know that angles opposite to equal sides of an isosceles triangle are equal

So, B=C

Since A=90 , therefore sum of remaining two angles =90

Hence B=C=45

 

Question 8:

Show that the angles of an equilateral triangle are 60 each.

15

Solution:

As ΔABC  is an equilateral

So, AB=BC=AC

Now, AB=AC

ACB=ABC——-(ii)[ angles sum property of a triangle]

Again BC=AC

BAC=ABC—-(ii)    [same reason]

Now in ΔABC

ABC+ACB+BAC=180  [angle sum property of a triangle are equal]

ABC+ACB+BAC=180=180 [from (i) (ii)]

3ABC=180

ABC =180/3=60°

Also from (i) and (ii)

ACB=60 and BAC

Hence each angle of an equilateral triangle is 60°  Proved

 

Exercise 7.3

Question 1:

On the same base BC XYZ and DYZ are two isosceles triangles and vertices X and D are on the same side of YZ. If XD is extended to intersect YZ at P, show that

(i) XYD XZD

(ii) XYP XZP

(iii) XP bisects X as well as D.

(iv) XP is the perpendicular and bisector of YZ.

16

Solution:

17

(i) In ⧍XYD and ⧍XZD, we have

XY = XZ [Given]

YD = ZD [Given]

XD = XD [Common]

⧍XYD ⩬ ⧍XZD [SSS congruence].

(ii) In ⧍XYP and ⧍XZP, we have

XY = XZ

∠YXP = ∠ZXP

XP = XP

⧍XYP ⩬ ⧍XZP [SAS congruence].

(iii)⧍XYD ⩬ ⧍XDY

∠XDY = ∠XDZ

=> 180° — ∠XDY = 180° — ∠XDZ
Also, from part (ii), ∠YXPD = ∠ZXP

XP bisects DX as well as ∠D.

(iv) Now, YP = ZP and ∠YPX = ∠ZPX

But ∠YPX + ∠ZPX = 180°

So, 2∠BPA = 180°

Or, ∠YPX = 90°

Since YP = ZP, therefore XP is perpendicular bisector of YZ.

 

Question 2:

Isosceles triangle XYZ has an altitude XD, in which XY = XZ. Show that

(i) XD bisects YZ

(ii) XD bisects X

18

Solution:

(i) In ⧍ABD and ⧍ACD, we have ∠ADB = ∠ADC

XY = XZ

XD = XD

⧍XYD ⩬ ⧍XZD.

YD = ZD

Hence, XD bisects YC.

(ii) Also, ∠YXD = ∠ZXD

Hence XD bisects ∠X.

 

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. Show that:

(i) ABM   APQN (ii) ABC PQR

19

Solution:

(i) In ⧍ABM and ⧍PQN,

we have

BM = QN

12BC=12QR

AB = PQ

AM = PN

⧍ABM = ⧍PQN [ SSS Congruence ]

∠ABM = ∠PQN.

(ii) Now, in ⧍ABC and ⧍PQR, we have

AB = PQ

∠ABC = ∠PQR

BC = QR

⧍ABC ⩬  ⧍PQR [SAS congruence]

 

Question 4:

In triangle ABC, BE and CF are two equal altitudes. By Using RHS congruence rule, prove that the triangle ABC is isosceles.

20

Solution:

BE and CF are altitudes of a ⧍ABC.

∠BEC = ∠CFB = 90°

Now, in right triangles BEB and CFB, we have

Hyp. BC = Hyp. BC

Side BE= Side CF

⧍BEC ⩬  ⧍CFB

∠BCE = ∠CBF

Now, in ⧍ABC, ∠B = ∠C

AB = AC

Hence, ⧍ABC is an isosceles triangle.

 

Question 5:

AB = AC in an isosceles triangle ABC. Draw AP perpendicular BC to show that ∠B= ∠C.

22

Solution:

Draw AP perpendicular BC. In ⧍ABP and ⧍ACP, we have

AB = AC

∠APB = ∠APC

AB = AP [Common]

⧍ABP = ⧍ACP [By RHS congruence rule]

Also, ∠B = ∠C Proved. [CPCT]

 

Exercise 7.4

Question 1:

Show that in a right-angled triangle, the hypotenuse is the longest side.

23

Solution:

            ABC is a right angle triangle, right angled at B.

Now A+C=90o

Angles A and C are less than 90o

Now,    B>A

AC > BC ….(i)

(Side opposite to greater angle is longer)

Again, B>C

AC > AB …(ii)

(Side opposite to greater angle are longer)

Hence, from (i) and (ii), we can say that AC (hypotenuse) is the longest side.

Hence proved.

 

Question 2:

In the figure, sides AB and AC of ABC are extended to points P and Q  respectively. Also, PBC<QCB . Show that AC > AB.

24

Solution:

ABC+PBC=180o(Linearpair) ABC=180oPBC

Similarly, ACB=180oQCB

It is given that PBC<QCB

180oQCB<180oPBC

Or ACB<ABC [From (i) and (ii)]

AB<AC AC>AB

Hence proved.

 

Question 3:

In the figure, B<A and C<D. Show that AD < BC.

25

Solution:

B<A (Given)

BO > AO …(i)

(Side opposite to greater angle is longer)

C<D (Given)

CO > DO …(ii)

(Same reason)

Adding (i) and (ii)

BO + CO > AO + DO

BC > AD

AD < BC

Hence Proved.

 

Question 4:

AB and CD are respectively the smallest and longest side of a quadrilateral ABCD (see fig.) Show that A>C andB>D .

26

27                      

Solution:

Join AC.

Mark the angles as shown in the figure..

In ABC ,

BC > AB (AB is the shortest side)

2>4 …(i)

[Angle opposite to longer side is greater]

In ADC ,

CD > AD (CD is the longest side)

1>3 …(ii)

[Angle opposite to longer side is greater]

Adding (i) and (ii), we have

2+1>4+3 A>C

Similarly, by joining BD, we can prove that

B>D

 

Question 5:

In the figure, PR > PQ and PS bisects QPR . Prove that PSR>PSQ

28

Solution:

PR > PQ

PQR>PRQ …(i)

[Angle opposite to longer side is greater]

QPS>RPS (PS bisects QPR) …(ii)

In PQS , PQS+QPS+PSQ=180o

PSQ=180o(PQS+QPS) …(iii)

Similarly in PRS , PSR=180o(PRS+RPS)

PSR=180o(PSR+QPS) [from (ii) … (iv)

From (i), we know that PQS<PSR

So from (iii) and (iv), PSQ<PSR

PSR>PSQ

Hence proved.

 

Question 6:

Show that of all the segments drawn from a given point not on it, the perpendicular line segment is the shortest.            

29

Solution:

We have a line I and O is the point not on I

OPi

We have to prove that OP < OQ, OP < OR and OP < OS.

OP < OS

In OPQ , P=90o

Therefore, Q is an acute angle (i.e, Q<90o )

Q<P

Hence, OP < OQ (Side opposite to greater angle is longer)

Similarly, we can prove that OP is shorter than OR, OS, etc.

Hence proved

 

Exercise 7.5

 Question 1:

Find a point in the interior of DEF which is at an equal distance or equidistant from all the vertices of DEF.

30

Solution:        

Draw perpendicular bisectors of sides DE,

EF and FD, which meets at O.

Hence, O is the required point.

 

Question 2:

Find a point in the interior of a triangle such that it is at equal distances from all the sides of the triangle.

 Solution:

 30

Question 3:

People are concentrated at three points in a park namely A,B and C. (see Fig.).

34

            A: is where swings and slides for children are present

            B: is where a lake is present

            C: is where there is a large parking lot and exit

Where do you think an ice – cream parlor should be set up such that the maximum number of people can access it? Draw bisectors A,B and C of ABC. Let these angle bisectors meet at O. O is the required point.

Solution:         Join AB, BC and CA to get a triangle ABC. Draw the perpendicular bisector of AB and BC. Let them meet at O. Then O is equidistant from A, B and C. Hence, the parlor should be set up at O so that all the other points are equidistant from it.

 

Question 4:

Fill the star shaped and hexagonal rangolies [see fig.(i) and (ii)] by filling them with as many equilateral triangles as you can of side 1 cm. What is the number of triangles in both the cases? Which one has the most number of triangles?

 32

 

Solution:

33