NCERT Solutions For Class 9 Maths Chapter 7

NCERT Solutions Class 9 Maths Triangles

Ncert Solutions For Class 9 Maths Chapter 7 PDF Download

A triangle is a closed, polygon with three vertices and three edges. It is one of the basic shapes in geometry. It mainly comprises three line segments which intersect in their endpoints. In general, the triangle has three angles. The total sum of the measures of the angles is always 180° in a triangle.

There are 8 types of triangles and are categorized by its total number of congruent sides and its angles.

  • Acute triangle.
  • Right triangle.
  • Obtuse triangle.
  • Equilateral triangle.
  • Isosceles triangle.
  • Scalene triangle.
  • Acute triangle.
  • Obtuse triangle.

Geometry of Triangles is an important chapter for class 9 maths. It holds higher weightage in both finals and other competitive exams like JEE, Olympiads and other aptitude test.

NCERT solutions for Class 9 Maths Chapter 7 – Geometry of 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 Geometry of 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 Maths Solutions for class 9 chapter 7.

NCERT solutions for Class 9 Chapter 7 Maths – Geometry of 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 Geometry of 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 Maths Solutions for chapter 7 class 9.

NCERT Solutions For Class 9 Maths Chapter 7 Exercises

 

Exercise 7.1

Question 1:

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

1

Solution:

In \(\triangle PQR\) and \(\triangle PQS\) , we have

PR = PS

\(\angle RPQ = \angle SPQ\) (PQ bisects \(\angle P\))

PQ = PQ (common)

\(\triangle PQR = \triangle PQS\) (By SAS congruence)

Hence Proved.

Therefore, QR = QS (CPCT)

 

Question 2:

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

(i) \(\triangle ABD = \triangle BAC\)

(ii) BD = AC

(iii) \(\angle ABD = \angle BAC\)

2

Solution:

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

In \(\triangle DAB\ and\ \triangle BAC\), we have

AD = BC [Given]

\(\angle DAB = \angle CBA\) [Given]

AB = AB [Common]

\(\triangle ABD = \triangle BAC\) [By SAS congruence]

BD = AC [CPCT]

and \(\angle ABD = \angle BAC\) [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 \(\triangle AOD = \triangle BOC\), we have,

\(\angle AOD = \angle BOC\) [Vertically opposite angles)

\(\angle CBO = \angle DAO (Each = 90^{o})\)

AD = BC [Given]

\(\triangle AOD = \triangle BOC\) [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 \(\triangle ABC + \triangle CDA\) .

4

Solution:

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

In \(\triangle ABC \, and\,  \triangle CDA\), we have,

\(\angle BAC = \angle DCA\) (Alternate angles)

\(\angle BCA = \angle DAC\) (Alternate angles)

AC = AC (Common)

\(\triangle ABC = \triangle CDA\) (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 \(\angle A\) (See fig.). Show that:

(i)  \(\triangle APB\ and\ \triangle AQB\)

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

5

Solution:

In \(\triangle APB\ and\ \triangle AQB \) , we have

\(\angle PAB = \angle AQB\) (l is the bisector of \(\angle A\))

\(\angle APB = \angle AQB\ (Each = 90^{o})\)

AB = AB

\(\triangle APB = \triangle AQB\) (By AAS congruence)

Also, BP = BQ (By CPCT)

i.e, B is equidistant from the arms of \(\angle A\)

Hence proved.

 

Question 6:

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

6

Solution:

\(\angle BAD = \triangle EAC\) (Given)

\(\angle BAD + \angle DAC = \angle EAC + \angle DAC (Adding \angle DAC\ to\ both\ sides)\)

\(\angle BAC = \angle EAC\)

Now, in \(\triangle ABC\, and\, \triangle ADE\) , We have

AB = AD

AC = AE

\(\angle BAC = \angle DAE\)

\(\triangle ABC= \triangle ADE\) (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 \(\angle BAD = \angle ABE\ and\ \angle EPA = \angle DPB\) (see fig.). Show that

            (i) \(\triangle DAP = \triangle EBP\)

            (ii) AD = BE

7

Solution:

In \(\triangle DAP\ and\ \triangle EBP\), we have

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

\(\angle BAD = \angle ABE \) (Given)

\(\angle EPB = \angle DPA\)

\((\angle EPA = \angle DPB \Rightarrow \angle EPA + \angle DPE = \angle DPB + \angle DPE)\)

\(\triangle DPA = \triangle EPB\)

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) \(\triangle AMC = \triangle BMD\)

            (II) \(\angle DBC\) is a right angle.

            (iii) \(\triangle DBC = \triangle ACB\)

            (iv) \( CM= \frac{1}{2} AB \)

8

Solution:

In \(\triangle BMB and \triangle DMC\) , we have

(i) DM = CM (given)

BM = AM (M is the midpoint of AB)

\(\angle DMB = \angle AMC\) (Vertically opposite angles)

\(\triangle AMC = \triangle BMD\) (By SAS)

Hence Proved.

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

\(\Rightarrow \angle DBC + \angle ACB = 180^{o}\) (Sum of co-interior angles)

\(\Rightarrow \angle DBC = 90^{o}\)

Hence proved.

(iii) In \(\triangle DBC\ amd\ \triangle ACB \), We have

DB = AC (CPCT)

BC = BC (Common)

\(\angle DBC = \angle ACB (Each = 90^{o})\)

\(\triangle DBC = \triangle ACB\) (By SAS)

Hence proved.

(iv) AB = CD

\(\Rightarrow \frac{1}{2} AB = \frac{1}{2} CD\) (CPCT)

Hence, \(\frac{1}{2} AB = CM\) Proved.

 

Exercise 7.2:

Question 1:

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

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

(1)OY=OZ    (2)XO bisects \(\angle A\)

9

Solution:

\(\frac{1}{2}\angle XYZ=\frac{1}{2}\angle XZY\)

\(\angle ZYO=\angle XZY\)

[OY and OZ are bisector of

\(\angle Y\) and \(\angle Z\)  respectively]

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

Again,     \(\frac{1}{2}\angle XYZ=\frac{1}{2}\angle XZY\)

\(\angle XYO=\angle XYO\)               [ \(∴\)OY and OZ are bisector  of \(\angle Y\) and  \(\angle Z\) respectively   ]

In \(\Delta XYO =\Delta XYO\) , we have

XY=XZ  [Given]

OY=OZ  [proved XYove]

\(\angle XYO=\angle XZO\)         [proved above]

\(\Delta XYO =\Delta XZO\)          [SAS congruence]

\(\Delta XYO\)= \(\angle ZXO\)   [CPCT]

XO bisects \(\angle X\)             proved    

 

Question 2:

In \(\Delta XYZ\)   ,XO is the perpendicular bisector of YZ .Prove that \(\Delta XYZ\) is an  isosceles triangle in which  XY=XZ

10

 Solution:

10

In \(\Delta XYO= \Delta XZO\) , we have

\(\angle XOZ =\angle XOZ\)    [] each =90°]

YO=ZO [ XO bisects YZ]

XO=XO [common]

\(∴ \Delta XYO=\Delta XZO\)   [SAS]

\(∴\) XY=XZ    [CPCT]

Hence, \(\Delta XYZ\) 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 \(\Delta ABC,\)

AB=AC   [Given]

\(\angle B=\angle C\)     [angles opposite to equal sides of a triangle are equal]

11

Now in right triangles BFC and CEB ,

\(\angle BFC=\angle CEB\)                  [Each =90° ]

\(\angle FBC=\angle ECB\)                 [proved above]

BC=BC

\(∴ \Delta BFC =\Delta CEB\)            [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) \(\Delta ABE \cong \Delta ACF\)

11

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

Solution(1) in \(\Delta ABE\) and ACF , we have

BE=CF         [Given]

\(\angle BAE=\angle CAF\)  [common]

\(\angle BEA=\angle CFA\)    [ Each =90°]

So, \(\Delta ABE = \angle ACF\)        [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

\(\angle ABD=\angle ACD\)

Solution. In isosceles \(\Delta ABC\), We have

AB= AC

12

\(\angle ABC=\angle ACB\)   ———-(i)

[Angles opposite to equal sides are equal]

Now , in isosceles \(\Delta DCB\), We have

BD=CD

\(\angle DBC=\angle DCB\) —————–(ii)

[Angle opposite to equal sides are equal]

Adding (i) and (ii)  , we have

\(\angle ABC+\angle DBC=\angle ACB+\angle DCB\)

\(\angle ABD=\angle ACD\) .proved

 

Question 6:

\(\Delta ABC\) is an isosceles triangle in which AB=AC side BA is produced to D such that AD =AB  .show that  \(\angle BCD\) is a right angle.

13

Solution:

AB=AC

\(\angle ACB= \angle ABC\)—————-(1)

[Angles opposite to equal sides are equal]

AB=AD          [Given]

13

AD=AC             [AB=AC]

\(∴ \angle ACD=\angle ADC\) ——————(ii)

[Angles opposite to equal sides are equal]

Adding (i) and (ii)

\(\angle ACB+\angle ACD=\angle ABC+\angle ADC\)

\(\angle BCD=\angle ABC+\angle ADC\)

Now in \(\Delta BCD ,We have\)

\(\angle BCD+\angle DBC+\angle BDC=\)= 180  [Angle sum property of a triangle]

\(∴ \angle BCD+\angle BCD\)=180°

\(2\angle BCD=180\)

\(2\angle BCD=180\)

 

Question 7:

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

14

\(\angle B and \angle C\)

Solution.  In \(\Delta ABC\), We have

\(\angle A\)=90

AB =AC

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

So, \(\angle B=\angle C\)

Since \(\angle A\)=90 , therefore sum of remaining two angles =90

Hence \(\angle B=\angle C=45\)

 

Question 8:

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

15

Solution:

As \(\Delta ABC\)  is an equilateral

So, AB=BC=AC

Now, AB=AC

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

Again BC=AC

\(\angle BAC=\angle ABC\)—-(ii)    [same reason]

Now in \(\Delta ABC\)

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

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

\(3\angle ABC\)=180

\(\angle ABC\) =180/3=60°

Also from (i) and (ii)

\(\angle ACB\)=60 and \(\angle BAC\)

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

\(\frac{1}{2}BC = \frac{1}{2}QR\)

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 \(\angle A + \angle C = 90^{o}\)

Angles A and C are less than 90o

Now,    \(\angle B > \angle A\)

AC > BC ….(i)

(Side opposite to greater angle is longer)

Again, \(\angle B > \angle C\)

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 \(\triangle ABC\) are extended to points P and Q  respectively. Also, \(\angle PBC < \angle QCB\) . Show that AC > AB.

24

Solution:

\(\angle ABC + \angle PBC = 180^{o} (Linear pair) \)

\(\Rightarrow \angle ABC = 180^{o} – \angle PBC\)

Similarly, \(\angle ACB = 180^{o} – \angle QCB\)

It is given that \(\angle PBC < \angle QCB\)

\(180^{o} – \angle QCB < 180^{o} – \angle PBC \)

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

\(\Rightarrow AB < AC\)

\(\Rightarrow AC > AB\)

Hence proved.

 

Question 3:

In the figure, \(\angle B < \angle A\ and\ \angle C < \angle D\). Show that AD < BC.

25

Solution:

\(\angle B < \angle A\) (Given)

BO > AO …(i)

(Side opposite to greater angle is longer)

\(\angle C < \angle D\) (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 \(\angle A > \angle  C\ and \angle B > \angle D\) .

26

27                      

Solution:

Join AC.

Mark the angles as shown in the figure..

In \(\triangle ABC\) ,

BC > AB (AB is the shortest side)

\(\angle 2 > \angle 4\) …(i)

[Angle opposite to longer side is greater]

In \(\triangle ADC\) ,

CD > AD (CD is the longest side)

\(\angle 1 > \angle 3\) …(ii)

[Angle opposite to longer side is greater]

Adding (i) and (ii), we have

\(\angle 2 + \angle 1 > \angle 4 + \angle 3\)

\(\Rightarrow \angle A > \angle C\)

Similarly, by joining BD, we can prove that

\(\angle B > \angle D\)

 

Question 5:

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

28

Solution:

PR > PQ

\(\angle PQR > \angle PRQ\) …(i)

[Angle opposite to longer side is greater]

\(\angle QPS > \angle RPS\ (PS\ bisects\ \angle QPR)\) …(ii)

In \(\triangle PQS\) , \(\angle PQS + \angle QPS + \angle PSQ = 180^{o}\)

\(\Rightarrow \angle PSQ = 180^{o} – (\angle PQS + \angle QPS)\) …(iii)

Similarly in \(\triangle PRS\) , \(\triangle PSR = 180^{o} – (\angle PRS + \angle RPS)\)

\(\Rightarrow \angle PSR = 180^{o} – (\angle PSR + \angle QPS)\) [from (ii) … (iv)

From (i), we know that \(\angle PQS < \angle PSR\)

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

\(\Rightarrow \angle PSR > \angle PSQ \)

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

\(OP \perp i\)

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

OP < OS

In \(\triangle OPQ\) , \(\angle P = 90^{o}\)

Therefore, \(\angle Q \) is an acute angle (i.e, \(\angle Q < 90^{o}\) )

\(\angle Q < \angle P\)

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 \(\triangle DEF\) which is at an equal distance or equidistant from all the vertices of \(\triangle DEF\).

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 \(\angle A, \angle B\ and\ \angle C\ of\ \triangle ABC  \). 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

 

BYJU’S, the number one online learning apps provides free NCERT solutions for all the classes and are available in chapter wise mainly to help students in their academics. NCERT solutions for class 9 maths chapter 7 – Geometry of Triangles are prepared by our subject experts under the guidelines of NCERT to assists students in their examination,and it includes detailed explanations for all the topics and are explained in a simple language, in an easier format for students to learn better and effectively. Those students who practice more problems from this NCERT Solutions can gain extra information on this topic as their are lot more solved examples for students benefits. Those students who find difficulties in learning these problems can either download pdf files or practice online by visiting our website at BYJU’S