NCERT Solutions for Class 9 Maths Chapter 7 Triangles

 

NCERT solutions for class 9 maths chapter 7 Triangles- A geometrical figure is made up of three straight lines is known as Triangles. A triangle has three vertices, three sides, and three angles. In the previous classes, you have studied the construction and types of triangles. You have studied briefly about triangles and their properties in the previous classes. In Class 9 Triangles, you will learn something of a higher level. Solutions of NCERT for class 9 maths chapter 7 Triangles can be a good tool whenever you are stuck in any of the problems. This chapter will be covering the properties of triangles like congruence of triangles, isosceles triangle, etc in detail. In this chapter, there are a total of 5 exercises with 31 questions in them. CBSE NCERT solutions for class 9 maths chapter 7 Triangles is covering the entire chapter including the optional exercises. The chapter is full of properties and theorems that's why the examples and theorems are as important as the practice exercises. There is another aspect to look at the importance of this chapter, apart from school exams this is an essential part of competitive examinations like- CAT, SSC, NTSE, INO, etc. By using NCERT solutions for class 9 maths chapter 7 Triangles, you can prepare 360 degree for your school as well as for the competitive examinations. If you need NCERT Solutions for other classes and subjects then you can download all these for free by clicking on the given link.

NCERT solutions for class 10 maths chapter 7 Triangles Excercise: 7.1

Q1 In quadrilateral \small ABCD, \small AC=AD and \small AB bisects \small \angle A (see Fig.). Show that  \small \Delta ABC\cong \Delta ABD. What can you say about \small BC and \small BD ?

               

Answer:

In the given triangles we are given that:-  

 (i)  \small AC=AD

(ii)  Further, it is given that AB bisects angle A.  Thus \angle BAC  =\ \angle BAD.  

(iii) Side AB is common in both the triangles.  AB=AB

Hence by SAS congruence, we can say that  :   \small \Delta ABC\cong \Delta ABD

By c.p.c.t. (corresponding parts of congruent triangles are equal) we can say thatBC\ =\ BD

Q2 (i) ABCD is a quadrilateral in which \small AD=BC and \small \angle DAB= \angle CBA  (see Fig. ). Prove that

\small \Delta ABD\cong \Delta BAC.

                               

Answer:

It is given that :- 

(i)  AD  =  BC 

(ii)  \small \angle DAB= \angle CBA

(iii)  Side AB is common in both the triangles.

So, by SAS congruence, we can write :

                 \small \Delta ABD\cong \Delta BAC       

Q2 (ii) \small ABCD is a quadrilateral in which \small AD=BC and \small \angle DAB=\angle CBA  (see Fig.). Prove that \small BD=AC

                                            

Answer:

In the previous part, we have proved that \small \Delta ABD\cong \Delta BAC.

Thus by  c.p.c.t. , we can write :  \small BD=AC

Q2 (iii) \small ABCD is a quadrilateral in which \small AD=BC and \small \angle DAB= \angle BAC   (see Fig.). Prove that \small \angle ABD= \angle BAC.

                                                

Answer:

In the first part we have proved that   \small \Delta ABD\cong \Delta BAC.

Thus by  c.p.c.t. , we can conclude :  

                                                                   \small \angle ABD= \angle BAC

Q3 \small AD and \small BC are equal perpendiculars to a line segment \small AB (see Fig.). Show that  \small CD bisects \small AB.

                                   

Answer:

In the given figure consider \DeltaAOD  and  \DeltaBOC.

(i)      AD  =  BC           (given)

(ii)   \angle A   =  \angle B          (given that the line AB is perpendicular to AD and BC)

(iii)    \angle AOD  =  \angle BOC       (vertically opposite angles).

Thus by AAS Postulate, we have   

                                                             \Delta AOD\ \cong \ \Delta BOC

Hence by c.p.c.t. we can write :                  AO\ =\ OB

And thus CD bisects AB.

Q4 \small l and \small m are two parallel lines intersected by another pair of parallel lines \small p and \small q (see Fig. ). Show that \small \Delta ABC\cong \Delta CDA.

                                              

Answer:

In the given figure, consider \Delta ABC and \Delta CDA :

(i)    \angle\ BCA\ =\ \angle DAC

(ii)    \angle\ BAC\ =\ \angle DCA

(iii)    Side AC is common in both the triangles.

Thus by ASA congruence, we have :

                                                      \Delta ABC\ \cong \ \Delta CDA 

Q5 (i) Line \small l is the bisector of an angle \small \angle A and B is any point on \small l. \small BP and \small BQ  are perpendiculars from \small B to the arms of \small \angle A (see Fig.). Show that: \small \Delta APB\cong \Delta AQB

                                    

Answer:

In the given figure consider \small \Delta APB   and   \small \Delta AQB,

(i)      \angle P\ =\ \angle Q                      (Right angle)

(ii)      \angle BAP\ =\ \angle BAQ       (Since it is given that I is bisector)

(iii)     Side AB is common in both the triangle.

Thus AAS congruence, we can write :

                                                     \small \Delta APB\cong \Delta AQB

Q5 (ii) Line \small l is the bisector of an angle \small \angle A and \small B  is any point on \small l. \small BP and \small BQ are perpendiculars from \small B to the arms of \small \angle A (see Fig. ). Show that: \small BP=BQ or \small B is equidistant from the arms of  \small \angle A.

Answer:

In the previous part we have proved that    \small \Delta APB\cong \Delta AQB.

Thus by c.p.c.t. we can write : 

      BP\ =\ BQ

Thus B is equidistant from arms of angle A.

Q6 In Fig, \small AC=AE,AB=AD and \small \angle BAD= \angle EAC. Show that \small BC=DE.

                                  

Answer:

From the given figure following result can be drawn :-  

                     \angle BAD\ =\ \angle EAC

Adding \angle DAC to the both sides, we get :

                     \angle BAD\ +\ \angle DAC\ =\ \angle EAC\ +\ \angle DAC

                                               \angle BAC\ =\ \angle EAD

Now consider \Delta ABC  and   \Delta ADE ,    :-

  (i)               AC\ =\ AE                          (Given)

  (ii)   \angle BAC\ =\ \angle EAD                      (proved above)

 (iii)          AB\ =\ AD                              (Given)

Thus by SAS congruence we can say that :

                                           \Delta ABC\ \cong \ \Delta ADE

Hence by c.p.c.t., we can say that :                BC\ =\ DE

Q7 (i) \small AB is a line segment and \small P is its mid-point. \small D and \small E are points on the same side of \small AB such that \small \angle BAD=\angle ABE and  \small \angle EPA=\angle DPB (see Fig). Show that \small \Delta DAP\cong \Delta EBP

            

Answer:

From the figure, it is clear that :
                                     \angle EPA\ =\ \angle DPB

Adding \angle DPE both sides, we get :

                                     \angle EPA\ +\ \angle DPE =\ \angle DPB\ +\ \angle DPE

or                                                            \angle DPA =\ \angle EPB

Now, consider \Delta DAP and   \Delta EBP :

(i)   \angle DPA =\ \angle EPB                 (Proved above)

(ii)    AP\ =\ BP                             (Since P is the midpoint of line AB)                               

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

Hence by ASA congruence, we can say that :

                                                 \small \Delta DAP\cong \Delta EBP

Q7 (ii) AB is a line segment and P is its mid-point. D and E are points on the same side of AB  such that \small \angle BAD=\angle ABE and \small \angle EPA=\angle DPA (see Fig). Show that \small AD=BE

                

Answer:

In the previous part we have proved that \small \Delta DAP\cong \Delta EBP.

Thus by c.p.c.t., we can say that :

     \small AD=BE

Q8 (i) In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to  M and produced to a point D such that \small DM=CM. Point D is joined to point B (see Fig.). Show that: \small \Delta AMC\cong \Delta BMD

                

Answer:

Consider \Delta AMC   and   \Delta BMD  ,

(i)    AM\ =\ BM          (Since M is the mid-point)

(ii)    \angle CMA\ =\ \angle DMB             (Vertically opposite angles are equal)

(iii)    CM\ =\ DM           (Given)

Thus by SAS congruency, we can conclude that :

   \small \Delta AMC\cong \Delta BMD

Q8 (ii) In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that \small DM=CM. Point D is joined to point B (see Fig.). Show that: \small \angle DBC is a right angle.

                  

Answer:

In the previous part, we have proved that   \small \Delta AMC\cong \Delta BMD.

By c.p.c.t. we can say that    :            \angle ACM\ =\ \angle BDM

This implies side AC is parallel to BD.

Thus we can write :            \angle ACB\ +\ \angle DBC\ =\ 180^{\circ}          (Co-interior angles)

and,                                            90^{\circ}\ +\ \angle DBC\ =\ 180^{\circ}

or                                                                 \angle DBC\ =\ 90^{\circ}

Hence \small \angle DBC is a right angle.

Q8 (iii) In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that \small DM=CM. Point D is joined to point B (see Fig.). Show that: \small \Delta DBC\cong \Delta ACB

                            

Answer:

Consider \Delta DBC  and  \Delta ACB,

(i)          BC\ =\ BC          (Common in both the triangles)

(ii)   \angle ACB\ =\ \angle DBC                    (Right angle)

(iii)         DB\ =\ AC            (By c.p.c.t. from the part (a) of the question.)

Thus SAS congruence we can conclude that :  

                                                      \small \Delta DBC\cong \Delta ACB

Q8 (iv) In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that \small DM=CM. Point D is joined to point B (see Fig.). Show that: \small CM=\frac{1}{2}AB

                                  

Answer:

In the previous part we have proved that  \Delta DBC\ \cong \ \Delta ACB.

Thus by c.p.c.t., we can write :        DC\ =\ AB

                                     DM\ +\ CM\ =\ AM\ +\ BM

or                                  CM\ +\ CM\ =\ AB                                              (Since M is midpoint.)

or                                                      \small CM=\frac{1}{2}AB.

Hence proved.

NCERT solutions for class 9 maths chapter 7 Triangles Excercise: 7.2

Q1 (i) In an isosceles triangle ABC, with \small AB=AC, the bisectors of \small \angle B and \small \angle C intersect each other at O. Join A to O. Show that : \small OB=OC

Answer:

In the triangle ABC,

Since AB  =  AC, thus        \angle B\ =\ \angle C

or                                        \frac{1}{2}\angle B\ =\ \frac{1}{2}\angle C

or                                      \angle OBC\ =\ \angle OCB      (Angles bisectors are equal)

Thus       \small OB=OC   as sides opposite to equal are angles are also equal.

Q1 (ii) In an isosceles triangle ABC, with \small AB=AC, the bisectors of \small \angle B and \small \angle C intersect each other at O. Join A to O. Show that : AO bisects \small \angle A 

Answer:

Consider \Delta AOB  and  \Delta AOC,

(i)   AB\ =\ AC                  (Given)

(ii)   AO\ =\ AO            (Common in both the triangles)

(iii)   OB\ =\ OC          (Proved in previous part)

Thus by  SSS congruence rule, we can conclude that  :

                                      \Delta AOB\ \cong \ \Delta AOC

Now, by c.p.c.t., 

                                      \angle BAO\ =\ \angle CAO

Hence  AO bisects \angle A.

Q2 In  \Delta ABC, AD is the perpendicular bisector of BC (see Fig). Show that \small \Delta ABC  is an isosceles triangle in which \small AB=AC.

                                     

Answer:

Consider \DeltaABD  and  \DeltaADC,

(i)  AD\ =\ AD                       (Common in both the triangles)

(ii)   \angle ADB\ =\ \angle ADC       (Right angle)

(iii)  BD\ =\ CD                      (Since AD is the bisector of BC)

Thus by SAS congruence axiom, we can state :

                                              \Delta ADB\ \cong \ \Delta ADC

Hence by c.p.c.t., we can say that :           \small AB=AC

Thus   \Delta ABC is an isosceles triangle with AB and AC as equal sides.

Q3 ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig.). Show that these altitudes are equal.

                

Answer:

Consider \Delta AEB   and  \Delta AFC,

(i)     \angle A    is common in both the triangles.

(ii)   \angle AEB\ =\ \angle AFC           (Right angles)

(iii)    AB\ =\ AC              (Given)

Thus by AAS congruence axiom, we can conclude that :

                                 \Delta AEB\ \cong \Delta AFC

Now, by c.p.c.t. we can say :     BE\ =\ CF

Hence these altitudes are equal.

Q4 (i) ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig). Show that \small \Delta ABE \cong \Delta ACF

             

Answer:

Consider \Delta ABE  and  \Delta ACF ,

(i)    \angle A  is common in both the triangles.

(ii)    \angle AEB\ =\ \angle AFC              (Right angles)

(iii)   BE\ =\ CF                             (Given)

Thus by  AAS congruence, we can say that :

                                              \small \Delta ABE \cong \Delta ACF   

Q4 (ii) ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig.). Show that \small AB=AC, i.e., ABC is an isosceles triangle.

                   

Answer:

From the prevoius part of the question we found out that  :        \Delta ABE\ \cong \Delta ACF

Now, by c.p.c.t. we can say that :           AB\ =\ AC

Hence \Delta \ ABC  is an isosceles triangle.

Q5 ABC and DBC are two isosceles triangles on the same base BC (see Fig.). Show that \small \angle ABD\ \cong \ \angle ACD .

                             

Answer:

Consider \Delta ABD  and   \Delta ACD,

(i)   AD\ =\ AD    (Common in both the triangles)

(ii)   AB\ =\ AC         (Sides of isosceles triangle)

(iii)   BD\ =\ CD       (Sides of isosceles triangle)

Thus by SSS congruency, we can conclude that :

 \small \angle ABD\ \cong \ \angle ACD

Q6 \Delta ABC is an isosceles triangle in which  AB=AC. Side BA is produced to D  such that AD=AB (see Fig.). Show that \angle BCD is a right angle.

               

Answer:

Consider \DeltaABC, 
It is given that   AB = AC


So, \angle ACB = \angle ABC    (Since angles opposite to the equal sides are equal.) 

Similarly in \DeltaACD, 

We have       AD = AB 
and       \angle ADC = \angle ACD  
So, 

                                       \angle CAB + \angle ACB + \angle ABC = 180^{\circ} 

                                        \angle CAB\ +\ 2\angle ACB = 180^{\circ}
or                                                                 \angle CAB\ = 180^{\circ}\ -\ 2\angle ACB                  ...........................(i)

And in \DeltaADC, 
                                                                     \angle CAD\ = 180^{\circ}\ -\ 2\angle ACD              ..............................(ii) 

Adding (i) and (ii), we get : 
                                                            \angle CAB\ +\ \angle CAD\ = 360^{\circ}\ -\ 2\angle ACD\ -\ 2\angle ACB 

 or                                                                              180^{\circ}\ = 360^{\circ}\ -\ 2\angle ACD\ -\ 2\angle ACB

and                                                                              \angle BCD\ =\ 90^{\circ}

Q7 ABC is a right angled triangle in which \small \angle A =90^{\circ} and \small AB=AC. Find \small \angle B and \small \angle C.

Answer:

In the triangle ABC, sides AB and AC are equal. 

We know that angles opposite to equal sides are also equal.

Thus,            \angle B\ =\ \angle C

Also, the sum of the interior angles of a triangle is 180^{\circ}.

So, we have :

                                      \angle A\ +\ \angle B\ +\ \angle C\ =\ 180^{\circ}

or                                                   90^{\circ} +\ 2\angle B\ =\ 180^{\circ}

or                                                                    \angle B\ =\ 45^{\circ}

Hence                        \angle B\ =\ \angle C\ =\ 45^{\circ}

Q8 Show that the angles of an equilateral triangle are \small 60^{\circ} each.

Answer:

Consider a triangle ABC which has all sides equal.

We know that angles opposite to equal sides are equal.

Thus we can write :                    \angle A\ =\ \angle B\ =\ \angle C

Also, the sum of the interior angles of a triangle is 180 ^{\circ}.

Hence,                                        \angle A\ +\ \angle B\ +\ \angle C\ =\ 180^{\circ}

or                                                                                3\angle A\ =\ 180^{\circ}

or                                                                                   \angle A\ =\ 60^{\circ}

So, all the angles of the equilateral triangle are equal (60^{\circ}).

NCERT solutions for class 9 maths chapter 7 Triangles Excercise: 7.3

Q1 (i) \small \Delta ABC and \small \Delta DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig.). If AD is extended to intersect BC at P, show that \small \Delta ABD\cong \Delta ACD

                                   

Answer:

Consider \Delta ABD  and   \Delta ACD ,

(i)  AD\ =\ AD             (Common)

(ii)   AB\ =\ AC            (Isosceles triangle)

(iii)   BD\ =\ CD          (Isosceles triangle)

Thus by SSS congruency we can conclude that : 

  \small \Delta ABD\cong \Delta ACD

Q1 (ii)  Triangles ABC and Triangle DBC are two isosceles triangles on the same base BC and vertices A  and D are on the same side of BC (see Fig). If AD is extended to intersect BC at P, show that \small \Delta ABP \cong \Delta ACP         

Answer:

Consider \Delta ABP  and   \Delta ACP,

(i)  AP  is common side in both the triangles.

(ii)  \angle PAB\ =\ \angle PAC                       (This is obtained from the c.p.c.t. as proved in the previous part.)

(iii)   AB\ =\ AC                                    (Isosceles triangles)

Thus by SAS axiom, we can conclude that :

    \small \Delta ABP \cong \Delta ACP

Q1 (iii) \small \Delta ABC and \small \Delta DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig.). If AD is extended to intersect BC at P, show that AP bisects \small \angle A  as well as \small \angle D.

                 

Answer:

In the first part, we have proved that \small \Delta ABD\cong \Delta ACD.

So, by c.p.c.t.   \angle PAB\ =\ \angle PAC.

Hence AP bisects \angle A.

Now consider \Delta BPD  and  \Delta CPD,

(i)  PD\ =\ PD                         (Common)

(ii)  BD\ =\ CD                         (Isosceles triangle)

(iii)  BP\ =\ CP                         (by c.p.c.t. from the part (b))

Thus by SSS congruency we have   :

                                         \Delta BPD\ \cong \ \Delta CPD

Hence by c.p.c.t. we have :      \angle BDP\ =\ \angle CDP

or  AP bisects \angle D.

Q1 (iv) \small \Delta ABC and  \small \Delta DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig.). If AD is extended to intersect BC at P, show that AP is the perpendicular bisector of BC.

               

Answer:

In the previous part we have proved that   \Delta BPD\ \cong \ \Delta CPD.

Thus by c.p.c.t. we can say that  :       \angle BPD\ =\ \angle CPD

Also,                                                         BP\ =\ CP

SInce BC is a straight line, thus  :           \angle BPD\ +\ \angle CPD\ =\ 180^{\circ}

or                                                                                    2\angle BPD\ =\ 180^{\circ}

or                                                                                       \angle BPD\ =\ 90^{\circ}

Hence it is clear that  AP is a perpendicular bisector of line BC.

Q2 (i) AD is an altitude of an isosceles triangle ABC in which \small AB=AC. Show that AD bisects BC

Answer:

Consider \Delta ABD   and   \Delta ACD , 

(i)  AB\ =\ AC               (Given)

(ii)  AD\ =\ AD              (Common in both triangles)

(iii)  \angle ADB\ =\ \angle ADC\ =\ 90^{\circ}

Thus by RHS axiom we can conclude that :

    \Delta ABD\ \cong \ \Delta ACD

Hence by c.p.c.t. we can say that  :         BD\ =\ CD      or        AD bisects BC.  

Q2 (ii) AD is an altitude of an isosceles triangle ABC in which \small AB=AC. Show that AD bisects \small \angle A.

Answer:

In the previous part of the question we have proved that    \Delta ABD\ \cong \ \Delta ACD

Thus by c.p.c.t., we can write :             

 \angle BAD\ =\ \angle CAD

Hence AD  bisects  \angle A .

Q3 Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of \small \Delta PQR (see Fig). Show that:

(i) \small \Delta ABM \cong \Delta PQN

(ii) \small \Delta ABC \cong \Delta PQR

   

Answer:

(i)  From the figure we can say that :

                                             BC\ =\ QR

or                                          \frac{1}{2}BC\ =\ \frac{1}{2}QR

or                                            BM\ =\ QN

Now, consider \Delta ABM   and   \Delta PQN,

(a)  AM\ =\ PN                        (Given)

(b)   AB\ =\ PQ                          (Given)

(c)   BM\ =\ QN                        (Prove above)

Thus by SSS congruence rule, we can conclude that :

                                                           \small \Delta ABM \cong \Delta PQN

(ii)   Consider \Delta ABC  and   \Delta PQR :

(a)  AB\ =\ PQ                  (Given)

(b)   \angle ABC\ =\ \angle PQR   (by c.p.c.t. from the above proof)

(c)   BC\ =\ QR                 (Given)

Thus by SAS congruence rule,

                                                \small \Delta ABC \cong \Delta PQR

Q4 BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

Answer:

Using the given conditions, consider \Delta BEC   and     \Delta CFB,

(i)  \angle BEC\ =\ \angle CFB                 (Right angle)

(ii)    BC\ =\ BC                              (Common in both the triangles)

(iii)   BE\ =\ CF                             (Given that altitudes are of the same length. )

Thus by RHS axiom, we can say that  :                \Delta BEC\ \cong \Delta CFB

Hence by c.p.c.t.,     \angle B\ =\ \angle C

And thus  AB\ =\ AC      (sides opposite to equal angles are also equal). Thus ABC is an isosceles triangle.

Q5 ABC is an isosceles triangle with \small AB=AC. Draw \small AP\perp BC to show that \small \angle B=\angle C.

Answer:

Consider \Delta ABP  and  \Delta ACP,

(i)   \angle APB\ =\ \angle APC\ =\ 90^{\circ}       (Since it is given that AP is altitude.)

(ii)   AB\ =\ AC                                      (Isosceles triangle)

(iii)   AP\ =\ AP                                     (Common in both triangles)

Thus by RHS axiom we can conclude that :           

           \Delta ABP\ \cong \Delta ACP

Now, by c.p.c.t.we can say that : 

           \angle B\ =\ \angle C

NCERT solutions for class 9 maths chapter 7 Triangles Excercise: 7.4

Q1 Show that in a right-angled triangle, the hypotenuse is the longest side.

Answer:

Consider a right-angled triangle ABC with right angle at A.

We know that the sum of the interior angles of a triangle is 180.

So,                                    \angle A\ +\ \angle B\ +\ \angle C\ =\ 180^{\circ}

or                                       90^{\circ}\ +\ \angle B\ +\ \angle C\ =\ 180^{\circ}

or                                                       \angle B\ +\ \angle C\ =\ 90^{\circ}

Hence \angle B and \angle C are less than \angle A   (90^{\circ}).

Also, the side opposite to the largest angle is also the largest.

Hence the side BC is largest is the hypotenuse of the \Delta ABC.

Hence it is proved that in a right-angled triangle, the hypotenuse is the longest side.

Q2 In Fig, sides AB and AC of \small \Delta ABC are extended to points P and Q respectively. Also, \small \angle PBC < \angle QCB. Show that \small AC> AB.

             

Answer:

We are given that, 

                                    \small \angle PBC < \angle QCB                                                                                        ......................(i)

Also,                            \angle ABC\ +\ \angle PBC\ =\ 180^{\circ}                    (Linear pair of angles)          .....................(ii)

and                              \angle ACB\ +\ \angle QCB\ =\ 180^{\circ}                    (Linear pair of angles)          .....................(iii)

From (i), (ii) and (iii) we can say that :

                                     \angle ABC\ > \ \angle ACB

Thus   AC\ > AB   ( Sides opposite to the larger angle is larger.)

Q3 In Fig., \small \angle B <\angle A  and \small \angle C <\angle D. Show that \small AD <BC.

                

Answer:

In this question, we will use the property that sides opposite to larger angle are larger.

We are given \small \angle B <\angle A  and \small \angle C <\angle D . 

Thus,                     BO\ > AO              ..............(i)

and                        OC\ > OD             ...............(ii)

Adding (i) and (ii), we get :

                          AO\ +\ OD\ <\ BO\ +\ OC

or                                        AD\ <\ BC 

Hence proved.

Q4 AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Fig.). Show that \small \angle A>\angle C and \small \angle B>\angle D.

              

Answer:

                                           

Consider \Delta ADC in the above figure : 

AD\ <\ CD          (Given)

Thus         \angle CAD\ > \angle ACD                   (as angle opposite to smaller side is smaller)      

Now consider \Delta ABC,

We have :                BC\ > AB

and                            \angle BAC\ > \angle ACB

Adding the above result we get,

                       \angle BAC\ +\ \angle CAD > \angle ACB\ +\ \angle ACD

or                                                     \small \angle A>\angle C

Similarly, consider \Delta ABD,

we have               AB\ <\ AD  

Therefore               \angle ABD\ > \angle ADB

and in  \Delta BDC  we have,

                                             CD\ >\ BC

and                                \angle CBD\ >\ \angle CDB

from the above result we have,

                                \angle ABD\ +\ \angle CBD\ >\ \angle ADB\ +\ \angle CDB

or                                                                \small \angle B>\angle D

Hence proved.

Q5 In Fig ,  \small PR>PQ  and PS bisects \small \angle QPR. Prove that  \small \angle PSR>\angle PSQ.

                            

Answer:

We are given that     \small PR>PQ.

Thus                 \angle PQR\ =\ \angle PRQ

Also,  PS bisects \small \angle QPR , thus :

                          \angle QPS\ =\ \angle RPS

Now, consider \Delta QPS

                         \angle PSR\ =\ \angle PQR\ +\ \angle QPS                (Exterior angle)

Now, consider \Delta PSR

                         \angle PSQ\ =\ \angle PRQ\ +\ \angle RPS

Thus from the above the result we can conclude that :

                                 \small \angle PSR>\angle PSQ

Q6 Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

Answer:

Consider a right-angled triangle ABC with right angle at B.

Then              \angle B\ >\ \angle A\ or\ \angle C            (Since \angle B\ =\ 90^{\circ})

Thus the side opposite to largest angle is also largest.        AC\ >\ BC\ or\ AB

Hence the given statement is proved that all line segments are drawn from a given point, not on it, the perpendicular line segment is the shortest.

NCERT solutions for class 9 maths chapter 7 Triangles Excercise: 7.5

Q1 ABC is a triangle. Locate a point in the interior of \small \Delta ABC which is equidistant from all the vertices of \small \Delta ABC.

Answer:

We know that circumcenter of a triangle is equidistant from all the vertices. Also, circumcenter is the point of intersection of the perpendicular bisectors of the sides of a triangle.

Thus, draw perpendicular bisectors of each side of the triangle ABC. And let them meet at a point, say O. 

Hence O is the required point which is equidistant from all the vertices.

Q2 In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.

Answer:

The required point is called in-centre of the triangle. This point is the intersection of the angle bisectors of the interior angles of a triangle.

Hence the point can be found out in this case just by drawing angle bisectors of all the angles of the triangle.

Q3 In a huge park, people are concentrated at three points (see Fig.):

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 icecream parlour be set up so that maximum number of persons can approach it? (Hint : The parlour should be equidistant from A, B and C)

                                    

Answer:

The three main points form a triangle ABC. Now we have to find a point which is equidistant from all the three points.

Thus we need to find the circumcenter of the \Delta ABC

We know that circumcenter is defined as the point as the intersection point of the perpendicular bisectors of the sides of the triangle.

Hence the required point can be found out by drawing perpendicular bisectors of \Delta ABC.

Q4 Complete the hexagonal and star shaped Rangolies [see Fig. (i) and (ii)] by filling them with as many equilateral triangles of side \small 1 cm as you can. Count the number of triangles in each case. Which has more triangles?

                                

Answer:

For finding the number of triangles we need to find the area of the figure.

Consider the hexagonal structure :

                      Area of hexagon  =   6  \times   Area of 1 equilateral 

Thus area of the equilateral triangle :           

           =\ \frac{\sqrt{3}}{4}\times a^2

or        =\ \frac{\sqrt{3}}{4}\times 5^2

or        =\ \frac{25\sqrt{3}}{4}\ cm^2

So, the area of the hexagon is  : 

       =\6\times \frac{25\sqrt{3}}{4}\ =\ \frac{75\sqrt{3}}{2}\ cm^2

And the area of an equilateral triangle having 1cm as its side is :

       =\ \frac{\sqrt{3}}{4}\times 1^2

or    =\ \frac{\sqrt{3}}{4}\ cm^2

Hence a number of equilateral triangles that can be filled in hexagon are :     

    =\ \frac{\frac{75\sqrt{3}}{2}}{\frac{\sqrt{3}}{4}}\ =\ 150

Similarly for star-shaped rangoli :

 Area :             

   =\12\times \frac{\sqrt{3}}{4}\times 5^2 \ =\ 75\sqrt{3}\ cm^2

Thus the number of equilateral triangles are : 

      =\ \frac{75\sqrt{3}}{\frac{\sqrt{3}}{4}}\ =\ 300

Hence star-shaped rangoli has more equilateral triangles.

 

NCERT solutions for class 9 maths chapter wise

Chapter No.

Chapter Name

Chapter 1

NCERT solutions for class 9 maths chapter 1 Number Systems

Chapter 2

CBSE NCERT solutions for class 9 maths chapter 2 Polynomials

Chapter 3

Solutions of NCERT class 9 maths chapter 3 Coordinate Geometry

Chapter 4

NCERT solutions for class 9 maths chapter 4 Linear Equations In Two Variables

Chapter 5

CBSE NCERT solutions for class 9 maths chapter 5 Introduction to Euclid's Geometry

Chapter 6

Solutions of NCERT class 9 maths chapter 6 Lines And Angles

Chapter 7

NCERT solutions for class 9 maths chapter 7 Triangles

Chapter 8

CBSE NCERT solutions for class 9 maths chapter 8 Quadrilaterals

Chapter 9

Solutions of NCERT class 9 maths chapter 9 Areas of Parallelograms and Triangles

Chapter 10

NCERT solutions for class 9 maths chapter 10 Circles

Chapter 11

CBSE NCERT solutions for class 9 maths chapter 11 Constructions

Chapter 12

Solutions of NCERT class 9 maths chapter 12 Heron’s Formula

Chapter 13

NCERT solutions for class 9 maths chapter 13 Surface Area and Volumes

Chapter 14

CBSE NCERT solutions for class 9 maths chapter 14 Statistics

Chapter 15

Solutions of NCERT class 9 maths chapter 15 Probability

NCERT solutions for class 9 subject wise

CBSE NCERT Solutions for Class 9 Maths

Solutions of NCERT Solutions for Class 9 Science 

How to use NCERT solutions for class 9 maths chapter 7 Triangles

  • Learn some basic properties of triangles using some books of previous classes.

  • Go through all the theorems and examples given in the chapter.

  • Once you have memorized all the theorems and properties, then you can practice the questions from practice exercises

  • During the practice, you can use NCERT solutions for class 9 maths chapter 7 Triangles as a helpmate.

  • After doing all the exercises you can do some questions from past papers.

Keep Working Hard and Happy Learning!

 

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