# NCERT Solutions for Class 9 Maths Chapter 7 Triangles

NCERT Solutions for Class 9 Maths Chapter 7 Triangles – This article contains detailed NCERT Solutions for Class 9 Maths Chapter 7 Triangles curated by the team of subject matter experts at GFG, to help students understand how to solve the NCERT problems easily.

Chapter 7 Triangles ofÂ NCERT for Class 9 Maths helps students understand the basic concepts of congruence of triangles and the rules of congruence. It is also helpful in understanding a few more properties of triangles and the inequalities in a triangle.

NCERT Chapter 7 Triangles in Class 9 covers topics such as :

This article provides solutions to all the problems asked in Class 9 Maths Chapter 7 Triangles of your NCERT textbook in a step-by-step manner as per the latest CBSE Syllabus 2023-24 and guidelines. Solutions to all the exercises in the NCERT Class 9 Maths Chapter 7 Triangles are regularly revised to check errors and updated according to the latest CBSE Syllabus 2023-24 and guidelines.

Class 9 Maths NCERT Solutions Chapter 7 Exercises:

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.1 â€“ 8 Questions & Solutions (6 Short Answers, 2 Long Answers)
NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.2Â â€“ 8 Questions & Solutions (6 Short Answers, 2 Long Answers)
NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.3Â â€“ 5 Questions & Solutions (3 Short Answers, 2 Long Answers)
NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.4Â â€“ 6 Questions & Solutions (5 Short Answers, 1 Long Answer)
NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.5Â â€“ 4 Questions & Solutions (3 Short Answers, 1 Long Answer)

## NCERT Solutions for Class 9 Maths Chapter 7 Triangles: Exercise 7.1

### Question 1. In quadrilateral ACBDAC = AD and AB bisects âˆ  A (see Fig. 7.16). Show that âˆ† ABC â‰… âˆ† ABD What can you say about BC and BD?

Solution:

Given that AC and AD are equalÂ

i.e. AC = AD and the line AB bisects âˆ A.

Considering the two triangles Î”ABC and Î”ABD,

Where,Â

AC = AD { As given}………………………………………… (i)Â

âˆ CAB = âˆ DAB ( As AB Â bisects of âˆ A)……………. (ii)Â

AB { Common side of both the triangle} …….. …(iii)

From above three equation both the triangle satisfies “SAS” congruency criterion

So, Î”ABC â‰… Î”ABD.

Also,Â

BC and BD will be of equal lengths as they are corresponding parts of congruent triangles(CPCT).Â

So BC = BD.Â

### QuestionÂ 2. ABCD is a quadrilateral in which AD = BC and âˆ  DAB = âˆ  CBA (see Fig. 7.17). Prove that

(i) âˆ† ABD â‰… âˆ† BAC

(ii) BD = AC

(iii) âˆ  ABD = âˆ  BAC.

Solution:

(i) Given that AD = BC,Â

And âˆ  DAB = âˆ  CBA.Â

Considering two triangles Î”ABD and Î”BAC.Â

Where,Â

AD = BC { As given }………………………………………….. (i)Â

âˆ  DAB = âˆ  CBA { As given also}……………………….. (ii)Â

AB Â {Common side of both the triangle)…………. (iii)Â

From above three equation two triangles ABD and BAC satisfies “SAS” congruency criterion

So, Î”ABD â‰… Î”BAC

(ii) Also,Â

BD and AC Â will be equal as they are corresponding parts of congruent triangles(CPCT).Â

So BD = AC

(iii) Similarly,

âˆ ABD and âˆ BAC will be equal as they are corresponding parts of congruent triangles(CPCT).Â

So,

Â âˆ ABD = âˆ BAC.Â

### QuestionÂ 3. AD and BC are equal perpendiculars to a line segment AB (see Fig. 7.18). Show that CD bisects AB.Â

Solution:

Given that AD and BC are two equal perpendiculars to a line segment AB

Considering two triangles Î”AOD and Î”BOC

Where,Â

âˆ  AOD = âˆ  BOC {Vertically opposite angles}………………. (i)Â

âˆ  OAD = âˆ  OBC {Given that they are perpendiculars}…. (ii)Â

AD = BC {As given}………………………………………………… (iii)Â

From above three equation both the triangle satisfies “AAS” congruency criterion

So, Î”AOD â‰… Î”BOC

AO and BO will be equal as they are corresponding parts of congruent triangles(CPCT).Â

So, AO = BO

Hence, CD bisects AB at O.Â

### QuestionÂ 4. l and m are two parallel lines intersected by another pair of parallel lines p and q (see Fig. 7.19). Show that âˆ† ABC â‰… âˆ† CDA.

Solution:

Given that l and m are two parallel lines p and q are another pair of parallel lines

Considering two triangles Î”ABC and Î”CDA

Where,Â

âˆ  BCA = âˆ DAC Â {Alternate interior angles}…. (i)Â

âˆ  BAC = âˆ  DCA {Alternate interior angles}…. (ii)Â

Â AC Â {Common side of two triangles}………….(iii)

From above three equation both the triangle satisfies “ASA” congruency criterion

So, Î”ABC â‰… Î”CDA

### QuestionÂ 5. Line l is the bisector of an angle âˆ  A and B is any point on l. BP and BQ are perpendiculars from B to the arms of âˆ A (see Fig. 7.20). Show that:

(i) âˆ† APB â‰… âˆ† AQB

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

Solution:

Given that, Line l is the bisector of an angle âˆ  A and B

BP and BQ are perpendiculars from angle A.Â

Considering two triangles Î”APB and Î”AQB

Where,Â

âˆ  APB = âˆ  AQB { Two right angles as given }…… (i)Â

âˆ BAP = âˆ BAQ (As line l Â bisects Â angle A }……… (ii)Â

AB Â { Common sides of both the triangle }……… (iii)Â

From above three equation both the triangle satisfies “AAS” congruency criterion

So, Î”APBâ‰… Î”AQB.

(ii) Also we can say BP and BQ are equal as they are corresponding parts of congruent triangles(CPCT).

So, BP = BQ

### QuestionÂ 6. In Fig. 7.21, AC = AE, AB = AD and âˆ  BAD = âˆ  EAC. Show that BC = DE.

Solution:

Given that AC = AE, AB = ADÂ

And âˆ BAD = âˆ EAC

As given that âˆ BAD = âˆ EAC

Adding âˆ DAC on both the sidesÂ

We get,

âˆ BAD + âˆ DAC = âˆ EAC + âˆ  DAC

Â âˆ BAC = âˆ EAD

Considering two triangles Î”ABC and Î”ADEÂ

Where,Â

Â AC = AE Â { As given }…………………… (i)Â

âˆ BAC = âˆ EAD { Hence proven }…….. (ii)Â

AB = AD {As also given }……………….. (iii)Â

From above three equation both the triangle satisfies “SAS” congruency criterion

(ii) Also we can say BC and Â DE are equal as they are corresponding parts of congruent triangles(CPCT).

So that BC = DE.

### QuestionÂ 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 âˆ  BAD = âˆ  ABE and âˆ  EPA = âˆ  DPB (see Fig. 7.22). Show that:

(i) âˆ† DAP â‰… âˆ† EBP

Solution:

Given that P is the mid-point of line Â AB, So AP = BP

Also, âˆ  BAD = âˆ  ABE and âˆ  EPA = âˆ  DPB

Now adding âˆ DPE on both the sides of two equal angle âˆ  EPA = âˆ  DPB

âˆ  EPA + âˆ  DPE = âˆ  DPB + âˆ DPE

Which implies that two angles âˆ  DPA = âˆ  EPB

Considering two triangles âˆ† DAP and âˆ† EBP

âˆ  DPA = âˆ  EPB { Hence proven }…… (i)Â

AP = BP { Hence Given }……………… (ii)Â

âˆ  BAD = âˆ  ABE { As given }…………..(iii)

From above three equation both the triangle satisfies “ASA” congruency criterion

So, Î”DAP â‰… Î”EBP

(ii) Also we can say AD and BE are equal as they are corresponding parts of congruent triangles(CPCT).

### QuestionÂ 8. 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 DM = CM. Point D is joined to point B (see Fig. 7.23). Show that:

(i) âˆ† AMC â‰… âˆ† BMD

(ii) âˆ  DBC is a right angle.

(iii) âˆ† DBC â‰… âˆ† ACB

(iv) CM = Â 1/2 AB

Solution:

Given that M is the mid-point of Â ABÂ

So AM = BM

âˆ  ACB = 90Â°

and DM = CM

(i) Considering two triangles Î”AMC and Î”BMD:

AM = BM { As given }………………………………………. (i)Â

âˆ  CMA = âˆ  DMB { Vertically opposite angles }…. (ii)Â

CM = DM { As given also}……………………………….. (iii)Â

From above three equation both the triangle satisfies “SAS” congruency criterion

So, Î”AMC â‰… Î”BMD

(ii) From above congruency we can sayÂ

âˆ  ACD = âˆ  BDCÂ

Also alternate interior angles of two parallel lines AC and DB.Â

Since sum of two co-interiors angles results to 180Â°.

So, âˆ  ACB + âˆ  DBC = 180Â°

âˆ  DBC = 180Â° – âˆ  ACB

âˆ  DBC = 90Â° { As âˆ ACB =90Â° }

(iii) In Î”DBC and Î”ACB,

BC Â { Common side of both the triangle }……. (i)Â

âˆ  ACB = âˆ  DBC { As both are right angles }….(ii)Â

DB = AC (by CPCT)………………………………….. (iii)Â

From above three equation both the triangle satisfies “SAS” congruency criterion

So, Î”DBC â‰… Î”ACB

(iv) As M is the mid point so we can say

Â DM = CM = AM = BMÂ

Also we can say that AB = CD ( By CPCT )Â

As M is the mid point of CD we can write

CM + DM = AB

Hence, CM + CM = AB Â (As DM = CM )Â

Hence, CM = (Â½) AB

## NCERT Solutions for Class 9 Maths Chapter 7 Triangles : Exercise 7.2

### (i) OB = OC Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  (ii) AO bisects âˆ A

Solution:

Given: (i) An isosceles âˆ†ABC in which AB=AC

Â  Â  Â  Â  Â  Â  (ii) bisects of âˆ B and âˆ C each other at O.

Show: (i) OB=OC

Â  Â  Â  Â  Â  (ii) AO bisects âˆ A (âˆ 1=âˆ 2)

(i) In âˆ†ABC,

AB = AC

âˆ B =âˆ C Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â [angles opposite to equal sides are equal]

1/2 âˆ B = 1/2âˆ CÂ

âˆ OBC=âˆ OCB

âˆ´OB = OC Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â [sides opposite equal âˆ  are equal]

AB = AC Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  [given side]Â
1/2 âˆ B = 1/2âˆ CÂ
âˆ ABO = âˆ ACO Â  Â  Â  Â  Â  Â  Â  [Angle]
BO = OC Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â [proved above side]
âˆ´âˆ†AOB â‰… AOCÂ
Thus âˆ 1 = âˆ 2
Therefore, AO bisects âˆ A

### Question 2. In Î”ABC, AD is the perpendicular bisector of BC (see fig.). Show that Î”ABC is an isosceles triangle in which AB = AC.

Solution:

Given: AD is âŠ¥ bisector of BC

Show: AB=BC

In âˆ†ABD and âˆ†ACD

BD=DC Â  Â  Â  Â  Â  Â  Â  Â  Â  Â [AD is âŠ¥ bisector side]Â

AD=AD Â  Â  Â  Â  Â  Â  Â  Â  Â  Â [common side]Â

âˆ´âˆ†ABDâ‰…âˆ†ACD Â  Â  Â  Â [S.A.S]

AB=AC Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  [C.P.C.T]

### Question 3. 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.

Solution:

Given: AB=AC,BE and CF are altitudesÂ

Show: BE=CF

In âˆ†AEB and âˆ†AFC,

âˆ E=âˆ F Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  [Each 90Â° angle]Â

âˆ A=âˆ A Â  Â  Â  Â  Â  Â  Â  Â  Â  Â [common angle]Â

AB=AC Â  Â  Â  Â  Â  Â  Â  Â  Â  Â [given side]Â

âˆ´âˆ†AEBâ‰…âˆ†AFC Â  Â  Â  Â [A.A.S]

BE=CF Â  Â  Â  Â  Â  Â  Â  Â  Â  Â [C.P.C.T]

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

Solution:

Given: Altitudes BE and CF to sides AC and AB are equal

Show: (i) Î”ABE â‰… Î”ACF

Â  Â  Â  Â  Â  Â  Â (ii) AB = AC

(i) In âˆ†ABF and âˆ†ACF,

âˆ E=âˆ F Â  Â  Â  Â  Â  Â  Â  Â  Â  Â [Each 90Â° angle]Â

âˆ A=âˆ A Â  Â  Â  Â  Â  Â  Â  Â  Â  [common angle]Â

AB=AC Â  Â  Â  Â  Â  Â  Â  Â  Â  [given] S

âˆ´âˆ†AEBâ‰…âˆ†AFC Â  Â  Â  [A.A.S]

(ii) AB=AC Â  Â  Â  Â  Â  Â  [C.P.C.T]

### Question 5. ABC and DBC are two isosceles triangles on the same base BC (see Fig.). Show that âˆ ABD = âˆ ACD.

Solution:

Given: AB=AC,BD=DC

Show: âˆ ABD = âˆ ACD

In âˆ†ABD,

âˆ´âˆ 1=âˆ 2 Â  Â  Â  Â  Â  Â  Â  Â  Â [angle opposite to equal sides are equal] [1]

In âˆ†BDC,

BD=DC

âˆ´âˆ 3=âˆ 4 Â  Â  Â  Â  Â  Â  Â  Â  Â [angle opposite to equal sides are equal] [2]

âˆ 1+âˆ 2= âˆ 2+âˆ 4

âˆ ABD=âˆ ACD

### Question 6. Î”ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Fig.). Show that âˆ BCD is a right angle.

Solution:

In âˆ†ABC,

AB=AC

âˆ ACB=âˆ ABC Â  Â  Â  Â  Â  [1]

In âˆ†ACD,

âˆ ACD=âˆ ADC Â  Â  Â  Â  Â [2]

âˆ BCD=âˆ ABC+âˆ BDC

Adding âˆ BCD on both side

âˆ BCD+âˆ BCD=âˆ ABC+âˆ BDC+âˆ BCD

2âˆ BCD=180Â°

âˆ BCD=(180Â°)/2=90Â°

### Question 7. ABC is a right-angled triangle in which âˆ A = 90Â° and AB = AC. Find âˆ B and âˆ C.

Solution:

Find: âˆ B=? and âˆ C?

In âˆ†ABC,

AB=AC

âˆ´âˆ B=âˆ C Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  [angle opposite to equal side are equal]

âˆ A+âˆ B+âˆ C=180Â° Â  Â  Â  Â  Â  Â [angle sum property of triangle]

90Â°+âˆ B+âˆ B=180Â°

2âˆ B=180Â°-90Â°

âˆ B=(90Â°)/2=45Â°

Therefore, âˆ B=45Â° and âˆ C =45Â°

### Question 8. Show that the angles of an equilateral triangle are 60Â° each.

Solution:

Given: Let âˆ†ABC is an equilateral âˆ†Â

Show: âˆ A=âˆ B=âˆ C=60Â°

In âˆ†ABC,

AB=AC

âˆ B=âˆ C Â  Â  Â  Â  Â  Â  Â  Â  Â  Â [1]

Also

AC=BC

âˆ B=âˆ A Â  Â  Â  Â  Â  Â  Â  Â  Â  Â [2]

From 1,2 and 2

âˆ A=âˆ B=âˆ C

In âˆ†ABC,

âˆ A+âˆ B+âˆ C=180Â° Â  Â  Â [angle sum property of triangle]

âˆ A+âˆ A+âˆ A=180Â°

3âˆ A=180Â°

âˆ A=(180Â°)/3=60Â°

âˆ A=60Â°Â

âˆ´âˆ B=60Â° and âˆ C=60Â°

## NCERT Solutions for Class 9 Maths Chapter 7 Triangles: Exercise 7.3

### Question 1. Î”ABC and Î”DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Figure). If AD is extended to intersect BC at P, show that

(i) Î”ABD â‰… Î”ACD

(ii) Î”ABP â‰… Î”ACP

(iii) AP bisects âˆ A as well as âˆ D.

(iv) AP is the perpendicular bisector of BC.

Solution:

Given: âˆ†ABC and âˆ†DCB are isosceles âˆ†on the same base BC.

To show:Â

• Î”ABD â‰… Î”ACD
• Î”ABP â‰… Î”ACP
• AP bisects âˆ A as well as âˆ D.
• AP is the perpendicular bisector of BC.

i) in âˆ†ABD and âˆ†ACB

AB=AC

BD=CD

âˆ†ABDâ‰…âˆ†ACD ————-(S.S.S)

ii) in âˆ†ABP and âˆ†ACP

AB=AC

âˆ  BAPâ‰…âˆ CAP Â  Â [âˆ†ABDâ‰…âˆ†ACD Â  BY C.P.CT]

AP=AP Â  ———[common]

âˆ´[âˆ†ABDâ‰…âˆ†ACD Â  Â  Â  Â  ———–[S.A.S]

iii) [âˆ†ABDâ‰…âˆ†ACD Â  Â  Â  Â  ———–[S.A.S]

AP, bisects âˆ A Â  —————–1

In âˆ† BDP and âˆ†DPB

BD=CD Â  —————(GIVEN)

DP=PC Â  Â  ———-[âˆ†ABâ‰… âˆ†ACP Â  C.P.C.T]

DP=DP Â  Â  Â  Â  ———–[common]

âˆ´âˆ†BDPâ‰…âˆ†CDP Â  Â (S.S.S) Â

âˆ BDP=âˆ CDP Â (C.P.C.T)

DP bisects âˆ D

AP bisects âˆ D Â ——————-2

From 1 and 2, AP bisects âˆ  A as well as âˆ  D

iv) âˆ  AP +âˆ APC =180Â° Â  Â  Â  Â ————[linear pair]

âˆ APB=âˆ APC Â  Â ————-[âˆ†ABPâ‰…âˆ†ACP Â  Â  Â  C.P.CT]

âˆ APB + âˆ APC=180Â°

2 âˆ  APB=180Â°

âˆ APB=180/2=90Â°

BP=PC Â  Â  Â  Â  Â  Â (FROM ii)

âˆ´AP is âŠ¥ bisects of BC.

### (i) AD bisects BC Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â (ii) AD bisects âˆ A.

Solution:

To Show:Â

(i) AD bisects BC Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â (ii) AD bisects âˆ A.

AB=AC Â  Â  Â  Â  Â  Â  Â  ——————–[given]S

BD=DC Â  Â  Â  Â  Â  Â  ————-[c.p.c.t]

âˆ 1=âˆ 2 Â  Â  Â  Â  Â  Â  Â  ————-[c.p.c.t]

### 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 Î”PQR (see Fig. 7.40). Show that:

(i) Î”ABM â‰… Î”PQN

(ii) Î”ABC â‰… PQR

Solution:

Given: Â  Â  Â

AB=PQ

BC=QR

AM=PN

AM and PN are medians

To show:(i) Î”ABM â‰… Î”PQN Â  Â  Â  Â  Â  (ii) Î”ABC â‰… PQR

Solution: In Â Î”ABM and Î”PQN Â

AB=PQ

AM=PN

because AM and PN are medians BC=QR

therefore Â  =1/2BC=1/2QR Â

âˆ´BM=QN

âˆ´) Î”ABM â‰… Î”PQN Â  Â  Â  Â  Â  Â  Â  Â  Â  Â ———[S.S.S]

Â  Â  Â  Â  âˆ B=âˆ Q Â  Â  Â  Â  Â  Â  ——–[c.p.c.t]

ii)now in Â  Â  Â  Î”ABC and Â Î”PQR Â  Â

AB=PQ Â  Â  Â  Â  Â  Â  Â  Â ———-[given]

âˆ B=âˆ Q Â  Â  Â  Â  Â  Â  from (i)

BC=QR Â  Â  Â  Â  Â  Â  Â  —————-[given]

âˆ´ Î”ABC â‰… PQR Â  Â  Â  Â  Â  Â  Â  Â  [S.A.S]

### 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: altitude BE and CF are equal Â

To prove: Î”ABC is an isosceles Â Î”

In Î”BEC and Î”CEB

âˆ E=âˆ F Â  Â  Â  Â —————-[each 90Â°] R

BC=BC Â  Â  Â  —————–[common] Â H

BF=CF Â  Â  Â  —————-[given] Â S

# Î”BEC â‰… Î”CEB Â  Â  [R.H.S]

âˆ C=âˆ B Â  Â  Â  Â ————-[C.P.C.T]

Â In Î”ABC,

âˆ C=âˆ B Â  Â  Â  Â

### Â Question 5. ABC is an isosceles triangle with AB = AC. Draw AP âŠ¥ BC to show that âˆ B = âˆ C.

Solution:

Given: Â

In âˆ†ABC,

AB=BC

AP âŠ¥ BC

to show that: âˆ B = âˆ C.

Solution: Â

In âˆ†APB and âˆ†APC

âˆ APB = âˆ APC Â  Â  Â  Â  —————[ each 90Â°] Â R

AB=AC Â  Â  Â  Â  Â  Â  Â  ——————-[given] Â  Â  Â H

AP=AP Â  Â  Â  Â  ——————–[common] Â  Â Â S

âˆ´âˆ†APB â‰… âˆ†APC Â  Â  Â  Â  Â  Â ———-[R.H.S]

âˆ B = âˆ C Â —————[C.P.C.T] Â  Â  Â  Â  Â

## NCERT Solutions for Class 9 Maths Chapter 7 Triangles: Exercise 7.4

### Question 1. Show that in a right-angled triangle, the hypotenuse is the longest side.

Solution:

Given: Right angle triangle intersect âˆ B=90Â°

To show: AC>AB and AC>BC

Solution:âˆ A+âˆ B+âˆ C=180Â° Â  Â  Â  Â  Â  —————-[angle sum property]

âˆ A+90Â°+âˆ C=180Â° Â  Â  Â  Â

âˆ A+âˆ C=180Â°=90Â° Â

âˆ A+âˆ C=90Â° Â

âˆ´âˆ A<90Â° Â  Â  Â  Â  and Â âˆ C<90Â° Â

BC<AC Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  AB<AC Â  Â  Â  Â  Â  Â  Â  Â  Â  ———-[sides opposite to longer angle is greater]

âˆ´Hypotenuse Â is the longest side.

### Question 2. In Fig. 7.48, sides AB and AC of Î”ABC are extended to points P and Q respectively. Also, âˆ PBC < âˆ QCB. Show that AC > AB.

Solution:

Given : âˆ PBC < âˆ QCB Â  Â  Â  LET this be Â [âˆ 1 < âˆ 2]

To Show : AD < BC

Solution: âˆ 1 < âˆ 2 Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  ——–[given]

-âˆ 1 > -âˆ 2

180-âˆ 1>180-âˆ 2 Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â

âˆ 3>âˆ 4 Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â ———[linear pair]

In âˆ†ABC,

âˆ 3>âˆ 4 Â

AC>AB

### Question 3. In Fig. 7.49, âˆ B < âˆ A and âˆ C < âˆ D. Show that AD < BC.

Solution:

Given: âˆ B < âˆ A and âˆ C < âˆ D

Solution: In Â âˆ†BOA

âˆ B < âˆ A

âˆ´AO<BO————-1

In âˆ†COD

âˆ C < âˆ D

âˆ´OD<OC————-2

AO+OD+<BO+OC

### Question 4. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Fig. 7.50). Show that âˆ A > âˆ C and âˆ B > âˆ D.

Solution:

Given: AB is smaller side

CD is longest side

To show: âˆ A > âˆ C and âˆ B > âˆ D.

Solution : In âˆ†ABC

BC>AB Â

âˆ 1>âˆ 2 Â  Â  Â  Â  Â  Â  Â ———-[angle opposite to greater side is greatest]-1

In âˆ†ABC

âˆ 3>âˆ 4 Â  Â  Â  Â  Â  Â  ———[ angle opposite to greater side is greatest]-2

âˆ 1+âˆ 2+âˆ 3+âˆ 4

âˆ A>âˆ C

ii) In âˆ†ABD

âˆ 5>âˆ 6 Â  Â  Â  Â  Â  Â  ——————-[ angle opposite to greater side is greatest]-3

In âˆ†BCD

CD>BC

âˆ 7>âˆ 8 Â  Â  Â  Â  Â  Â  ——————-[ angle opposite to greater side is greatest]-4

âˆ 5+âˆ 6+âˆ 7+âˆ 8

âˆ B > âˆ D

### Question 5. In Fig 7.51, PR > PQ and PS bisects âˆ QPR. Prove that âˆ PSR > âˆ PSQ.

Solution:

Given: PR>PQ

PS is angle bisects âˆ QPR

To show: âˆ PSR > âˆ PSQ

Solution: PR>PQ

âˆ´âˆ 3+âˆ 4 Â  Â  Â  Â  Â ————[angle opposite to greater side is larger]

âˆ 3+âˆ 1+x=180Â° ————-[angle sum property of âˆ†] Â

âˆ 3=180Â°-âˆ 1-x ————1

In âˆ†PSR

4+âˆ 2+x=180Â° ————-[angle sum property of âˆ†] Â

âˆ 4=180Â°-âˆ 2-x ————2

Because âˆ 3>âˆ 4

180Â°-âˆ 1-x >180Â°-âˆ 2-x

-âˆ 1>-âˆ 2 Â

âˆ 1<âˆ 2 Â

âˆ PSQ<âˆ PSR Â OR âˆ PSR>âˆ PSQ

### Question 6. Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest

Solution:

Given: Let P be any point not lying on a line Â L.

PM âŠ¥ L

Now, âˆ  is any point another than M lying on line=L

In âˆ†PMN

âˆ M90Â° Â  Â  Â  Â  Â  Â  Â  Â  Â ———-[ PM âŠ¥ L]

âˆ L<90Â° Â  Â  Â  Â  Â  Â  Â  Â  ——-[âˆ´âˆ M90Â° âˆ L<90Â° Â âˆ L<90Â°] Â

âˆ L<âˆ M Â

AM<PL Â  Â  Â  Â  Â  Â  Â  Â ———[side opposite to greater is greater ]

## NCERT Solutions for Class 9 Maths Chapter 7 Triangles : Exercise 7.5 (Optional)

### Question 1. ABC is a triangle. Locate a point in the interior of Î”ABC which is equidistant from all the vertices of Î”ABC.

Solution:

To obtain a point which is equidistant from all vertices of a triangle we construct perpendicular bisectors of all sides (AB, BC, CA) of the triangle (Î”ABC). The point of intersection of these bisectors is known as Circumcenter(O) which is equidistant from all vertices.

### Question 2. In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.

Solution:

To obtain a point which is equidistant from all sides of a triangle we construct angle bisectors of all angles present in Î”ABC i.e âˆ BAC, âˆ ABC, âˆ ACB. The point of intersection of these bisectors is called Incentre(I) which is equidistant from all sides.

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

Solution:

The ice-cream parlour must be set somewhere so that it’s easily available for the public. So for such point it should be at a equal distance from point A, B, C & such point is termed as circumcenter.

### Question 4. Complete the hexagonal and star-shaped Rangolies 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:Â

We need to find the number of triangles that can get fit the above figures i.e the hexagon and the star.

So,

Area of hexagon = (Area of small triangle inside hexagon) * 6

Area of small equilateral triangle = âˆš3/4 * a2Â

= âˆš3/4 * 52

= âˆš3/4 * 25 = 25âˆš3/4

So,

Area of hexagon = 25âˆš3/4 * 6 Â  Â

= 150âˆš3/4 cm2Â  Â

Area of Star = Area of 6 triangles and 1 hexagon

= 6 * 25âˆš3/4 + 150âˆš3/4 Â

= 300âˆš3/4 cm2Â

Area of triangles of 1cm side that are to be fitted = âˆš3/4 * 12

= âˆš3/4 cm2

Number of triangles that can be accommodated inside hexagon and stars :

a. For Hexagon : Area of hexagon/ Area of 1cm side triangle

= 150âˆš3/4 cm2 / âˆš3/4 cm2Â

= 150 triangles Â  Â  Â  Â  Â  Â  Â

b. For Star : Area of star/ Area of 1cm triangle

= 300âˆš3/4 cm2 / âˆš3/4 cm2

= 300 triangles Â  Â  Â  Â

Hence, the star can accommodate 150 more triangles than the hexagon.

## Key Features of NCERT Solutions for Class 9 Maths Chapter 7 Triangles:

• These NCERT solutions are developed by the GfG team, with a focus on students’ benefit .
• NCERT Solutions are developed for each chapter of class 9 including Triangles.
• NCERT Solutions provides accurate and complete solutions for problems given in NCERT textbooks.

## FAQs on NCERT Solutions for Class 9 Maths ChapterÂ 7-Triangles

### 1. Why is it important to learn about Triangles?

One of the fundamental geometric forms is the triangle. Understanding polygonal characteristics and connections is based on them. Students build a solid geometric foundation by studying triangles, and they also gain understanding into the concepts of congruence, resemblance, and symmetry.

### 2. What topics are covered in NCERT Solutions for Class 9 Maths Chapter 7 Triangles?

NCERT Solutions for Class 9 Maths ChapterÂ 7-Triangles covers topics such as congruence of triangles, their rules, attributes, and triangle inequalities.

### 3. How can NCERT Solutions for Class 9 Maths Chapter 7 Triangleshelp me?

NCERT Solutions for Class 9 Maths ChapterÂ 7-Triangles can help you solve the NCERT exercise without any limitations. If you are stuck on a problem, you can find its solution in these solutions and free yourself from the frustration of being stuck on some question.

### 4. How many exercises are there in NCERT Solutions for Class 9 Maths Chapter 7 Triangles?

There are 3 exercises in the Class 9 Maths ChapterÂ 7-Triangles which covers all the important topics and sub-topics.

### 5. Where can I find NCERT Solutions for Class 9 Maths Chapter 7 Triangles?

You can find these NCERT Solutions in this article created by our team of experts at GeeksforGeeks.

Previous
Next