ML Aggarwal Solutions for Class 9 Maths Chapter 10 Triangles are provided here for students to practise and prepare for their exams. Practising ML Aggarwal Solutions is the ultimate need for students who aspire to score good marks in the Maths examination. Students facing trouble in solving problems from the Class 9 ML Aggarwal textbook can refer to our free ML Aggarwal Solutions for Class 9 provided below. Students are suggested to practise these solutions given in the PDF, which also helps them to understand the basic concepts of Maths easily. Chapter 10 of ML Aggarwal Solutions for Class 9 explains triangles, types of triangles and properties of triangles. Triangles are three-sided polygons that consist of three edges and three vertices. The most important property of a triangle is that the sum of the internal angles of a triangle is equal to 180 degrees. These solutions are really helpful in getting a better understanding of the concepts of Triangles. These solutions help to develop confidence in students who are preparing for the exams. Chapter 10 – Triangles contains five exercises, and the ML Aggarwal Class 9 Maths Solutions present on this page provide solutions to the questions discussed in these exercises.
ML Aggarwal Solutions for Class 9 Maths Chapter 10 – Triangles
Access answers to ML Aggarwal Solutions for Class 9 Maths Chapter 10 – Triangles
Exercise 10.1
1. It is given that ∆ABC ≅ ∆RPQ. Is it true to say that BC = QR? Why?
Solution:
Given ∆ABC ≅ ∆RPQ
Therefore, their corresponding sides and angles are equal.
Therefore BC = PQ
Hence it is not true to say that BC = QR
2. “If two sides and an angle of one triangle are equal to two sides and an angle of another triangle, then the two triangles must be congruent.” Is the statement true? Why?
Solution:
No, it is not a true statement as the angles should be included angle of their two given sides.
3. In the given figure, AB=AC and AP=AQ. Prove that
(i) ∆APC ≅ ∆AQB
(ii) CP = BQ
(iii) ∠APC = ∠AQB.
Solution:
(i) In △ APC and △AQB
AB=AC and AP=AQ [given]
From the given figure, ∠A = ∠A [common in both the triangles]
Therefore, using SAS axiom, we have ∆APC ≅ ∆AQB
(ii) In △ APC and △AQB
AB=AC and AP=AQ [given]
From the given figure, ∠A = ∠A [common in both the triangles]
By using corresponding parts of congruent triangles concept, we have
BQ = CP
(iii) In △ APC and △AQB
AB=AC and AP=AQ [given]
From the given figure, ∠A = ∠A [common in both the triangles]
By using corresponding parts of congruent triangles concept, we have
∠APC = ∠AQB.
4. In the given figure, AB = AC, P and Q are points on BA and CA respectively such that AP = AQ. Prove that
(i) ∆APC ≅ ∆AQB
(ii) CP = BQ
(iii) ∠ACP = ∠ABQ.
Solution:
(i) In the given figure AB = AC
P and Q are points on BA and CA produced, respectively, such that AP = AQ
Now we have to prove ∆APC ≅ ∆AQB
By using corresponding parts of the congruent triangles concept, we have
CP = BQ
∠ACP = ∠ABQ
(ii) CP = BQ
(iii) ∠ACP = ∠ABQ
In ∆ APC and ∆AQB
AC = AB (Given)
AP = AQ (Given)
∠PAC =∠QAB (Vertically opposite angle)
5. In the given figure, AD = BC and BD = AC. Prove that :
∠ADB = ∠BCA and ∠DAB = ∠CBA.
Solution:
Given: in the figure, AD = BC, BD = AC
To prove :
(i) ∠ADB = ∠BCA
(ii) ∠DAB = ∠CBA
Proof : in ∆ADB and ∆ACB
AB = AB (Common)
AD = BC (given)
DB = AC (Given)
∆ADB = ∆ACD (SSS axiom)
∠ADB = ∠BCA
∠DAB = ∠CBA
6. In the given figure, ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA. Prove that
(i) ∆ABD ≅ ∆BAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC.
Solution:
Given: in the figure, ABCD is a quadrilateral
In which AD = BC
∠DAB = ∠CBA
To prove :
(i) ∆ABD = ∆BAC
(ii) ∠ABD = ∠BAC
Proof: in ∆ABD and ∆ABC
AB = AB (common)
∠DAB = ∠CBA (Given)
AD = BC
(i) ∆ABD ≅ ∆ABC (SAS axiom)
(ii) BD = AC
(ii) ∠ABD = ∠BAC
7.In the given figure, AB = DC and AB || DC. Prove that AD = BC.
Solution :
Given: in the given figure.
AB = DC, AB ∥ DC
To prove: AD = BC
Proof: AB ∥ DC
∠ABD = ∠CDB (Alternate angles)
In ∆ABD and ∆CDB
AB = DC
∠ABD = ∠CDB (Alternate angles)
BD = BD (common)
∆ABD ≅ CDB (SAS axiom)
AD = BC
8. In the given figure. AC = AE, AB = AD and ∠BAD = ∠CAE. Show that BC = DE.
Solution:
Given: in the figure, AC = AE, AB = AD
∠BAD = ∠CAE
To prove: BC = DE
Proof: in ∆ABC and ∆ADE
AB = AD (given)
AC = AE (given)
∠BAD + ∠DAC + ∠CAE
∠BAC = ∠DAE
∆ABC = ∆ ADE (SAS axiom)
BC = DE
9. In the adjoining figure, AB = CD, CE = BF and ∠ACE = ∠DBF. Prove that
(i) ∆ACE ≅ ∆DBF
(ii) AE = DF.
Solution:
Given: in the given figure
AB = CD
CE = BF
∠ACE = ∠DBF
To prove : (i) ∆ACE ≅ ∆DBF
(i) ∆ACE ≅ ∆DBF (SAS axiom)
AE = DE
(ii) AE = DF
Proof : AB = CD
Adding BC to both sides
AB + BC = BC + CD
AC = BD
Now in ∆ACE and ∆DBF
AC = BD (Proved)
CE = BF (Given)
∠ACE = ∠DBF (SAS axiom)
1. In triangles ABC and PQR, ∠A= ∠Q and ∠B = ∠R. Which side of APQR should be equal to side AB of AABC so that the two triangles are congruent? Give reason for your answer.
Solution:
In triangle ABC and triangle PQR
∠A = ∠Q
∠B = ∠R
AB = QP
Because triangles are congruent of their corresponding two angles and included sides are equal
2. In triangles ABC and PQR, ∠A = ∠Q and ∠B = ∠R. Which side of APQR should be equal to side BC of AABC so that the two triangles are congruent? Give reason for your answer.
Solution:
In ∆ABC and ∆PQR
∠A = ∠Q
∠B = ∠R
Their included sides AB and QR will be equal for their congruency.
Therefore, BC = PR by corresponding parts of congruent triangles.
3. “If two angles and a side of one triangle are equal to two angles and a side of another triangle, then the two triangles must be congruent”. Is the statement true? Why?
Solution:
The given statement can be true only if the corresponding (included) sides are equal otherwise, it is not.
4. In the given figure, AD is median of ∆ABC, BM and CN are perpendiculars drawn from B and C respectively on AD and AD produced. Prove that BM = CN.
Solution:
Given in ∆ABC, AD is median, BM and CN are perpendicular to AD from B and C, respectively.
To prove:
BM = CN
Proof:
In ∆BMD and ∆CND
BD = CD (because AD is median)
∠M = ∠N
∠BDM = ∠CDN (vertically opposite angles)
∆BMD ≅ ∆CND (AAS axiom)
Therefore, BM = CN.
5. In the given figure, BM and DN are perpendiculars to the line segment AC. If BM = DN, prove that AC bisects BD.
Solution:
Given in the figure, BM and DN are perpendicular to AC
BM = DN
To prove:
AC bisects BD that is BE = ED
Construction:
Join BD, which intersects AC at E
Proof:
In ∆BEM and ∆DEN
BM = DN
∠M = ∠N (given)
∠DEN = ∠BEM (vertically opposite angles)
∆BEM ≅ ∆DEN
BE = ED
Which implies AC bisects BD
6. In the given figure, l and m are two parallel lines intersected by another pair of parallel lines p and q. Show that ∆ABC ≅ ∆CDA.
Solution:
In the given figure, two lines l and m are parallel to each other and lines p and q are also a pair of parallel lines intersecting each other at A, B, C and D. AC is joined.
To prove:
∆ABC ≅ ∆CDA
Proof:
In ∆ABC and ∆CDA
AC = AC (common)
∠ACB = ∠CAD (alternate angles)
∠BAC = ∠ACD (alternate angles)
∆ABC ≅ ∆ DCA (ASA axiom)
7. In the given figure, two lines AB and CD intersect each other at the point O such that BC || DA and BC = DA. Show that O is the mid-point of both the line segments AB and CD.
Solution:
In the given figure, lines AB and CD intersect each other at O such that BC ∥ AD and BC = DA
To prove:
O is the midpoint of Ab and Cd
Proof:
Consider ∆AOD and ∆BOC
AD = BC (given)
∠OAD = ∠OBC (alternate angles)
∠ODA = ∠OCB (alternate angles)
∆AOD ≅ ∆BOC (SAS axiom)
Therefore, OA = OB and OD = OC
Therefore O is the midpoint of AB and CD.
Exercise 10.3
1. ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.
Solution:
In the right angled triangle ABC, ∠A = 90o
∠B + ∠C = 180o – ∠A
= 180o – 90o = 90o
Because AB = AC
∠C = ∠B (Angles opposite to equal sides)
∠B + ∠B = 90o (2∠B = 90o
∠B = 90/2o = 45o
∠B = ∠C = 45o
∠B = ∠C = 45o
2. Show that the angles of an equilateral triangle are 60° each.
Solution:
∆ABC is an equilateral triangle
AB = BC = CA
∠A = ∠B = ∠C (opposite to equal sides )
But ∠A + ∠B + ∠C = 180o (sum of angles of a triangle)
3∠A = 180o (∠A = 180o/3 = 60o)
∠A = ∠B = ∠C = 60o
3. Show that every equiangular triangle is equilateral.
Solution:
∆ABC is an equiangular
∠A = ∠B = ∠C
In ∆ABC
∠B = ∠C
AC = AB (sides opposite to equal angles)
Similarly, ∠C = ∠A
BC = AB
From (i) and (ii)
AB = BC = AC
∆ABC is an equilateral triangle
4. In the following diagrams, find the value of x:
Solution:
(i) in the following diagram, given that AB =AC
That is ∠B = ∠ ACB (angles opposite to equal sides in a triangle are equal)
Now, ∠A + ∠B + ∠ACB = 180o
(sum of all angles in a triangle is 180o)
50 + ∠B + ∠B = 180o
(∠A = 50o (given) ∠B = ∠ACB)
50o + 2 ∠B = 180o (2 ∠B = 180o – 50o
2 ∠B = 130o ( ∠B = 130/2 = 65o
∠ACB = 65o
Also ∠ACB + xo = 180o (Linear pair)
65o + xo = 180o (xo = 180o – 65o
xo = 115o
Hence, Value of x = 115
(ii) in ∆PRS,
Given that PR = RS
∠PSR = ∠RPS
(Angles opposite in a triangle, equal sides are equal)
30o = ∠RPS (∠PPS = 30o……(1)
∠QPS = ∠QPR + ∠RPS
∠QPS = 52o + 30o
(Given, ∠QPR = 52o and from (i), ∠RPS = 30o
∠QPS = 82o
Now, In ∆PQS
∠QPS + ∠QSP + PQS = 180o
(sum of all angles in a triangle is 180o)
= 82o + 30o + xo = 180o
(from (2) ∠QPS = 82o and ∠QSP = 30o (given)
112o + xo = 180o (xo = 180o – 1120
Hence, Value of x = 68
(iii) In the following figure, Given
That, BD = CD = AC and ∠DBC = 27o
Now in ∆ BCD
BD = CD (Given)
∠DBC = ∠BCD …….(1)
(in a triangle sides opposite equal angles are equal)
Also,, ∠DBC = 27o (given) ………..(2)
From (1) and (2) we get
∠BCD = 27o
Now, ext ∠CDA = ∠DBC + ∠BCD
(exterior angles are equal to the sum of two interior opposite angles)
Ext ∠CDA = 27o + 270 (from (2) and (3)
∠CDA = 54o ( from (4)) ……..(5)
Also, in ∆ACD
∠CAD + ∠CDA + ∠ACD = 1800
(sum of all angles in a triangle is 180o)
54O + 54O + Y = 180O
108O + Y = 180O (Y = 180O – 108O
y = 72o
5. (a) In the figure (1) given below, ABC is an equilateral triangle. Base BC is produced to E, such that BC’= CE. Calculate ∠ACE and ∠AEC.
(b) In the figure (2) given below, prove that ∠ BAD : ∠ ADB = 3 : 1.
(c) In the figure (3) given below, AB || CD. Find the values of x, y and ∠.
Solution:
(a) in the following figure
Given. ABC is an equilateral triangle BC = CE
To find. ∠ACE and ∠AEC
As given that ABC is an equilateral triangle,
That is ∠BAC = ∠B = ∠ACB = 600 ……..(1)
(each angle of an equilateral triangle is 60o)
Now, ∠ACE = ∠BAC + CB
(Exterior angle is equal to sum of two interior opposite angles)
(∠ACE = 60o + 60o)
∠ACE = 1200
Then, in ∆ACE
Given, AC = CE … [because AC = BC = CE]
∠CAE = ∠AEC … (2)
We know that in a triangle, equal sides have equal angles opposite to them.
So, ∠CAE + ∠AEC + 120o = 80o
∠AEC + ∠AEC + 120o = 180o … [by equation (2) we get]
2∠AEC = 180o – 120o
2∠AEC = 60o
∠AEC = 60o/2
∠AEC = 30o
Therefore, ∠ACE = 120o and ∠AEC = 30o.
(b) In the given figure,
Given ∆ABD, AC meets BD in C. AB = BC, AC = CD.
We have to prove that ∠BAD : ∠ADB = 3: 1
Then, consider ∆ABC,
AB = BC … [given]
Therefore, ∠ACB = ∠BAC … (1)
(In a triangle, equal angles opposite to them)
In ∆ACD,
AC = CD … [given]
Therefore, ∠ADC = ∠CAD
(In a triangle, equal sides have equal angles opposite to them)
∠CAD = ∠ADC … (2)
By adding (1) and (2), we get
∠BAC + ∠CAD = ∠ACB + ∠ADC
∠BAD = ∠ACB + ∠ADC … (3)
Now, in ∆ACD
Exterior ∠ACB = ∠CAD + ∠ADC … (4)
(In a triangle, the exterior angle is equal to the sum of two interior opposite angles)
Therefore, ∠ACB = ∠ADC + ∠ADC … [from (2) and (4)]
∠ACB = 2∠ADC … (5)
Now, ∠BAD = 2∠ADC + ∠ADC … [from (3) and (4)]
∠BAD = 3∠ADC = (∠BAD/∠ADC) = 3/1
∠BAD: ∠ADC = 3: 1
(c) In the given figure,
Given AB parallel to CD, ∠ECD = 24o, ∠CDE = 42o
We have to find the value of x, y and z.
Consider, ∆CDE
Exterior, ∠CEA = 24o + 42o
(In a triangle exterior angle is equal to the sum of two interior opposite angles)
∠CEA = 66o … (1)
Then, in ∆ACE
AC = CE … (given)
Therefore, ∠CAE = ∠CEA
(In a triangle, the equal sides have equal angles opposite to them)
By equation (1),
Y = 66o … (2)
Also, y + z + ∠CEA = 180o
We know that the sum of all angles in a triangle is 180o
66o + z + 66o = 180o
z + 132o = 180o
z = 180o – 132o
z = 48o … (3)
Then it is given that AB is parallel to CD,
∠x = ∠ADC … [alternate angles]
x = 42o … (4)
Therefore, from (2), (3) and (4) equation gives x = 42o, y = 66o and z = 48o.
6. In the given figure, AD, BE and CF arc altitudes of ∆ABC. If AD = BE = CF, prove that ABC is an equilateral triangle.
Given in the figure,
AD, BE and CF are altitudes of ∆ABC and
AD = BE = CF
To prove: ∆ABC is an equilateral triangle
Proof: in the right ∆BEC and ∆BFC
Hypotenuse BC = BC (Common)
Side BE = CF (Given)
∆BEC ≅ ∆BFC (RHS axiom)
∠C = ∠B
AB = AC (sides opposite to equal angles)
Similarly, we can prove that ∆CFA ≅ ∆ADC
∠A = ∠C
AB = BC
From (i) and (ii)
AB = BC = AC
∆ABC is an equilateral triangle
7. In the given figure, D is mid-point of BC, DE and DF are perpendiculars to AB and AC respectively such that DE = DF. Prove that ABC is an isosceles triangle.
Solution:
In triangle ABC
D is the midpoint of BC
DE perpendicular to AB
And DF perpendicular to AC
DE = DE
To prove:
Triangle ABC is an isosceles triangle
Proof:
In the right-angled triangle BED and CDF
Hypotenuse BD = DC (because D is a midpoint)
Side DF = DE (given)
∆BED ≅ ∆CDF (RHS axiom)
∠C = ∠B
AB = AC (sides opposite to equal angles)
∆ABC is an isosceles triangle
Exercise 10.4
1. In ∆PQR, ∠P = 70° and ∠R = 30°. Which side of this triangle is longest? Give reason for your answer.
Solution:
In ∆PQR, ∠P = 70o, ∠R = 30o
But ∠P + ∠Q + ∠R = 180o
100o + ∠Q = 180o
∠Q = 180o – 100o = 180o
∠Q = 80o the greatest angle
Its opposite side PR is the longest side
(side opposite to the greatest angle is the longest)
2. Show that in a right angled triangle, the hypotenuse is the longest side.
Solution:
Given: in right angled ∆ABC, ∠B = 90o
To prove: AC is the longest side
Proof: in ∆ABC,
∠B = 90o
∠A and ∠C are acute angles
That is less then 90o
∠B is the greatest angle
Or ∠B> ∠C and ∠B> ∠A
AC > AB and AC > BC
Hence AC is the longest side
3. PQR is a right angle triangle at Q and PQ : QR = 3:2. Which is the least angle.
Solution:
Here, PQR is a right-angle triangle at Q. Also, given that
PQ : QR = 3:2
Let PQ = 3x, then, QR = 2x
It is clear that QR is the least side,
Then, we know that the least angle has the least side
Opposite to it.
Hence ∠P is the least angle
4. In ∆ ABC, AB = 8 cm, BC = 5.6 cm and CA = 6.5 cm. Which is (i) the greatest angle ? (ii) the smallest angle ?
Solution:
Given that AB = 8 cm, BC = 5.6 cm, CA = 6.5 cm.
Here AB is the greatest side
Then ∠C is the least angle
The greatest side has the greatest angle opposite to it)
Also, BC is the shortest side
Then ∠A is the least angle
(the least side has least angle opposite to it)
Chapter test
1. In triangles ABC and DEF, ∠A = ∠D, ∠B = ∠E and AB = EF. Will the two triangles be congruent? Give reasons for your answer.
Solution:
In ∆ABC and ∆DEF
∠A = ∠D
∠B = ∠E
AB = EF
In ∆ABC, two angles and included side is given. In ∆DEF, corresponding angles are equal, but the side is not included in their angle.
Triangles Cannot be congruent.
2. In the adjoining figure, ABCD is a square. P, Q and R are points on the sides AB, BC and CD respectively such that AP= BQ = CR and ∠PQR = 90°. Prove that (a) ∆PBQ ≅ ∆QCR (b) PQ = QR (c) ∠PRQ = 45°
Solution:
Given: in the given figure, ABCD is a square
P,Q and R are the Points on the sides AB,
BC and CD respectively such that
AP = BQ = CR, ∠PQR = 90o
To prove: (a) ∆PBQ = ∆QCR
(b) PQ = QR
(c) ∠PQR = 45o
Proof: AB = BC = CD (Sides of Square)
And AP = BQ = CR (Given)
Subtracting, we get
AB – AP = BC – BQ = CD – CR
(PB = QC = RD)
Now in ∆PBQ and ∠QCR
PB = QC (Proved)
BQ = CR (Given)
∠B = ∠C (Each 90o)
∆PBQ ≅ ∆QCR
PQ = QR
But ∠PQR = 90o
∠RPQ = ∠PQR
(Angles opposite to equal angles)
But ∠RPQ + ∠PRQ = 90o
∠RPQ = ∠PQR = 90o
∠RPQ = ∠PRQ = 90o/2 = 45o
3. In the given figure, OA ⊥ OD, OC X OB, OD = OA and OB = OC. Prove that AB = CD.
Solution :
Given : in the figure OA ⊥ OD, OC ⊥ OB.
To prove: AB = CD
Proof : ∠AOD = ∠COB (each 90o)
Adding ∠AOC
∠AOD + ∠AOC = ∠AOC + ∠COB
∠COD = ∠AOB
Now, in ∆AOB and ∆DOC
OA = OD (Given)
OB = OC (Given)
∠AOB = ∠COD (Proved)
∆AOB ≅ ∆DOC
AB = CD
4. In the given figure, PQ || BA and RS CA. If BP = RC, prove that: (i) ∆BSR ≅ ∆PQC (ii) BS = PQ (iii) RS = CQ.
Solution:
PQ ∥ BA, RS ∥ CA
BP = RC
To prove :
(i) ∆BSR ≅ ∆PQC
(ii) RS = CQ
Proof : BP = RC
BC – RC = BC – BP
BR = PC
Now, in ∆BSR and ∆PQC
∠B = ∠P (Corresponding angles)
∠R = ∠C (Corresponding angles)
BR = PC (Proved)
∆BSR ≅ ∆PQC
BS = PQ
RS = CQ
