ML Aggarwal Triangles Exe-10.3 Class 9 ICSE Maths APC Understanding Solutions. Solutions of  Exercise-10.3. This post is the Solutions of  ML Aggarwal  Chapter 10- Triangles for ICSE Maths Class-9.  APC Understanding ML Aggarwal Solutions (APC) Avichal Publication Solutions of Chapter-10 Triangles for ICSE Board Class-9Visit official website CISCE for detail information about ICSE Board Class-9.

## ML Aggarwal Triangles Exe-10.3 Class 9 ICSE Maths Solutions

 Board ICSE Publications Avichal Publishig Company (APC) Subject Maths Class 9th Chapter-10 Triangles Writer ML Aggarwal Book Name Understanding Topics Solution of Exe-10.3 Questions Edition 2021-2022

### Exe-10.3 Solutions of ML Aggarwal for ICSE Class-9 Ch-10, Triangles

Note:- Before viewing Solutions of Chapter -10 Triangles Class-9 of ML AggarwaSolutions .  Read the Chapter Carefully. Then solve all example given in Exercise-10.1, Exercise-10.2, Exercise-10.3, Exercise-10.4, MCQs, Chapter Test.

### Triangles Exe-10.3

ML Aggarwal Class 9 ICSE Maths Solutions

Page 210

#### Question 1. ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.

In 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 = 90

(2∠B = 90o)

⇒ ∠B = 90/2= 45o

⇒ ∠B = ∠C = 45o

⇒ ∠B = ∠C = 45

#### Question 2. Show that the angles of an equilateral triangle are 60° each.

∆ABC is an equilateral triangle

AB = BC = CA

∠A = ∠B = ∠C (opposite to equal sides )

But ∠A + ∠B + ∠C = 180(sum of angles of a triangle)

3∠A = 180(∠A = 180o/3 = 60o)

⇒ ∠A = ∠B = ∠C = 60o

#### Question 3. Show that every equiangular triangle is equilateral.

∆ABC is an equiangular

∠A = ∠B = ∠C

In ∆ABC

∠B = ∠C …(i)

AC = AB …(ii) (sides opposite to equal angles)

∠C = ∠A

BC = AB

From (i) and (ii)

AB = BC = AC

∆ABC is an equilateral triangle

#### Question 4.  In the following diagrams, find the value of x:

##### (i) In following diagram given that AB =AC

That is ∠B = ∠ ACB (angles opposite to equal sides in a triangle are equal)

In a triangle are equal)

Now, ∠A + ∠B + ∠ACB = 180(sum of all angles in a triangle is 180o)

50 + ∠B + ∠B = 180o(∠A = 50(given) ∠B = ∠ACB)

⇒ 50+ 2 ∠B = 180(2∠B = 180– 50o)

⇒ 2∠B = 130o (∠B = 130/2 = 65o)

⇒ ∠ACB = 65o

Also ∠ACB + x= 180(Linear pair)

⇒ 65o + x= 180o (x= 180– 65o)

⇒ x= 115o

Hence, Value of x = 115

##### (ii) in ∆PRS,

Given that PR = RS

∠PSR = ∠RPS (Angles opposite in a triangle, equal sides are equal)

30= ∠RPS …(i) (∠PPS = 30o)

⇒ ∠QPS = ∠QPR + ∠RPS

⇒ ∠QPS = 52o + 30(Given, ∠QPR = 52and from (i), ∠RPS = 30o)

⇒ ∠QPS = 82o

Now, In ∆PQS

∠QPS + ∠QSP + PQS = 180(sum of all angles in a triangles is 180o)

= 82+ 30+ x= 180(from (2) ∠QPS = 82and ∠QSP = 30o (given)

⇒ 112o + x= 180(x= 180o – 112o)

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 = 27….(2) (given)

From (1) and (2) we get

∠BCD = 27o

Now,

∠CDA = ∠DBC + ∠BCD (exterior angles is equal to sum of two interior opposite angles)

Ext ∠CDA = 27+ 27o [from (2) and (3)]

∠CDA = 54o

Also, in ∆ACD

∠CAD + ∠CDA + ∠ACD = 180(sum of all angles in a triangle is 180o)

⇒ 54o + 54o+ Y = 180o

⇒ 108o + Y = 180o (Y = 180o – 108o)

⇒ y = 72o

Update soon..

#### Question 6.

(a) In the figure (1) given below, AB = AD, BC = DC. Find ∠ ABC.
(b)In the figure (2) given below, BC = CD. Find ∠ACB.
(c) In the figure (3) given below, AB || CD and CA = CE. If ∠ACE = 74° and ∠BAE =15°, find the values of x and y.

Update soon..

### Triangles Exe-10.3

ML Aggarwal Class 9 ICSE Maths Solutions

Page 211

#### (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 ∠.

##### (a) in 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 = 60o …(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 given figure,

Given, ∆ABD, AC meets BD in C. AB = BC, AC = CD.

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)

From, adding (1) and (2), we get

Now, in ∆ACD

Exterior ∠ACB = ∠CAD + ∠ADC …(4) (In an triangle, exterior angle is equal to sum of two interior opposite angles)

(c) In 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 + 42(In a triangle exterior angle is equal to sum of two interior opposite angles)

∠CEA = 66o …(1)

Then, in ∆ACE

AC = CE (given)

Therefore, ∠CAE = ∠CEA (In a triangle equal side have equal angles opposite to them)

By equation (1),

Y = 66o …(2)

Also, y + z + ∠CEA = 180o

We know that, 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 = 42o …(4)

Hence, from (2), (3) and (4) equation gives x = 42o, y = 66o and z = 48o.

#### Question 9. 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.

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

#### Question 10. In the given figure, AD, BE and CF arc altitudes of ∆ABC. If AD = BE = CF, prove that ABC is an equilateral triangle.

AD, BE and CF are altitudes of ∆ABC and

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

#### Question 11. In a triangle ABC, AB = AC, D and E are points on the sides AB and AC respectively such that BD = CE. Show that:

(i) ∆DBC ≅ ∆ECB
(ii) ∠DCB = ∠EBC
(iii) OB = OC, where O is the point of intersection of BE and CD.

#### Question 12. ABC is an isosceles triangle in which AB = AC. P is any point in the interior of ∆ABC such that ∠ABP = ∠ACP. Prove that

(a) BP = CP
(b) AP bisects ∠BAC.

(a)

Proof:

In  tr. APB and tr. APC,

AB = AC [given]

∠ ABP = ∠ ACP [given]

AP = AP [common]

So Tr ABP and Tr ACP are congruent  by (SSA)

So BP = PC   c.p.c.t.

(b) To Prove: AP bisects angle BAC

angle PAB = angle PAC [corresponding angles of congruent triangles]

thus, AP bisects angle BAC.

hence proved.

#### Question 13. In the adjoining figure, D and E are points on the side BC of ∆ABC such that BD = EC and AD = AE. Show that ∆ABD ≅ ∆ACE.

Point D and E are on side BC of ∆ABC,

such that BD = CE ,and AD = AE .

To prove : ∆ABD congruent to ∆ACE

proof :

AD = AE ( given )

[ Angles opposite to equal sides ]

[ Linear pair ]

ADB = 180° – x —–( 1 )

(iii) Similarly ,

AEC = 180° – x ——–( 2 )

(iv) In ∆ABD and ∆ACE ,

BD = EC ( S ) given ,

ADE = AEC ( A ) from ( 1 ) and ( 2 )]

AD = AE ( S ) given ,

Hence ,

∆ABD is congruent to ∆ACE.

[ ASA congruence Rule ]

### Triangles Exe-10.3

ML Aggarwal Class 9 ICSE Maths Solutions

Page 212

#### (b) In the figure (ii) given below, O is a point in the interior of a square ABCD such that OAB is an equilateral triangle. Show that OCD is an isosceles triangle.

##### (b) △ OAB is an equilateral triangle
So it can be written as
∠ OAB = ∠ OBA = AOB = 60o
From the figure we know that ABCD is a square
So we get
∠ A = ∠ B = ∠ C = ∠ D = 90o
In order to find the value of ∠ DAO
We can write it as
∠ A = ∠ DAO + ∠ OAB
By substituting the values we get
90o = ∠ DAO + 60o
On further calculation
∠ DAO = 90o – 60o
By subtraction
∠ DAO = 30o
We also know that ∠ CBO = 30o
Considering the △ OAD and △ OBC
We know that the sides of a square are equal
We know that the sides of an equilateral triangle are equal
OA = OB
By SAS congruence criterion
So we get OD = OC (c. p. c. t)
Hence, it is proved that △ OCD is an isosceles triangle.

#### Question 15.  In the given figure, ABC is a right triangle with AB = AC. Bisector of ∠A meets BC at D. Prove that BC = 2AD.

In ∆ABD and ∆ACD

AB = AC (given)

⇒ ∆DAB = ∆DAC (by SAS congruency rule)

and BD = DC

In ∆ABD,

AD2 + BD2 = AB2 …(i)

⇒ AD2 + DC2 = AC2 …(ii)

Adding (i) and (ii), we get

2 AD2 + BD2 + DC2 = AB2 + AC2

⇒ 2AD2 + BD2 + DC2 = BC2

⇒ 2 AD2 + 2BD2 = BC2

⇒ 2 (AD2 + BD2) = BC2