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

**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-9****. **Visit 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 Aggarwa**l **Solutions . 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.**

**Answer :**

In right angled triangle ABC, ∠A = 90^{o}

∠B + ∠C

= 180^{o} – ∠A

= 180^{o} – 90^{o} = 90^{o}

Because AB = AC

∠C = ∠B **(Angles opposite to equal sides)**

∠B + ∠B = 90^{o }

**(2∠B = 90 ^{o})**

⇒ ∠B = 90/2^{o }= 45^{o}

⇒ ∠B = ∠C = 45^{o}

⇒ ∠B = ∠C = 45^{o }

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

**Answer :**

∆ABC is an equilateral triangle

AB = BC = CA

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

But ∠A + ∠B + ∠C = 180^{o }(sum of angles of a triangle)

3∠A = 180^{o }(∠A = 180^{o/}3 = 60^{o})

⇒ ∠A = ∠B = ∠C = 60^{o}

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

**Answer :**

∆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:**

**Answer :**

**(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^{o }**(sum of all angles in a triangle is 180 ^{o})**

50 + ∠B + ∠B = 180^{o}**(∠A = 50 ^{o }(given) ∠B = ∠ACB)**

⇒ 50^{o }+ 2 ∠B = 180^{o }**(2∠B = 180 ^{o }– 50^{o})**

⇒ 2∠B = 130^{o} **(∠B = 130/2 = 65 ^{o})**

⇒ ∠ACB = 65^{o}

Also ∠ACB + x^{o }= 180^{o }**(Linear pair)**

⇒ 65^{o} + x^{o }= 180^{o} **(x ^{o }= 180^{o }– 65^{o})**

⇒ x^{o }= 115^{o}

Hence, Value of x = 115

**(ii) in ∆PRS,**

Given that PR = RS

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

30^{o }= ∠RPS **…(i) (∠PPS = 30 ^{o})**

⇒ ∠QPS = ∠QPR + ∠RPS

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

⇒ ∠QPS = 82^{o}

Now, In ∆PQS

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

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

⇒ 112^{o} + x^{o }= 180^{o }(x^{o }= 180^{o} – 112^{o})

Hence, Value of x = 68

**(iii)** **In the following figure, Given**

That, BD = CD = AC and ∠DBC = 27^{o}

Now in ∆BCD,

BD = CD (Given)

⇒ ∠DBC = ∠BCD …(1) (in a triangle sides opposite equal angles are equal)

Also,, ∠DBC = 27^{o }**….(2) (given)**

From (1) and (2) we get

∠BCD = 27^{o}

Now,

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

Ext ∠CDA = 27^{o }+ 27^{o}^{ }[from (2) and (3)]

∠CDA = 54^{o}

Also, in ∆ACD

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

⇒ 54^{o}^{ }+ 54^{o}+ Y = 180^{o}

⇒ 108^{o}^{ }+ Y = 180^{o}^{ }(Y = 180^{o}^{ }– 108^{o})

⇒ y = 72^{o}

**Question 5. ****In the following diagrams, find the value of x:**

**Answer :**

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.

**Answer :**

Update soon..

**Triangles Exe-10.3**

ML Aggarwal Class 9 ICSE Maths Solutions

Page 211

**Question 7. ****In ∆ABC, AB = AC, ∠A = (5x + 20)° and each of the base angle is the of ∠A. Find the measure of ∠A.**

**Answer :**

**Question 8. ****(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 ∠.**

**Answer :**

**(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 = 60^{o} **…(1) (each angle of an equilateral triangle is 60 ^{o})**

Now, ∠ACE = ∠BAC + CB **(Exterior angle is equal to sum of two interior opposite angles) (∠ACE = 60 ^{o} + 60^{o})**

∠ACE = 120^{0}

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 + 120^{o} = 80^{o}

⇒ ∠AEC + ∠AEC + 120^{o} = 180^{o}** [by equation (2) we get]**

⇒ 2∠AEC = 180^{o} – 120^{o}

⇒ 2∠AEC = 60^{o}

⇒ ∠AEC = 60^{o}/2

⇒ ∠AEC = 30^{o}

Therefore, ∠ACE = 120^{o} and ∠AEC = 30^{o}.

**(b)** **In 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)**

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

∠BAC + ∠CAD = ∠ACB + ∠ADC

⇒ ∠BAD = ∠ACB + ∠ADC **…(3)**

Now, in ∆ACD

Exterior ∠ACB = ∠CAD + ∠ADC **…(4) (In an triangle, exterior angle is equal to 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 given figure,

Given, AB parallel to CD, ∠ECD = 24^{o}, ∠CDE = 42^{o}

We have to find the value of x, y and z.

Consider, ∆CDE

Exterior, ∠CEA = 24^{o} + 42^{o }**(In a triangle exterior angle is equal to sum of two interior opposite angles)**

∠CEA = 66^{o} **…(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 = 66^{o} **…(2)**

Also, y + z + ∠CEA = 180^{o}

We know that, sum of all angles in a triangle is 180^{o}

66^{o} + z + 66^{o} = 180^{o}

⇒ z + 132^{o} = 180^{o}

⇒ z = 180^{o} – 132^{o}

⇒ z = 48^{o} **…(3)**

Then it is given that, AB is parallel to CD,

∠x = ∠ADC **[alternate angles]**

⇒ x = 42^{o} **…(4)**

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

**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.**

**Answer :**

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.**

**Answer :**

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

**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.

**Answer :**

**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.

**Answer :**

**(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.**

**Answer :**

Point D and E are on side BC of ∆ABC,

such that BD = CE ,and AD = AE .

To prove : ∆ABD congruent to ∆ACE

proof :

(i) In ∆ADE ,

AD = AE ( given )

**∠**ADE = **∠**AED = x

ii ) **∠**ADB + **∠**ADC= 180°

**∠**ADB + x = 180°

**∠**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

**Question 14.**

**(a) In the figure (i) given below, CDE is an equilateral triangle formed on a side CD of a square ABCD. Show that ∆ADE ≅ ∆BCE and hence, AEB is an isosceles triangle.**

**(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.**

**Answer :**

**(a) According to figure,**

$ADC=BCD=$

$CDE=DCE=$$º$

$∴ADE=BCE=$$º$

$InADEandBCE$

$AD=BC$

$DE=CE$

$ADE=BCE$

$ADE≅BCE$

**(b) △ OAB is an equilateral 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.**

**Answer :**

In ∆ABD and ∆ACD

AB = AC (given)

∠BAD = ∠CAD

As AD is bisector of ∠A and AD = AD

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

⇒ ∠ADB = ∠ADC (by c.p.c.t)

⇒ ∠ADB = ∠ADC = 90°

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

⇒ 4 AD2 = BC2

⇒ 2 AD = BC

i.e. BC = 2AD

Hence proved.

— : End of ML Aggarwal Triangles Exe-10.3 Class 9 ICSE Maths Solutions :–

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