**NCERT Solutions Class 9 Maths exercise 7.2 of the chapter 7-Triangles**

**NCERT Solutions Class 9 Maths exercise 7.2 of the chapter 7-Triangles** is based on the** solutions** of the questions related to congruency of **triangles**.All questions of the **exercise 7.2 -Triangles** are solved by an expert of CBSE **Maths**.All students of** class 9** are required to go through each solutions for clearing their concept on** Triangles**.You can study here science and **maths** **NCERT solutions** and notes of each** chapters** from** class** **9-12**, sample papers, **solutions** of previous year’s question papers**,solutions** of important questions of science and **maths** ,articles on science and **maths**,government entrance exams and other competitive entrance exams and online jobs.

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**NCERT Solutions of Class 9 Science : Chapter 1 to Chapter 15**

**Q1.In an isosceles triangle ABC, with AB = AC, the bisector of angle ∠B and ∠C, intersect each other at O. Join A to O. Show that:**

**(i) OB = OC (ii) AO bisects ∠A**

Ans.

**GIVEN**: In ΔABC

AB = AC

BO is bisector of ∠B and CO is bisector of ∠C

**TO PROVE**:(i) OB = OC (ii) AO bisects ∠A

**PROOF**:In ΔABC

AB = AC (given)

∠ABC = ∠ACB (angles opposite to equal sides)

It is given to us that BO is bisector of ∠B and CO is bisector of ∠C

1/2(∠ABC) = 1/2(∠ACB)

∠OBC = ∠OCB

OB = OC (sides opposite to equal angles)

Hence proved

(ii) OB = OC (proved above)

AO = AO (common)

AB = AC (given)

ΔABO ≅ΔACO (SSS rule)

∠OAB = ∠OCA (by CPCT)

Therefore AO is the bisector of ∠A.

Hence proved

**Q2. In ΔABC , AD is perpendicular bisector of BC (see the given figure). Show that ΔABC is an isoscles triangle in which AB = AC.**

Ans.

**GIVEN:** In ΔABC , AD is perpendicular bisector of BC

∴ BD = DC

∠ADB = ∠ADC = 90°

**TO PROVE:** AB = AC

**PROOF: ** In ΔABD and ΔACD

∠ADB = ∠ADC = 90°(given)

AD = AD (common)

BD = DC(given)

ΔABD ≅ ΔACD (SAS rule)

AB = AC(by CPCT)

Hence ΔABC is an isosceles triangle in which AB = AC

**Q3. ABC is an isoscles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see the given figure).Show that these altitudes are equal.**

Ans.

**GIVEN:** ΔABC in which AC = AB

CF ⊥ AB and CE ⊥ AC

∴∠AFC = ∠AEB = 90°

**TO PROVE:** CF = BE

**PROOF:** AC = AB (given)

∠AFC = ∠AEB = 90°(given)

∠A = ∠A (common)

ΔAEB ≅ AFC (AAS rule)

BE = CF(by CPCT)

Hence proved

**Q4.ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see the given figure). Show that.**

**(i) ΔABE ≅ ACF**

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

Ans.

**GIVEN:** ΔABC in which BE = CF

CF ⊥ AB and CE ⊥ AC

∴∠AFC = ∠AEB = 90°

**TO PROVE:** AC = AB

**(i) PROOF:**BE = CF (given)

∠AFC = ∠AEB = 90°(given)

∠A = ∠A (common)

ΔAEB ≅ AFC (AAS rule)

(ii) AC = AB (by CPCT)

Hence proved

**Q5. ABC and DBC are two isosceles triangles on the same base BC (see the given figure). Show that ∠ABD = ∠ACD.**

Ans.

**GIVEN:** ΔABC in which AB = AC

ΔDBC in which BD = DC

**TO PROVE**: ∠ABD = ∠ACD

**PROOF: **In Δ ABC

AB = AC (given)

∠ABC = ∠ACB ….(i) (angles opposite to equal sides)

BD = DC

∠DBC = ∠DCB….(ii) (angles opposite to equal sides)

Adding equation (i) and (ii)

∠ABC + ∠DBC = ∠ACB + ∠DCB

∠ABD = ∠ACD

Hence proved

**Q6. ΔABC is an isosceles triangle in which AB = AC, side BA is produced to D such that AD = AB(see the given figure). Show that ∠BCD is a right angle.**

Ans.

**GIVEN:**AB = AC

AD = AB

**TO PROVE:** ∠BCD = 90°

**PROOF :** AB = AC (given)

∠ABC = ∠ACB (angles opposite to equal sides)

Applying the angle sum property in ΔABC

∠ABC + ∠BAC + ∠ACB = 180°

∠BAC = 180°- (∠ABC + ∠ACB)

∠BAC and ∠DAC are lienear pair

So, ∠BAC + ∠DAC = 180°

180°- (∠ABC + ∠ACB) + ∠DAC = 180°

∠DAC = ∠ABC + ∠ACB =∠ACB + ∠ACB =2∠ACB

∠ACB = 1/2 ∠DAC…..(i)

AD = AC (given)

∠ADC = ∠ACD (angles opposite to equal sides)

Applying the angle sum property in ΔABC

∠ADC + ∠ACD + ∠DAC = 180°

∠ACD + ∠ACD + ∠DAC = 180°

2∠ACD + ∠DAC = 180°

∠DAC = 180° – 2∠ACD

∠ACD = 90° – 1/2 ∠DAC……..(ii)

Adding equation (i) and (ii)

∠ACB + ∠ACD = 1/2 ∠DAC+ 90° – 1/2 ∠DAC

∠BCD = 90°

Hence ΔBCD is right triangle

**Q7. ABC is a right triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.**

Ans.

We are given triagle ABC in which

∠A = 90°

AB = AC

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

∠B + ∠C + ∠A = 180°

∠B + ∠B +90° = 180°

2∠B + 90° = 180°

2∠B = 90°

∠B = 45°

∠C = 45°

Hence both of the angles ∠B and ∠C are of 45°

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**Q8. Show that the angles of an equilateral triangle are 60° each.**

Ans.

We are given that ΔABC is an equilateral triangle

AB = AC (sides of equilateral triangle)

∠C = ∠B …..(i)(angles opposite to equal sides)

BC = AC (sides of equilateral triangle)

∠B = ∠A …..(ii)(angles opposite to equal sides)

From equation (i) and (ii)

∠A = ∠B = ∠C

Applying angle sum property of the triangle in ΔABC

∠A +∠B +∠C = 180°

∠A + ∠A + ∠A = 180°

3∠A = 180°

∠A = 180°/3

∠A = 60°

∴∠A = ∠B =∠C=60°

Therefore angles of an equilateral triangle are 60° each.

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Chapter 1- Number System | Chapter 9-Areas of parallelogram and triangles |

Chapter 2-Polynomial | Chapter 10-Circles |

Chapter 3- Coordinate Geometry | Chapter 11-Construction |

Chapter 4- Linear equations in two variables | Chapter 12-Heron’s Formula |

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Chapter 1-Sets | Chapter 9-Sequences and Series |

Chapter 2- Relations and functions | Chapter 10- Straight Lines |

Chapter 3- Trigonometry | Chapter 11-Conic Sections |

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Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |

Chapter 7- Permutations and Combinations | Chapter 15- Statistics |

Chapter 8- Binomial Theorem | Chapter 16- Probability |

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Chapter 1-Relations and Functions | Chapter 9-Differential Equations |

Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |

Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |

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Chapter 6- Application of Derivation | CBSE Class 12- Question paper of maths 2021 with solutions |

Chapter 7- Integrals | |

Chapter 8-Application of Integrals |

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