**NCERT Solutions of class 9 Maths exercise 7.1 – Triangles**

**NCERT Solutions of class 9 Maths exercise 7.1 of chapter 7- Triangles** are the **solutions** of unsolved questions of **class 9 NCERT maths** textbook, these** NCERT** **solutions** are solved by an expert of **maths** as per the CBSE norms.** The NCERT solutions** of **exercise 7.1 of chapter 7-Triangles** are based on the congruency of the **triangle**. Every year one or two questions of 3 to 4 marks are always asked from this **exercise 7.1**, so it is very important for **class 9** students to study **NCERT solutions of exercise 7.1.**

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

**Q1.In quadrilateral ACBD, AC = AD, and AB bisect ∠A (see the given figure). Show that ΔABC≅ΔABD. What can you say about BC and BD?**

Ans.

**GIVEN**:In quadrilateral ACBD, AC = AD

AB is the bisector of ∠A

**So, ∠BAC = ∠DAB**

**TO PROVE**: ΔABC≅ΔABD

**PROOF:** In ΔABC and ΔABD

We have

AC = AD (given)

AB = AB (common)

∠BAC = ∠DAB (given)

ΔABC≅ΔABD (SAS rule)

BC = BD (by CPCT)

**Q2. ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see the given figure ).Prove that**

(i) ΔABD ≅ΔBAC

(ii) BD = AC

(iii) ∠ABD = ∠BAC

Ans.

(i) **GIVEN:** ABCD is a quadrilateral

AD = BC and ∠DAB = ∠CBA

**TO PROVE:** ΔABD ≅ΔBAC

**PROOF:** In ΔABD and ΔBAC

AD = BC (given)

AB = AB (common)

∠DAB = ∠CBA

ΔABD ≅ΔBAC (SAS)

Hence proved

(ii) BD = AC (By CPCT)

(iii) ∠ABD = ∠BAC(By CPCT)

**Q3. AD and BC are equal perpendiculars to a line segment AB (see the given figure).Show that CD bisects AB.**

Ans.

**GIVEN:** AD = BC

AB ⊥ BC and AB ⊥ BC

∴∠OBC = ∠OAD = 90°

**TO PROVE:** CD bisects AB i.e AO = BO

**PROOF:** In ΔAOD and ΔBOD

Since BOA and COD is a line

∠DOA = ∠BOC (vertically opposite angle)

AD = BC (Given)

∠OBC = ∠OAD = 90°(given)

ΔBOC ≅ ΔAOD(AAS)

AO = BO (By CPCT)

Hence Proved

Q4. l and m are two parallel lines intersected by another pair of parallel lines p and q(see the given figure). Show that ΔABC ≅ CDA.

Ans.

**GIVEN:** p ∥ q and l∥ m

**TO PROVE:** ΔABC ≅ ΔCDA

**PROOF:**In the figure

l∥ m and AC is the transversal(given)

So, ∠DAC = ∠ACB (alternate angles)

p ∥ q and AC is the transversal(given)

∠BAC = ∠ACD(alternate angles)

AC = AC(given)

ΔABC ≅ Δ CDA(ASA rule)

Hence proved

**Q5.Line l is the bisector of an angle ∠A and B is any point on l ,BP and BQ are perpendiculars from B to the arms of ∠A(see the given figure). Show that**

**(i) ΔAPB ≅ΔAQB**

**(ii) BP = BQ or B is equidistant from the arms of ∠A**

Ans.

**(i) GIVEN:** ,BP and BQ are perpendiculars from B to the arms of ∠A

∴∠AQB = ∠APB = 90°

Line l is the bisector of an angle ∠A and

∴∠QAB = ∠PAB

**TO PROVE:** ΔAPB ≅ΔAQB

**PROOF:**∠AQB = ∠APB = 90°(given)

AB = AB (common)

∠QAB = ∠PAB (given)

ΔAPB ≅ΔAQB (AAS rule)

(ii) BP = BQ(by CPCT)

Hence proved

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**Q6. In the given figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC =DE.**

Ans.

**GIVEN:** AC = AE

AB = AD

∠BAD = ∠EAC

**TO PROVE:** BC =DE

**PROOF:** We have to prove BC =DE, which are sides of the triangle ΔABC and ΔADE

In triangles ΔABC and ΔADE

AB = AD(given)

AC = AE(given)

∠BAD = ∠EAC (given)

Adding both sides ∠DAC

∠BAD + ∠DAC = ∠EAC + ∠DAC

∠BAC = ∠DAE(proved)

ΔABC ≅ ΔADE (SAS rule)

BC = DE (by CPCT )

Hence proved

**Q7. AB is a line segment and p is its midpoint. D and E are points on the same sides of AB such that ∠BAD = ∠ABE and ∠EPA=∠DPB(see the given figure). Show that**

**(i) ΔDAP≅ ΔEBP**

**(ii) AD = BE**

Ans.(i)

**GIVEN:** ∠DAP = ∠EBP

∠EPA = ∠DPB

P is the mid point of AB

∴AP = BP

**TO PROVE:**

(i) ΔDAP≅ ΔEBP

(ii) AD = BE

PROOF: In ΔDAP and ΔEBP

∠BAD = ∠ABE(given)

∠EPA=∠DPB(given)

Adding ∠EPD both sides

∠EPA +∠EPD =∠DPB+ ∠EPD

∠APD = ∠EPB

AP = BP(given)

ΔDAP≅ ΔEBP(ASA rule)

**Q8.In right triangle ABC ,right angled at C , M is th 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 the given figure). Show that**

**(i) ΔAMC ≅ BMD**

**(ii) ∠DBC is a right angle**

**(iii) ΔDBC ≅ ΔACB**

Ans.

**(i) GIVEN:**ΔABC is a right triagle ,∠C = 90°

M is the midpoint of AB

∴BM = AM

AM = AB/2

DM = CM

**TO PROVE**: ΔAMC ≅ BMD

**PROOF:**DM = CM (given)

BM = AM (given)

∠BMD = ∠AMC (Vertically opposite angle)

ΔAMC ≅ BMD(SAS rule)

Hence proved

(ii) Since ΔAMC ≅ BMD(proved)

∠DBM = ∠CAM (By CPCT)

∠DBM and ∠CAM are alternate angles

∴DB ∥ AC

∠DBC + ∠ACB = 180° (co-interior angles)

∠ACB = 90°

∠DBC + 90° = 180°

∠DBC = 90°

Hence proved

(iii) Since ΔAMC ≅ BMD(proved)

BD = AC (By CPCT)

BC = BC(common)

∠DBC = ∠ACB =90°(Proved above)

ΔDBC ≅ ΔACB (SAS rule)

proved above

(iv) CM = DM(given)

CM + DM = DC

CM + CM = DC

2CM = DC

CM = 1/2 DC

Since we have proved

ΔDBC ≅ ΔACB

So, DC = AB (by CPCT)

Therefore

CM = 1/2 AB

Hence proved

**Class 9-Sample papers and question papers**

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**NCERT Solutions of class 9 maths**

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 |

Chapter 5- Introduction to Euclid’s Geometry | Chapter 13-Surface Areas and Volumes |

Chapter 6-Lines and Angles | Chapter 14-Statistics |

Chapter 7-Triangles | Chapter 15-Probability |

Chapter 8- Quadrilateral |

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**NCERT solutions of class 11 maths**

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

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

Chapter 3- Trigonometry | Chapter 11-Conic Sections |

Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |

Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |

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

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

Chapter 8- Binomial Theorem | Chapter 16- Probability |

**CBSE Class 11-Question paper of maths 2015**

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**NCERT solutions of class 12 maths**

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 |

Chapter 4-Determinants | Chapter 12-Linear Programming |

Chapter 5- Continuity and Differentiability | Chapter 13-Probability |

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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