**NCERT Solutions class 9 maths exercise 9.3 of chapter 9-Areas of parallelogram and triangles**

NCERT Solutions class 9 maths exercise 9.3 of chapter 9-Areas of parallelogram and triangles are created here for the purpose of helping the students of 9 class students to boost their preparation of exams. All questions of class 9 maths exercise 9.3 of chapter 9-Areas of parallelogram and triangles are solved by an expert of maths by a step by step method. Here you can study NCERT solutions of maths and science from class 9 to 12, sample papers, solutions of previous years’ question papers. carrier oriented articles.

**NCERT Solutions class 9 maths of chapter 9-Areas of parallelogram and triangles**

**Exercise 9.1 and 9.2- Areas of Parallelogram and Triangles**

**Exercise 9.3-Areas of Parallelogram and Triangles**

**Download pdf of NCERT solutions of class 9 maths chapter 9-Areas of Parallelogram and Triangle**

**PDF-NCERT solutions of class 9 maths chapter 9-Areas of Parallelogram**

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**Exercise 4.3- Linear equations in two variables**

**Exercise 4.1 & 4.2- Linear equation in two variables**

**Exercise.9.3- Areas of parallelograms and triangles**

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

**NCERT Solutions class 9 maths exercise 9.3 of chapter 9-Areas of parallelogram and triangles**

**Q1.In the given figure,E is any point on the median AD of a ΔABC .Show that ar(ABE) = ar(ACE).**

Ans. The median of triagle divides the triangle in two equal parts

AD is given us the median of ΔABC

∴ar ΔABD = arΔACD……(i)

ED is the median of ΔCBE

∴arΔDBE =arDCE……..(ii)

Subtracting equation (ii) from (i)

ar ΔABD – arΔDBE = arΔACD -arDCE

ar(ABE) = ar(ACE), Hence proved

**Q2. In a triangle ABC, E is the midpoint of median AD .Show that ar(ΔBED) =1/4 ar(ΔABC).**

Ans.

**GIVEN**: A ΔABC, AD is the median and E is the midpoint of AD

**TO PROVE:**ar(ΔBED) =1/4 ar(ΔABC)

**PROOF: **AD is the median of ΔABC (given)

∴ ar(ΔABD) = ar(ΔACD)

⇒ar(ΔABD) =1/2 ar((ΔABC)….(i)

AE = DE (E is the mid point of AD)

∴BE will be the median of ΔABD

ar(BED) = 1/2 ar (ABD)…..(ii)

From equation (i) and equation (ii)

ar(BED) = 1/2 × 1/2 ar((ΔABC)

ar(BED) = 1/4 ar((ΔABC),Hence proved

**Q3.Show that diagonal of a parallelogram divide it into four triangles of equal area.**

Ans.

**GIVEN:** ABCD is a parallelogram and AC and BC are diagonals

**TO PROVE: **ar(AOB) = ar(AOD) =ar(DOC) =ar(BOC)

**PROOF:** In ΔADB

DO = BO (diagonal of parallelogram bisect each other)

∴ AO is the median of ΔADB

since the median of the triangle divides the triangle into two equal parts

∴ ar(AOD) = ar(AOB)

DO is the median of ΔADC

∴ar(AOD) =ar(DOC)

BO is the median of ΔABC

∴ar(AOB) =ar(BOC)

It is clear that

ar(AOB) = ar(AOD) =ar(DOC) =ar(BOC),Hence proved

**Q4. In the given figure ,ABC and ABD are two triangles on the same base AB .If line segment CD is bisected by AB at O ,show that ar(ABC) = ar(ABD)**

**Ans.**

**GIVEN: In **ABCD ,CD is bisected by AB

CO = DO

**TO PROVE:** ar(ABC) = ar(ABD)

**PROOF: **CO = DO (given)

Therefore AO is the median of ΔADC and BO is the median of ΔBDC.

since the median of the triangle divides the triangle into two equal parts

∴ar(AOC)=ar(AOD)…(i)

ar(BOC) = ar(BOD)….(ii)

Adding both equations (i) and (ii)

ar(AOC)+ ar(BOC) =ar(AOD) + ar(BOD)

ar(ABC) = ar(ABD),Hence proved

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**Q5. D,E and F are the mid points of the sides BC ,CA and AB of a ΔABC. Show that **

**(i) BDEF is a parallelogram**

**(ii) ar(DEF) = 1/4 ar(ABC)**

**(iii) ar(BDEF) = 1/2ar(ABC)**

Ans.

**GIVEN:**D,E and F are the mid points of the sides BC ,CA and AB of a ΔABC

**TO PROVE:**

(i) BDEF is a parallelogram

(ii) ar(DEF) = 1/4 ar(ABC)

(iii) ar(BDEF) = 1/2ar(ABC)

**PROOF:** (i) F is the midpoint of AB and E is the mid point of AC

According to mid point theorem

FE∥BC, FE = 1/2(BC)

⇒FE∥BD….(i)

D is the midpoint of BC and E is the mid point of AC

According to the midpoint theorem

DE∥AB, DE = 1/2(AB)

DE∥BF…..(ii)

Equation (i) and equation (ii) shows that

BDEF is a parallelogram

(ii) BDEF is a parallelogram

ar(DEF) = ar(BDF)…..(i)

DFAE and DFAC will also be parallelograms

ar(DEF) = ar(AFE)….(ii)

ar(DEF) = ar(DEC)….(iii)

From the figure we have

ar(DEF) + ar(AFE) + ar(DEC) +ar(BDF) = ar(ABC)

From (i),(ii) and (iii) we have

ar(DEF) +ar(DEF)+ar(DEF)+ar(DEF) = ar(ABC)

4 ar(DEF) = ar(ABC)

ar(DEF) = 1/4(ABC), Hence proved

(iii) ar(BDEF) = 1/2ar(ABC)

ar(DEF) = 1/4 ar(ABC) [Proved above]

4ar(DEF) = ar(ABC)

2 ar(DEF) + 2(DEF) = ar(ABC)

ar(BDEF) + ar(BDEF) = ar(ABC)

2ar(BDEF) = ar(ABC)

ar(BDEF) = 1/2(arABC),Hence proved

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**Q6. In the given figure ,diagonal AC and BD of quadrilateral ABCD intersect at O such that OB = OD, if AB = CD,then show that **

**(i) ar(DOC) = ar(AOB)**

**(ii) ar(DCB) = ar(ACB)**

**(iii) DA ∥ CB or ABCD is a parallelogram**

**[ Hint: from D and B , draw perpendiculars to AC]**

Ans.

**GIVEN: **Diagonal AC and BD of quadrilateral ABCD intersect at O such that OB = OD

AB = CD

**CONSTRUCTION:** Drawing DE ⊥ AC, BF ⊥ AC

**TO PROVE:**

(i) ar(DOC) = ar(AOB)

(ii) ar(DCB) = ar(ACB)

(iii) DA ∥ CB or ABCD is a parallelogram

**PROOF: **In ΔDOE and ΔBOF

∠DOE = ∠BOF (vertically opposite angles)

∠DEO = ∠BFO =90°(DE ⊥ AC, BF ⊥ AC)

DO = BO (given)

ΔDOE ≅ ΔBOF (AAS rule)

ar(ΔDOE) = ar(ΔBOF)…..(i)

DE = BF (by CPCT)

In ΔDCE and ΔBFA

DE = BF (proved above)

AB = CD(given)

∠DEC = ∠BFA = 90°(DE ⊥ AC, BF ⊥ AC)

ΔDEC≅ ΔBFA(RHS rule)

ar(ΔDEC) = ar(ΔBFA)….(ii)

Adding equation (i) and (ii)

ar(ΔDOE)+ ar(ΔDEC) = ar(ΔBFA). + ar(ΔBOF)

ar(DOC) = ar(AOB),Hence proved

(ii) ar(DOC) = ar(AOB) [Proved above]

Adding ar(BOC) in both sides

ar(DOC)+ ar(BOC) = ar(AOB) + ar(BOC)

ar(DCB) = ar(ACB)

**Q7.D and E are points on sides AB and AC respectively of ΔABC such that ar(DBC) = ar(EBC).Prove that DE||BC.**

Ans.

**GIVEN:** D and E are points on sides AB and AC respectively of ΔABC such that ar(DBC) = ar(EBC)

**TO PROVE:**DE||BC

**PROOF: **We know the area of two triangles on the base and between the same parallels are equal,so its converse must also be true

Since ΔDBC and ΔEBC are on the same base BC and between the lines BC and DE

Therefore DE||BC,Hence proved

Q8.XY is a line parallel to side BC of a triangle ABC. If BE||AC and CF||AB meet XY at E and F respectively. show that ar(ABE) = ar(ACF)

Ans.

**GIVEN: **In figure, XY|| BC,BE||AC and CF||AB

**TO PROVE:** ar(ABE) = ar(ACF)

**PROOF: **BE||AC (given)

⇒ BE||YC…..(i)

CF||AB (given)

⇒BX ||AB ….(ii)

XY|| BC (given)

XY is extended upto E and F

⇒EY|| BC….(iii) and XF|| BC …(iv)

From (i) and (ii) it is clear that EYCB is a parallelogram

The parallelogram EYCB and triangle ABE are on the same base BE and between the same parallels BE||AC

ar(ΔABE) = 1/2(ar EYCB)…..(v)

From (ii) and (iii) it is clear that XFCB is a parallelogram

The parallelogram XFCB and triangle ACF are on the same base CF and between the same parallels BE||AC

ar(ΔACF) = 1/2(ar XFCB)…..(vi)

Since parallelogram EYCB and XFCB are on the same base BC and between the same parallels XF|| BC

ar(EYCB) = ar(XFCB)

From (vi),ar(ΔACF) = 1/2(ar EYCB)….(vii)

From (v) and (vii) ,we have

ar(ABE) = ar(ACF) ,Hence Proved

**Q9.The side AB of a parallelogram ABCD is produced to any point P.A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see the following figure). Show that ar(ABCD) = ar(PBQR)**

**[Hint: Join AC and PQ. Now compare area(ACQ) ad area(APQ)]**

Ans.

**GIVEN:** ABCD is a parallelogram and AQ||CP ,PBQR is a parallelogram

**TO PROVE**: ar(ABCD) = ar(PBQR)

**CONSTRUCTION**: Joining A to C and P to Q

**PROOF:** The ΔACQ and ΔAPQ are on the same base AQ and between the same parallels AQ||CP

Therefore

ar ΔACQ = ΔAPQ …..(i)

ABC is the common figure in both the triangles ΔACQ and ΔAPQ

Subtracting arABC from both sides of equation (i)

ar ΔACQ – arABC = ΔAPQ- arABC

arΔABC = ar ΔPBQ

Since AC and PQ are the diagonals of parallelograms ABCD and PBQR

Therefore

2arΔABC = 2ar ΔPBQ (diagonals of parallelogram bisects it into two equal parts)

**ar(ABCD) = ar(PBQR), Hence Proved**

**NCERT Solutions of Science and Maths for Class 9,10,11 and 12**

**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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**CBSE Class 9-Question paper of science 2020 with solutions**

**CBSE Class 9-Sample paper of science**

**CBSE Class 9-Unsolved question paper of science 2019**

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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**

**CBSE Class 11 – Second unit test of maths 2021 with solutions**

**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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