**NCERT Solutions Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles** – Here are all the NCERT solutions for Class 9 Maths Chapter 9. This solution contains questions, answers, images, explanations of the complete Chapter 9 titled Parallelograms and Triangles of Maths taught in class 9. If you are a student of class 9 who is using NCERT Textbook to study Maths, then you must come across Chapter 9 Areas of Parallelograms and Triangles. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles in one place.

## NCERT Solutions Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles

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Class | 9 |

Subject | Maths |

Book | Mathematics |

Chapter Number | 9 |

Chapter Name |
Areas Of Parallelograms And Triangles |

### NCERT Solutions Class 9 Maths chapter 9 Areas Of Parallelograms And Triangles

Class 9, Maths chapter 9, Areas Of Parallelograms And Triangles solutions are given below in PDF format. You can view them online or download PDF file for future use.

### Areas Of Parallelograms And Triangles

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### Question & Answer

Q.1:Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and the two parallels.

Ans :(i) Yes. It can be observed that trapezium ABCD and triangle PCD have a common base CD and these are lying between the same parallel lines AB and CD. (ii) No. It can be observed that parallelogram PQRS and trapezium MNRS have a common base RS. However, their vertices, (i.e., opposite to the common base) P, Q of parallelogram and M, N of trapezium, are not lying on the same line. (iii) Yes. It can be observed that parallelogram PQRS and triangle TQR have a common base QR and they are lying between the same parallel lines PS and QR. (iv) No. It can be observed that parallelogram ABCD and triangle PQR are lying between same parallel lines AD and ac. However, these do not have any common base. (v) Yes. It can be observed that parallelogram ABCD and parallelogram APQD have a common base AD and these are lying between the same parallel lines AD and 3Q. (vi) No. It can be observed that parallelogram PBCS and PQRS are lying on the same base PS. However, these do not lie between the same parallel lines.

Q.2:In Figure, ABCD is a parallelogram, \( A E \perp D C \) and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.

Ans :In parallelogram ABCD, CD = AB = 16 cm [Opposite sides of a parallelogram are equal] We know that Area of a parallelogram = Base x Corresponding altitude Area of parallelogram ABCD = CD x AE = AD x CF Area of parallelogram = Base x Corresponding altitude Area of parallelogram ABCD = CD x AE = AD x CF \( 16 \mathrm{cm} \times 8 \mathrm{cm}=\mathrm{AD} \times 10 \mathrm{cm} \) \( \mathrm{AD}=\frac{16 \times 8}{10} \mathrm{cm}=12.8 \mathrm{cm} \) Thus, the length of AD is 12.8 cm.

Q.3:If E,F,G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that \( \operatorname{ar}(\mathrm{EFG} \mathrm{H})=\frac{\mathrm{I}}{2} \operatorname{ar}(\mathrm{ABCD}) \).

Ans :Let us join HF. In parallelogram ABCD, AD = BC and \( \mathrm{AD}\|\mathrm{BC} \) ( Opposite sides of a parallelogram are equal and parallel) AB = CD ( Opposite sides of a parallelogram are equal) \( \Rightarrow \frac{1}{2} \mathrm{AD}=\frac{1}{2} \mathrm{BC} \) and \( \mathrm{AH}\|\mathrm{BF} \) Therefore, ABFH is a parallelogram. Since \( \Delta \mathrm{HEF} \) and parallelogram ABFH are on the same base HF and between the same parallel lines AB and HF, \( \therefore \text { Area }(\Delta \mathrm{HEF})=\frac{1}{2} \text { Area (ABFH) } \ldots \) (1) Similarly, it can be proved that Area \( (\Delta H G F)=\frac{1}{2} \text { Area (HDCF) } \ldots \) (2) On adding equations (1) and (2), we obtain Area \( (\Delta \mathrm{HEF})+\text { Area }(\Delta \mathrm{HGF})=\frac{1}{2} \text { Area }(\mathrm{ABFH})+\frac{1}{2} \text { Area }(\mathrm{HDCF}) \) \( =\frac{1}{2}[\text { Area }(\mathrm{ABFH})+\text { Area }(\mathrm{HDCF})] \) \( \Rightarrow \text { Area }(\mathrm{EFGH})=\frac{1}{2} \text { Area }(\mathrm{ABCD}) \)

Q.4:P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (APB) = ar (BQC).

Ans :It can be observed that \( \triangle B Q C \) and paralleogram ABCD lie on the same base BC and these are between the same parallel lines AD and BC. \( \therefore \text { Area }(\Delta B Q C)=\frac{1}{2} \text { Area (ABCD) } \dots \) (1) Similarly, \( \triangle \mathrm{APB} \) and parallelogram ABCD lie on the same base AB and between the same parallel lines AB and DC. \( \therefore \text { Area }(\Delta \mathrm{APB})=\frac{1}{2} \text { Area (ABCD) } \ldots \) (2) From equation (1) and (2), we obtain Area \( (\Delta B Q C)=\text { Area }(\Delta A P B) \)

Q.5:In Figure, P is a point in the interior of a parallelogram ABCD. Show that (i) \( \operatorname{ar}(\mathrm{APB})+\operatorname{ar}(\mathrm{PCD})=\frac{1}{2} \operatorname{ar}(\mathrm{ABCD}) \) (ii) \( \operatorname{ar}(\mathrm{APD})+\operatorname{ar}(\mathrm{PBC})=\operatorname{ar}(\mathrm{APB})+\operatorname{ar}(\mathrm{PCD}) \) [Hint : Through P, draw a line parallel to AB.]

Ans :(i) Let us draw a line segment EF, passing through point P and parallel to line segment AB. In parallelogram ABCD, \( \mathrm{AB}\|\mathrm{EF}(\mathrm{By} \text { construction }) \ldots(1) \) \( \begin{array}{l}{\text { ABCD is a parallelogram. }} \\ {\therefore \text { AD }\|\text { BC (Opposite sides of a parallelogram) }}\end{array} \) \( \Rightarrow \mathrm{AE}\|\mathrm{BF} \ldots(2) \) From equations (1) and (2), we obtain \( \mathrm{AB}\|\mathrm{EF} \text { and } \mathrm{AE}\| \mathrm{BF} \) Therefore, quadrilateral ABFE is a parallelogram It can be observed that \( \triangle \mathrm{APB} \) and parallelogram ABFE are lying on the same base AB and between the same parallel lines AB and EF. \( \therefore \text { Area }(\Delta \mathrm{APB})=\frac{1}{2} \text { Area (ABFE) } \ldots \) (3) Similarly, for \( \Delta \mathrm{PCD} \) and parallelogram EFCD, Area \( (\triangle \mathrm{PCD}) \) = \( \frac{1}{2}{\text { Area }(E F C D) \ldots(4)} \) Adding equation (3) and (4), we obtain \( (\Delta \mathrm{APB})+\text { Area }(\Delta \mathrm{PCD})=\frac{1}{2}[\text { Area }(\mathrm{ABFE})+\text { Area }(\mathrm{EFCD})] \) \( (\Delta \mathrm{APB})+\text { Area }(\Delta \mathrm{PCD})=\frac{1}{2} \text {Area }(\mathrm{ABCD}) \) …(5) (ii) Let us draw a line segment MN, passing through point P and parallel to line segment AD. In parallelogram ABCD, MN || AD (By construction)...(6) ABCD is a parallelogram. \( \therefore \text { AB }\|\text { DC (Opposite sides of a parallelogram) } \) \( \Rightarrow \mathrm{AM}\|\mathrm{DN} \dots(7) \) From equations (6) (7), we obtain \( \mathrm{MN}\|\mathrm{AD} \text { and } \mathrm{AM}\| \mathrm{DN} \) Therefore, quadrilateral AMND is a parallelogram. It can be observed that \( \triangle A P D \) and parallelogram AMND are lying on the same base AD and between the same parallel lines AD and MN. \( \therefore \text { Area }(\Delta A P D)=\frac{1}{2} \text { Area (AMND) } \dots \) (8) Similarly, for \( \triangle \mathrm{PCB} \) and parallelogram MNCB, Area \( (\Delta \mathrm{PCB})=\frac{1}{2} \text { Area (MNCB) } \ldots \) (9) Adding equations (8) and (9), we obtain \( (\Delta \mathrm{APD})+\text { Area }(\Delta \mathrm{PCB})=\frac{1}{2}[\text { Area }(\mathrm{AMND})+\text { Area }(\mathrm{MNCB})] \) \( (\Delta \mathrm{APD})+\text { Area }(\Delta \mathrm{PCB}) \) = \( \frac{1}{2} \operatorname{Area}(\mathrm{ABCD}) \quad \ldots(10) \) On comparing equations (5) and (10), we obtain \( (\Delta \mathrm{APD})+\text { Area }(\Delta \mathrm{PBC})=\text { Area }(\Delta \mathrm{APB})+\text { Area }(\Delta \mathrm{PCD}) \)

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