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## NCERT Class 9 Maths Textbook Chapter 9 With Answer Book PDF Free Download

### Chapter 9: Areas of Parallelograms and Triangles

#### 9.1 Introduction

In Chapter 5, you have seen that the study of Geometry, originated with the measurement of the earth (lands) in the process of recasting boundaries of the fields and dividing them into appropriate parts.

For example, a farmer Budhia had a triangular field and she wanted to divide it equally among her two daughters and one son.

Without actually calculating the area of the field, she just divided one side of the triangular field into three equal parts and joined the two points of division to the opposite vertex.

In this way, the field was divided into three parts and she gave one part to each of her children.

Do you think that all the three parts so obtained by her were, in fact, equal in area? To get answers to this type of questions and other related problems, there is a need to have a relook at areas of plane figures, which you have already studied in earlier classes

#### 9.5 Summary

In this chapter, you have studied the following points :

- Area of a figure is a number (in some unit) associated with the part of the plane enclosed by that figure.
- Two congruent figures have equal areas but the converse need not be true.
- If a planar region formed by a figure T is made up of two non-overlapping planar regions formed by figures P and Q, then ar (T) = ar (P) + ar (Q), where ar (X) denotes the area of figure X.
- Two figures are said to be on the same base and between the same parallels, if they have a common base (side) and the vertices, (or the vertex) opposite to the common base of each figure lie on a line parallel to the base.
- Parallelograms on the same base (or equal bases) and between the same parallels are equal in area.
- Area of a parallelogram is the product of its base and the corresponding altitude.
- Parallelograms on the same base (or equal bases) and having equal areas lie between the same parallels.
- If a parallelogram and a triangle are on the same base and between the same parallels, then area of the triangle is half the area of the parallelogram.
- Triangles on the same base (or equal bases) and between the same parallels are equal in area.
- Area of a triangle is half the product of its base and the corresponding altitude.
- Triangles on the same base (or equal bases) and having equal areas lie between the same parallels.
- A median of a triangle divides it into two triangles of equal areas

Author | NCERT |

Language | English |

No. of Pages | 16 |

PDF Size | 603 KB |

Category | Mathematics |

Source/ Credits | ncert.nic.in |

### NCERT Solutions Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles

**1. Which of the following figures lie on the same base and in-between the same parallels? In such a case, write the common base and the two parallels.**

Solution:

(i) Trapezium ABCD and ΔPDC lie on the same DC and in-between the same parallel lines AB and DC.

(ii) Parallelogram PQRS and trapezium SMNR lie on the same base SR but not in-between the same parallel lines.

(iii) Parallelogram PQRS and ΔRTQ lie on the same base QR and in-between the same parallel lines QR and PS.

(iv) Parallelogram ABCD and ΔPQR do not lie on the same base but in-between the same parallel lines BC and AD.

(v) Quadrilateral ABQD and trapezium APCD lie on the same base AD and in-between the same parallel lines AD and BQ.

(vi) Parallelogram PQRS and parallelogram ABCD do not lie on the same base SR but in-between the same parallel lines SR and PQ.

## Exercise 9.2 Page: 159

**1. In Fig. 9.15, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.**

Solution:

Given,

AB = CD = 16 cm (Opposite sides of a parallelogram)

CF = 10 cm and AE = 8 cm

Now,

Area of parallelogram = Base × Altitude

= CD×AE = AD×CF

⇒ 16×8 = AD×10

⇒ AD = 128/10 cm

⇒ AD = 12.8 cm

NCERT Class 9 Maths Textbook Chapter 9 With Answer Book PDF Free Download