Chapter 9 – Areas of Parallelograms and Triangles

Maths

Exercise 9.1

Question 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.

(i)

(iv)

(ii)

(iii)

(v)

(vi)

Answer:

(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)

Page 1 of 41

Page 1 of 41

Class IX

Chapter 9 – Areas of Parallelograms and Triangles

Maths

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 BC. 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 BQ. Page 2 of 41

Page 2 of 41

Class IX

Chapter 9 – Areas of Parallelograms and Triangles

Maths

(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.

Page 3 of 41

Page 3 of 41

Class IX

Chapter 9 – Areas of Parallelograms and Triangles

Maths

Exercise 9.2

Question 1:

In the given figure, ABCD is parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.

Answer:

In parallelogram ABCD, CD = AB = 16 cm

[Opposite sides of a parallelogram are equal]

We know that

Area of a parallelogram = Base × Corresponding altitude

Area of parallelogram ABCD = CD × AE = AD × CF

16 cm × 8 cm = AD × 10 cm

Thus, the length of AD is 12.8 cm.

Question 2:

If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD show that

ar (EFGH)

ar (ABCD)

Page 4 of 41

Page 4 of 41

Class IX

Chapter 9 – Areas of Parallelograms and Triangles

Maths

Answer:

Let us join HF.

In parallelogram ABCD,

AD = BC and AD || BC (Opposite sides of a parallelogram are equal and parallel) AB = CD (Opposite sides of a parallelogram are equal)

and AH || BF

⇒ AH = BF and AH || BF (

H and F are the mid-points of AD and BC)

Therefore, ABFH is a parallelogram.

Since ∆HEF and parallelogram ABFH are on the same base HF and between the same parallel lines AB and HF,

∴ Area (∆HEF) =

Area (ABFH) ... (1)

Similarly, it can be proved that

Area (∆HGF) =

Area (HDCF) ... (2)

On adding equations (1) and (2), we obtain

Page 5 of 41

Page 5 of 41

Class IX

Chapter 9 – Areas of Parallelograms and Triangles

Maths

Question 3:

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).

Answer:

It can be observed that ∆BQC and parallelogram ABCD lie on the same base BC and these are between the same parallel lines AD and BC.

∴Area (∆BQC) =

Area (ABCD) ... (1)

Similarly, ∆APB and parallelogram ABCD lie on the same base AB and between the same parallel lines AB and DC.

∴ Area (∆APB) =

Area (ABCD) ... (2)

From equation (1) and (2), we obtain

Area (∆BQC) = Area (∆APB)

Question 4:

In the given figure, P is a point in the interior of a parallelogram ABCD. Show that

(i) ar (APB) + ar (PCD) =

ar (ABCD)

(ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD)

[Hint: Through. P, draw a line parallel to AB]

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Page 6 of 41

Class IX

Chapter 9 – Areas of Parallelograms and Triangles

Maths

Answer:

(i) Let us draw a line segment EF, passing through point P and parallel...