 # NCERT Solutions for class 9 Maths Chapter 9- Area of Parallelograms and Triangles Exercise 9.2

Section 9.3 of the textbook under which Exercise 9.2 appears, deals with the concept of “Parallelograms on the same Base and Between the same Parallels.” It also explains the “Theorem 9.1 : Parallelograms on the same base and between the same parallels are equal in area” with proof. The questions and exercises given in this section try to explain this concept to you with the help of simple examples.

These questions from the textbook are designed as per the NCERT syllabus and guidelines. Subject experts have designed the NCERT Solutions for Class 9 Maths Chapter 9 – Area of Parallelograms and Triangles with the aim to help students understand the concepts well and score high marks.

### Download PDF of NCERT Solutions for Class 9 Maths Chapter 9-Area of Parallelograms and Triangles Exercise 9.2    ### Access Other Exercise Solutions of Class 9 Maths Chapter 9 Area of Parallelograms and Triangles

Exercise 9.1 Solutions 1 Short Type Answer with Reasoning

Exercise 9.3 Solutions 12 Short Type Answer and 4 Long Type Answer

Exercise 9.4 Solutions 4 Short Type Answer, 1 Long Type Answer and 3 Very Long Type Answers.

## NCERT Solutions for Class 9 Maths Chapter 9 Area of Parallelograms and Triangles Exercise 9.2

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

2. If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar (EFGH) = 1/2 ar(ABCD).

Solution: Given,

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

To Prove,

ar (EFGH) = ½ ar(ABCD)

Construction,

H and F are joined.

Proof,

AD || BC and AD = BC (Opposite sides of a parallelogram)

⇒ ½ AD = ½ BC

Also,

AH || BF and and DH || CF

⇒ AH = BF and DH = CF (H and F are mid points)

∴, ABFH and HFCD are parallelograms.

Now,

We know that, ΔEFH and parallelogram ABFH, both lie on the same FH the common base and in-between the same parallel lines AB and HF.

∴ area of EFH = ½ area of ABFH — (i)

And, area of GHF = ½ area of HFCD — (ii)

area of ΔEFH + area of ΔGHF = ½ area of ABFH + ½ area of HFCD

⇒ area of EFGH = area of ABFH

∴ ar (EFGH) = ½ ar(ABCD)

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).

Solution: ΔAPB and parallelogram ABCD lie on the same base AB and in-between same parallel AB and DC.

ar(ΔAPB) = ½ ar(parallelogram ABCD) — (i)

Similarly,

ar(ΔBQC) = ½ ar(parallelogram ABCD) — (ii)

From (i) and (ii), we have

ar(ΔAPB) = ar(ΔBQC)

4. In Fig. 9.16, P is a point in the interior of a parallelogram ABCD. Show that

(i) ar(APB) + ar(PCD) = ½ ar(ABCD)

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

[Hint : Through P, draw a line parallel to AB.] Solution: (i) A line GH is drawn parallel to AB passing through P.

In a parallelogram,

AB || GH (by construction) — (i)

∴,

AD || BC ⇒ AG || BH — (ii)

From equations (i) and (ii),

ABHG is a parallelogram.

Now,

ΔAPB and parallelogram ABHG are lying on the same base AB and in-between the same parallel lines AB and GH.

∴ ar(ΔAPB) = ½ ar(ABHG) — (iii)

also,

ΔPCD and parallelogram CDGH are lying on the same base CD and in-between the same parallel lines CD and GH.

∴ ar(ΔPCD) = ½ ar(CDGH) — (iv)

ar(ΔAPB) + ar(ΔPCD) = ½ [ar(ABHG)+ar(CDGH)]

⇒ ar(APB)+ ar(PCD) = ½ ar(ABCD)

(ii) A line EF is drawn parallel to AD passing through P.

In the parallelogram,

AD || EF (by construction) — (i)

∴,

AB || CD ⇒ AE || DF — (ii)

From equations (i) and (ii),

AEDF is a parallelogram.

Now,

ΔAPD and parallelogram AEFD are lying on the same base AD and in-between the same parallel lines AD and EF.

∴ar(ΔAPD) = ½ ar(AEFD) — (iii)

also,

ΔPBC and parallelogram BCFE are lying on the same base BC and in-between the same parallel lines BC and EF.

∴ar(ΔPBC) = ½ ar(BCFE) — (iv)

ar(ΔAPD)+ ar(ΔPBC) = ½ {ar(AEFD)+ar(BCFE)}

⇒ar(APD)+ar(PBC) = ar(APB)+ar(PCD)

5. In Fig. 9.17, PQRS and ABRS are parallelograms and X is any point on side BR. Show that
(i) ar (PQRS) = ar (ABRS)
(ii) ar (AXS) = ½ ar (PQRS) Solution:

(i) Parallelogram PQRS and ABRS lie on the same base SR and in-between the same parallel lines SR and PB.

∴ ar(PQRS) = ar(ABRS) — (i)

(ii) ΔAXS and parallelogram ABRS are lying on the same base AS and in-between the same parallel lines AS and BR.

∴ ar(ΔAXS) = ½ ar(ABRS) — (ii)

From (i) and (ii), we find that,

ar(ΔAXS) = ½ ar(PQRS)

6. A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the fields is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?

Solution: The field is divided into three parts each in triangular shape.

Let, ΔPSA, ΔPAQ and ΔQAR be the triangles.

Area of (ΔPSA + ΔPAQ + ΔQAR) = Area of PQRS — (i)

Area of ΔPAQ = ½ area of PQRS — (ii)

Here, the triangle and parallelogram are on the same base and in-between the same parallel lines.

From (i) and (ii),

Area of ΔPSA +Area of ΔQAR = ½ area of PQRS — (iii)

From (ii) and (iii), we can conclude that,

The farmer must sow wheat or pulses in ΔPAQ or either in both ΔPSA and ΔQAR.

NCERT Solutions for Class 9 Maths covers all the questions from Exercise 9.2 of chapter “Area of Parallelograms and Triangles” from the textbook. Exercise 9.2 contains totally 6 questions. Of this, 5 are short answer questions and 1 is a long answer question. In this exercise, the questions may require you to prove or provide the answer to the questions, as is specified.

Meanwhile, here we also give some benefits to solving these exercises:

• Find it easier to face the exams
• Gain more practice solving questions and be more confident
• Learn the concept more easily
• Understand the topic thoroughly