# Class 9 RD SHARMA Solutions Maths Chapter 14 - Areas of Parallelograms and Triangles

## Areas of Parallelograms and Triangles Exercise Ex. 14.1

### Solution 1

(i) ΔAPB and trapezium ABCD are on the same base AB and between the same parallels AB and CD.

(ii) Parallelograms ABCD and APQD are on the same base AD and between the same parallels AD and BQ.

(iii) Parallelogram ABCD and ΔPQR are between the same parallels AD and BC but they are not on the same base.

(iv) ΔQRT and parallelogram PQRS are on the same base QR and between the same parallels QR and PS.

(v) Parallelogram PQRS and trapezium SMNR are on the same base SR but they are not between the same parallels.

(vi) Parallelograms PQRS, AQRD, BQRC are between the same parallels. Also, parallelograms PQRS, BPSC and APSD are between the same parallels.

## Areas of Parallelograms and Triangles Exercise Ex. 14.2

### Solution 1

We know that,

Area of parallelogram = Base x corresponding attitude

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

16 cm x 8 cm = AD x 10 cm

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## Areas of Parallelograms and Triangles Exercise Ex. 14.3

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Draw a line *l* through A parallel to BC.

Given that, BD = DE = EC.

We observe that the triangles ABD, ADE and AEC are on the equal bases and between the same parallels *l* and BC. Therefore, their areas are equal.

Hence, ar (ΔABD) = ar (ΔADE) = ar(ΔAEC).

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(i)

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Given, ABC and BDE are two equilateral triangles.

Let AB = BC = CA = x. Then, BD = = DE = BE

(i) We have,

ar(ABC) = x^{2}

ar (BDE) =

ar(BDE) = ar (ABC)

(ii) It is given that triangles ABC and BED are equilateral triangles.

ACB = DBE = 60^{o}

BE||AC(Since, alternate angles are equal)

Triangles BAE and BEC are on the same base BE and between the same parallels BE and AC.

ar (BAE) = ar(BEC)

ar (BAE) =2 ar (BDE)

[ ED is a median of EBC ar(BEC) = 2ar(BDE)]

ar (BDE) = ar(BAE)

(iii) Since ABC and BDE are equilateral triangles.

ABC = 60^{o} and BDE = 60^{o}

ABC = BDE

AB||DE(Since, alternate angles are equal)

Triangles BED and AED are on the same base ED and between the same parallels AB and DE.

ar (BED) = ar(AED)

ar (BED) ar(EFD) = ar(AED) ar(EFD)

ar(BEF) = ar(AFD)

(iv) Since ED is a median of BEC

ar (BEC) = 2 ar (BDE)

ar (BEC) = ar (ABC)[From (i), ar (BDE) = ar (ABC)]

ar(BEC) = ar (ABC)

ar (ABC) = 2 ar (BEC)

(v) Let h be the height of vertex E, corresponding to the side BD in triangle BDE.

Let H be the height of vertex A, corresponding to the side BC in triangle ABC.

From part (i),

ar(BDE) = ar (ABC)

From part (iii),

ar (BFE) = ar (AFD)

(vi) ar (AFC) = ar (AFD) + ar (ADC)

= ar (BFE) + ar (ABC)

(Using part (iii); and AD is the median of ABC)

= ar (BFE) + 4 ar (BDE)(Using part (i))

= ar (BFE) + 2 ar (BDE) (2)

Now, from part (v),

ar (BFE) = 2ar (FED) (3)

ar (BDE) = ar (BFE) + ar (FED)

= 2 ar (FED) + ar (FED)

= 3 ar (FED) (4)

From (2), (3) and (4), we get,

ar (AFC) = 2ar (FED) + 2 3 ar (FED) = 8 ar (FED)

Hence, ar (FED) = ar(AFC)

### Solution 30

(i) In MBC and ABD, we have

MB = AB

BC = BD

And MBC = ABD

[MBC and ABD are obtained by adding ABC to a right angle]

So, by SAS congruence criterion, we have

MBC ABD

ar (MBC) = ar(ABD) (1)

(ii) Clearly, triangle ABD and rectangle BYXD are on the same base BD and between the same parallels AX and BD.

ar(ABD) = ar (rect. BYXD)

ar (rect. BYXD) = 2 ar(ABD)

ar (rect. BYXD) = 2 ar (MBC)...(2)

[ ar (ABD) = ar (MBC), from (1)]

(iii) Since triangle MBC and square MBAN are on the same base MB and between the same parallels MB and NC.

2 ar (MBC) = ar (MBAN) (3)

From (2) and (3), we have

ar (sq. MBAN) = ar(rec. BYXD)

(iv) In triangles FCB and ACE, we have

FC = AC

CB = CE

And, FCB = ACE

[FCB and ACE are obtained by adding ACB to a right angle]

So, by SAS congruence criterion, we have

FCB ACE

(v) We have,

FCB ACE

ar (FCB) = ar (ACE)

Clearly, ACE and rectangle CYXE are on the same base CE ad between the same parallels CE and AX.

2 ar (ACE) = ar (CYXE)

2 ar (FCB) = ar (CYXE) (4)

(vi) Clearly, FCB and rectangle FCAG are on the same base FC and between the same parallels FC and BG.

2ar (FCB) = ar(FCAG) (5)

From (4) and (5), we get

ar(CYXE) = ar (ACFG)

(vii) Applying Pythagoras theorem in ACB, we have

BC^{2} = AB^{2} + AC^{2}

## Areas of Parallelograms and Triangles Exercise 14.62

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## Areas of Parallelograms and Triangles Exercise 14.60

### Solution 1

Area of parallelogram = Base × height

Base = Length of base

Height = distance between Base and Side parallel to it

In figure, there are two Parallelograms.

Base of both is same, and because both lie under same parallels that's why height is also same.

Thus, the Ratio of Areas of both parallelogram = 1 : 1

Hence, correct option is (c).

### Solution 2

### Solution 3

When a triangle is formed by joining the mid-points of sides of a triangle, the triangle formed is Congruent to triangles formed around that.

i.e. ΔPQR is congruent to ΔRPA, ΔQBP & ΔCQR.

Hence, Area of all four triangles formed inside ΔABC is same.

So (4 × Area of any one Δ) = Area of ΔABC

4 × (Area of ΔPQR) = 24 sq. units

Area of ΔPQR = 6 sq. units

Hence, correct option is (b).

### Solution 4

A median divides the base in two equal parts but height of a triangle remains the same.

Now, since bases and heights are equal, areas of both Δs are equal.

Hence, correct option is (d).

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## Areas of Parallelograms and Triangles Exercise 14.61

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### Solution 12

E is the mid-point of AD and CE is median of ΔACD.

Hence Ar(ΔAEC) = Ar(ΔCED) = 4 cm^{2} ....(1)

(Median divides a Δ in two two equal Areas)

Also AD is median of ΔABC and and ED is median of ΔBEC.

So Ar(ΔBED) = Ar(CED) = 4 cm^{2} [From eq (1)]

So Ar(ΔBEC) = Ar(ΔBED) + Ar(ΔCED) = 4 + 4 = 8 cm^{2}

Hence, correct option is (c).

### Solution 13

Area of parallelogram = AD × FC = AB × AE

Thus,

AD × 15 = 12 × 7.5

AD = 6 cm

Hence, correct option is (b).

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### Solution 15

ΔABD & ΔABC have same base AB and are between same parallels.

Then,

Ar(ΔABD) = Ar(ΔABC)

But Ar(ΔAOB) is common in both.

Thus, Ar(ΔAOD) = Ar(ΔBOC)

Hence, correct option is (b).

### Solution 16