# Class 9 RD Sharma Solutions – Chapter 15 Areas of Parallelograms and Triangles- Exercise 15.3 | Set 3

Last Updated : 19 Apr, 2021

### Question 21. In figure, PSDA is a parallelogram in which PQ = QR = RS and AP || BQ || CR. Prove that ar (Î”PQE) = ar(Î”CFD).

Solution:

According to the question

PSDA is a parallelogram

So, AP || BQ || CR || DS and AD || PS

PQ = CD       …..(i)

Prove that (Î”PQE) = ar(Î”CFD)

Proof:

In Î”BED,

C is the midpoint of BD and CF || BE

So, F is the midpoint of ED

Hence, EF = PE

Similarly, EF = PE

Therefore, PE = FD             …..(ii)

In Î”PQE and CFD,

PE = FD

âˆ  EPQ = âˆ FDC                            (Alternate angles)

PQ = CD

So, by SAS congruence, we have

Î”PQE â‰… Î”DCF

Hence, ar(Î”PQE) = ar(Î”DCF)

### Question 22. In figure, ABCD is a trapezium in which AB || DC and DC = 40 cm and AB = 60 cm. If X and Y are, respectively, the mid-points of AD and BC, prove that:

(i) XY = 50 cm

(ii) DCYX is a trapezium

(iii) ar(trap. DCYX) = (9/11) ar(XYBA).

Solution:

(i) Join DY and produce it to meet AB produced at P.

In Î”BYP and CYD

âˆ BYP = âˆ  CYD          (Vertically opposite angles)

âˆ DCY = âˆ  PBY           (As, DC || AP)

BY = CY

So, by ASA congruence,

Î”BYP â‰… Î”CYD

By C.P.C.T

DY = YP and DC = BP

As we know that, Y is the midpoint of DP

also, X is the midpoint of AD

So, XY || AP and XY || (1/2) AP

XY = (1/2) (AB + BP)

XY = (1/2) (AB + DC)

XY = (1/2) (60 + 40) = 50 cm

(ii) Given that,

XY || AP

XY || AB and AB || DC

XY || DC

Hence, DCYX is a trapezium

(iii) As we know that X and Y are the mid-points of Ad and BC.

So, trapezium DCYX and ABYX are of the same height

Let us assume, the height of the trapezium is h cm

Now,

ar(trap. DCXY) = (1/2)(DC + XY) Ã— h

ar(trap. DCXY) = (1/2) (50 + 40) Ã— h cm2 = 45 h cm2

ar(trap. ABYX) = (1/2)(AB + XY) Ã— h

ar(trap. ABYX) = (1/2)(60 + 50) Ã— h cm2 = 55h cm2

ar(trap. DCYX) ar(trap. ABYX) = 45h/55h = 9/11

Hence, ar(trap. DCYX) = 9/11 ar(trap. ABYX)

### Question 23. In figure, ABC and BDE are two equilateral triangles such that D is the midpoint of BC. AE intersects BC in F. Prove that:

(i) ar(Î”BDE) = (1/4) ar(Î”ABC)

(ii) ar(Î”BDE) = (1/2) ar(Î”BAE)

(iii) ar(Î”BFE) = ar(Î”AFD)

(iv) ar(Î”ABC) = 2 ar(Î”BEC)

(v) ar(Î”FED) = 1/8 ar(Î”AFC)

(vi) ar(Î”BFE) = 2 ar(Î”EFD)

Solution:

According to the question

ABC and BDE are two equilateral triangles.

Let us considered AB = BC = CA = x. Then, BD = x/2 = DE = BE

(i) Given that,

ar(Î”ABC) = âˆš3/4 x2 ….(i)

ar(Î”BDE) = âˆš3/4 (x/2)2 = 1/4 x âˆš3/4 x2

Now put the value of âˆš3/4 x2 from eq(i), we get

Hence, ar(Î”BDE) = 1/4 ar(Î”ABC)

Hence proved

(ii) Given that,

Î”ABC and BED are equilateral triangles

So, âˆ  ACB = âˆ DBE = 60Â°

BE || AC                                    (Alternative angles are equal)

As we know that Î”BAF and Î”BEC are on the same base BE

and between same parallels BF and AC.

So, ar(Î”BAE) = ar(Î”BEC)

ED is a median of Î”EBC

So, ar(Î”BAE) = 2ar(Î”BDE)

Hence, ar(Î”BDE) = (1/2) ar(Î”BAE)

(iii) Given that,

Î”ABC and BDE are equilateral triangles

So, âˆ ABC = 60Â° and âˆ BDE = 60Â°

âˆ ABC = âˆ BDE

AB || DE                   (Alternate angles are equal)

As we know that Î”BED and Î”AED are on the same base ED and

between same parallels AB and DE.

So, ar(Î”BED) = ar(Î”AED)

ar(Î”BED) âˆ’ ar(Î”EFD) = ar(Î”AED) âˆ’ ar(Î”EFD)

Hence, ar(Î”BEF) = ar(Î”AFD)

(iv) Given that,

ED is the median of tÎ”BEC

So, ar(Î”BEC) = 2ar(Î”BDE)

ar(Î”BEC) = 2 Ã— (1/2) ar(Î”ABC)    (Proved above)

ar(Î”BEC) = (1/2) ar(Î”ABC)

Hence, ar(Î”ABC) = 2ar(Î”BEC)

(v) ar(Î”AFC) = ar(Î”AFD) + ar(Î”ADC)

ar(Î”BFE) + (1/2) ar(Î”ABC)

ar(Î”BFE) + (1/2) Ã— 4ar(Î”BDE)

ar(Î”BFE) = 2ar(Î”FED)  …..(iii)

ar(Î”BDE) = ar(Î”BFE) + ar(Î”FED)

2ar(Î”FED) + ar(Î”FED)

3ar(Î”FED)         ….(iv)

From eq (ii), (iii) and (iv), we get

ar(Î”AFC) = 2ar(Î”FED) + 2 Ã— 3 ar(Î”FED) = 8 ar(Î”FED)

Hence, ar(Î”FED) = (1/8) ar(Î”AFC)

(vi) Let’s assume that h be the height of vertex E, corresponding to the side BD in Î”BDE and

H be the height of vertex A, corresponding to the side BC in Î”ABC

As we proved above

ar(Î”BDE) = (1/4) ar(Î”ABC)

(1/2) Ã— BD Ã— h = (1/4) (1/2 Ã— BC Ã— h)

BD Ã— h = (1/4)(2BD Ã— H)

h = (1/2) H    ….(i)

From part (iii), we get

ar(Î”BFE) = ar(Î”AFD)

ar(Î”BFE) = (1/2) Ã— FD Ã— H

ar(Î”BFE) = (1/2) Ã— FD Ã— 2h

ar(Î”BFE) = 2((1/2) Ã— FD Ã— h)

Hence, ar(Î”BFE) = 2ar(Î”EFD)

### Question 24. D is the midpoint of side BC of Î”ABC and E is the midpoint of BD. If O is the midpoint of AE, Prove that ar(Î”BOE) = (1/8) ar(Î”ABC).

Solution:

According to the question

D is the midpoint of sides BC of Î” ABC,

E is the midpoint of BD and O is the midpoint of AE,

So, AD and AE are the medians of Î”ABC and Î”ABD

Hence, ar(Î”ABD) = (1/2) ar(Î”ABC)   …..(i)

and ar(Î”ABE) = (1/2) ar(Î”ABD)   …….(ii)

Also, OB is the median of triangle ABE

So, ar(Î”BOE) = (1/2) ar(Î”ABE)

From eq (i), (ii) and (iii), we conclude that

ar(Î”BOE) = (1/8) ar(Î”ABC)

Hence proved

### Question 25. In figure, X and Y are the mid points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. Prove that ar(Î”ABP) = ar(Î”ACQ).

Solution:

According to the question

X and Y are the mid-points of AC and AB

So, XY || BC

Prove: ar(Î”ABP) = ar(Î”ACQ)

Proof:

From the figure, we conclude that Î”BYC and BXC are on the same base BC and

between the same parallels XY and BC

So, ar(Î”BYC) = ar(Î”BXC)

ar(Î”BYC) âˆ’ ar(Î”BOC) = ar(Î”BXC) âˆ’ ar(Î”BOC)

ar(Î”BOY) = ar(Î”COX)

ar(Î”BOY) + ar(Î”XOY) = ar(Î”COX) + ar(Î”XOY)

ar(Î”BXY) = ar(Î”CXY)             …..(i)

From the figure we conclude that the quadrilaterals XYAP and XYAQ are on

the same base XY and between same parallels XY and PQ.

So, ar(quad. XYAP) = ar(quad XYQA)   ….(ii)

Now, add eq(i) and (ii), we get

ar(Î”BXY) + ar(quad. XYAP) = ar(Î”CXY) + ar(quad XYQA)

ar(Î”ABP) = ar(Î”ACQ)

Hence proved

### Question 26. In figure, ABCD and AEFD are two parallelograms. Prove that

(i) PE = FQ

(ii) ar(Î”APE) : ar(Î”PFA) = ar(Î”QFD) : ar(Î”PFD)

(iii) ar(Î”PEA) = ar(Î”QFD)

Solution:

According to the question

ABCD and AEFD are two parallelograms

(i) Prove that PE = FQ

Proof:

In Î”EPA and FQD

âˆ PEA = âˆ QFD         (Corresponding angles)

âˆ EPA=âˆ FQD           (Corresponding angles)

PA = QD                   (Opposite sides of parallelogram)

So, by AAS congruence

Î”EPA â‰… Î”FQD

Hence, by C.P.C.T

EP = FQ

Hence proved

(ii) Prove that ar(Î”APE) : ar(Î”PFA) = ar(Î”QFD) : ar(Î”PFD)

From the figure we conclude that Î”PEA and Î”QFD stand on equal bases

PE and FQ lies between the same parallels EQ and AD

So, ar(Î”PEA) = ar(Î”QFD) ……(i)

From the figure we conclude that Î”PEA and Î”PFD stand on the same

base PF and between same parallels PF and AD

Therefore, ar(Î”PFA) = ar(Î”PFD)      ……(ii)

Now divide eq(i) by eq(ii), we get

Hence, ar(Î”PEA) : ar(Î”PFA) = ar(Î”QFD) : ar(Î”PFD)

Hence proved

(iii) Prove that ar(Î”PEA) = ar(Î”QFD)

Proof:

As we proved above that

Î”EPA â‰… Î”FQD

Hence, ar(Î”PEA) = ar(Î”QFD)

Hence proved

### Question 27. In figure, ABCD is a parallelogram. O is any point on AC. PQ || AB and LM || AD. Prove that: ar(||gm DLOP) = ar(||gm BMOQ).

Solution:

According to the question

ABCD is a parallelogram and PQ || AB and LM || AD

Prove that ar(||gm DLOP) = ar(||gm BMOQ)

Proof:

Here, the diagonal AC of a parallelogram divides it into two triangles of equal area

So, ar(Î”ADC) = ar(Î”ABC)

ar(Î”APO) + ar(||gm DLOP) + ar(Î”OLC)

ar(Î”AOM) + ar(||gm BMOQ) + ar(Î”OQC)           ……(i)

From the figure, AO and OC are diagonals of parallelograms AMOP and OQCL

So, ar(Î”APO) = ar(Î”AMO)             ……(ii)

And ar(Î”OLC) = ar(Î”OQC)            ……(iii)

Now subtracting eq(ii) and (iii) from (i), we get

ar(||gm DLOP) = ar(||gm BMOQ)

Hence proved

### Question 28. In a triangle ABC, if L and M are points on AB and AC respectively such that LM || BC. Prove that:

(i) ar(Î”LCM) = ar(Î”LBM)

(ii) ar(Î”LBC) = ar(Î”MBC)

(iii) ar(Î”ABM) = ar(Î”ACL)

(iv) ar(Î”LOB) = ar(Î”MOC)

Solution:

According to the question

ABC is a triangle and L and M are points on AB and AC respectively such that LM || BC

(i) Here, Î” LMB and LMC are on the same base LM and between the same parallels LM and BC.

Hence, ar(Î”LMB) = ar(Î”LMC)

(ii) Here, Î”LBC and MBC are on the same base BC and between same parallels LM and BC.

Hence, ar(Î”LBC) = ar(Î”MBC)

(iii) Here, we have

ar(Î”LMB) = ar(Î”LMC)            (Proved above)

ar(Î”ALM) + ar(Î”LMB) = ar(Î”ALM) + ar(Î”LMC)

Hence, ar(Î”ABM) = ar(Î”ACL)

(iv) Here, we have

ar(Î”LBC) = ar(Î”MBC)              (Proved above)

ar(Î”LBC) âˆ’ ar(Î”BOC) = ar(Î”MBC) âˆ’ ar(Î”BOC)

Hence, ar(Î”LOB) = ar(Î”MOC).

### Question 29. In figure, D and E are two points on BC such that BD = DE = EC. Show that ar(Î”ABD) = ar(Î”ADE) = ar(Î”AEC).

Solution:

In the triangle ABC, draw a line || through A parallel to BC.

BD = DE = EC  (Given)

Now from the figure, triangles ABD and AEC are on the same base AC and

between the same parallels l and BC. Hence, the area of triangles ABD and AEC is equal.

Hence, ar(Î”ABD) = ar(Î”ADE) = ar(Î”AEC).

### Question 30. In figure, ABC is a right angled triangle at A, BCED, ACFG and ABMN are squares on the sides BC, CA, and AB respectively. Line segment AX âŠ¥ DE meets BC at Y. Show that

(i) Î”MBC â‰… Î”ABD

(ii) ar(BYXD) = 2ar(Î”MBC)

(iii) ar(BYXD) = ar(ABMN)

(iv) Î”FCB â‰… Î”ACE

(v) ar(CYXE) = 2ar(Î”FCB)

(vi) ar(CYXE) = ar(ACFG)

(vii) ar(BCED) = ar(ABMN) + ar(ACFG)

Solution:

According to the question

ABC is a right-angled triangle at A, BCED, ACFG, and ABMN are squares on the sides BC, CA, and AB

(i) In Î”MBC and Î”ABD,

MB = AB,

BC = BD,

âˆ MBC = âˆ ABD

So, by SAS congruence, we have

Î”MBC â‰… Î”ABD

Hence, ar(Î”MBC) = ar(Î”ABD)

(ii) From the figure, triangle ABC and rectangle BYXD are on the same base BD

and between the same parallels AX and BD.

So, ar(Î”ABD) = (1/2) ar(rect BYXD)

ar(rect BYXD) = 2ar(Î”ABD)

From part (i)

Hence, ar(rect BYXD) = 2ar(Î”MBC)   ……(i)

(iii) From the figure, triangles MBC and square MBAN are on the same base MB and

between the same parallels MB and NC.

So, 2ar(Î”MBC) = ar(MBAN)  ……(ii)

From eq(i) and (ii), we get

Hence, ar(sq. MBAN) = ar(rect BYXD)

(iv) In Î”FCB and ACE,

FC = AC

CB = CE

âˆ FCB = âˆ ACE

So, by SAS congruence,

Hence, Î”FCB â‰… Î”ACE

(v) As we proved above

Î”FCB â‰… Î”ACE

So, ar(Î”FCB) = ar(Î”ACE)

From the figure, triangle ACE and rectangle CYXE are on the same base CE

and between same parallels CE and AX.

So, 2ar(Î”ACE) = ar(CYXE)

Hence, 2ar(Î”FCB) = ar(Î”CYXE)        …..(iii)

(vi) From the figure, triangle FCb and rectangle FCAG are on the same base FC

and between the same parallels FC and BG.

So, 2ar(Î”FCB) = ar(FCAG)            ……(iv)

From eq(iii) and (iv), we get

ar(CYXE) = ar(ACFG)

(vii) In Î”ACB, we have

Using Pythagoras theorem

BC2 = AB2 + AC2

BC Ã— BD = AB Ã— MB + AC Ã— FC

Hence, ar(BCED) = ar(ABMN) + ar(ACFG)

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