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

### 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 questionPSDA 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 cm

^{2}= 45 h cm^{2}ar(trap. ABYX) = (1/2)(AB + XY) Ã— h

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

^{2}= 55h cm^{2}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 questionABC 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 x

^{2}….(i)ar(Î”BDE) = âˆš3/4 (x/2)

^{2}= 1/4 x âˆš3/4 x^{2}Now put the value of âˆš3/4 x

^{2 }from eq(i), we getHence, 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 andH 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 questionD 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 questionX 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 questionABCD and AEFD are two parallelograms

(i)Prove that PE = FQProof:

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 questionABCD 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 andL 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 havear(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 havear(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 questionABC 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 BDand 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 andbetween 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 aboveÎ”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 FCand 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 haveUsing Pythagoras theorem

BC

^{2}= AB^{2}+ AC^{2}BC Ã— BD = AB Ã— MB + AC Ã— FC

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

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