RS Aggarwal Solutions Class 9 Maths Chapter 11 Areas of Parallelograms and Triangles (Updated For 2021-22)

RS Aggarwal Solutions Class 9 Maths Chapter 11 Areas of Parallelograms and Triangles

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Page No 387:

Question 1:

Which of the following figures lie on the same base and between the same parallels. In such a case, write the comon base and the two parallels.

ANSWER:

(i) No, it doesnt lie on the same base and between the same parallels.
(ii) No, it doesnt lie on the same base and between the same parallels.
(iii) Yes, it lies on the same base and between the same parallels. The same base is AB and the parallels are AB and DE.
(iv) No, it doesnt lie on the same base and between the same parallels.
(v) Yes, it lies on the same base and between the same parallels. The same base is BC and the parallels are BC and AD.
(vi) Yes, it lies on the same base and between the same parallels. The same base is CD and the parallels are CD and BP.

Page No 387:

Question 2:

In the adjoining figure, show that ABCD is a parallelogram.
Calculate the area of || gm ABCD.

ANSWER:



Given: A quadrilateral ABCD and BD is a diagonal.
To proveABCD is a parallelogram.
Construction: Draw AM ⊥ DC and CL ⊥ AB   (extend DC and AB). Join AC, the other diagonal of ABCD.

Proof: ar(quad. ABCD) = ar(∆ABD) + ar(​∆DCB)
                                      = 2 ar(​∆ABD)                    [∵ ar​(∆ABD) = ar(​∆DCB)]
∴ ar(​∆ABD) = 1212ar(quad. ABCD)                 …(i)

Again, ar(quad. ABCD) = ar(∆ABC) + ar(​∆CDA)
                                    = 2 ar(​∆ ABC)                    [∵ ar​(∆ABC) = ar(​∆CDA)]
∴ ar(​∆ABC) = 1212ar(quad. ABCD)                …(ii)
From (i) and (ii), we have:
 ar(​∆ABD) = ar(​∆ABC) = 1212 AB ⨯ BD = 1212 AB ⨯ CL
 ⇒ CL = BD
 ⇒ DC |​​| AB
Similarly, AD |​​| BC.
Hence, ABCD is a paralleogram.
∴ ar(​|​| gm ABCD) = base ​⨯ height = 5 ​⨯ 7 = 35 cm2

Page No 387:

Question 3:

In a parallelogram ABCD, it is being given that AB = 10 cm and the altitudes corresponding to the sides AB and AD are DL = 6 cm and BM = 8 cm, respectively. Find AD.

ANSWER:

ar(parallelogram ABCD) = base ​⨯ height
⇒ AB ​⨯DL = AD ​⨯ BM
⇒ 10 ​​⨯ 6 = AD ​⨯ BM
⇒ AD ​⨯ 8 = 60 cm2
⇒ AD =  7.5 cm
∴ AD = 7.5 cm



Page No 388:

Question 4:

Find the area of a figure formed by joining the midpoints of the adjacent sides of a rhombus with diagonals 12 cm and 16 cm.

ANSWER:


Let ABCD be a rhombus and P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively. 
Join the diagonals, AC and BD.
In ∆ ABC, we have:
PQ ∣∣ AC and PQ = 1212AC                    [By midpoint theorem]
PQ=12×16=8 cmPQ=12×16=8 cm
Again, in ∆DACthe points S and R are the midpoints of AD and DC, respectively.
∴ SR ∣∣ AC and SR = 1212AC                    [By midpoint theorem]
SR=12×12=6 cmSR=12×12=6 cm
Area of PQRS=length×breadth=6×8=48 cm2Area of PQRS=length×breadth=6×8=48 cm2

Page No 388:

Question 5:

Find the area of a trapezium whose parallel sides are 9 cm and 6 cm respectively and the distance between these sides is 8 cm.

ANSWER:

ar(trapezium) = 1212 ⨯ (sum of parallel sides) ⨯ (distance between them)
 = 1212 ⨯ (9 + 6) ⨯ 8
= 60 cm2
Hence, the area of the trapezium is 60 cm2.

Page No 388:

Question 6:

(i) Calculate the area of quad. ABCD, given in Fig. (i).
(ii) Calculate the area of trap. PQRS, given in Fig. (ii).

ANSWER:

(i) In BCD, 
DB2+BC2=DC2DB2=17282=225DB=15 cmDB2+BC2=DC2⇒DB2=172-82=225⇒DB=15 cm
Ar(BCD) = 12×b×h=12×8×15=60 cm212×b×h=12×8×15=60 cm2
In BAD,
DA2+AB2=DB2AB2=15292=144AB=12 cmDA2+AB2=DB2⇒AB2=152-92=144⇒AB=12 cm
Ar(DAB) = 12×b×h=12×9×12=54 cm212×b×h=12×9×12=54 cm2

Area of quad. ABCD = Ar(DAB) + Ar(BCD) = 54 + 60 = 114 cm.

(ii) Area of trap(PQRS) = 12(8+16)×8=96 cm2128+16×8=96 cm2

Page No 388:

Question 7:

In the adjoining figure, ABCD is a trapezium in which AB || DCAB = 7 cm; AD = BC = 5 cm and the distance between AB and DC is 4 cm. Find the length of DC and hence, find the area of trap. ABCD.

ANSWER:

ADL is a right angle triangle.
So, DL = (52  42 )−−−−−−−−−√ = 9–√ = 3 cm52 – 42  = 9 = 3 cm
Similarly, in ∆BMC, we have:
MC = (52  42 )−−−−−−−−−√ = 9–√ = 3 cm52 – 42  = 9 = 3 cm
∴ DC =  DL + LM + MC =  3 + 7 + 3 = 13 cm
Now, ar(trapezium. ABCD) = 1212⨯ (sum of parallel sides) ⨯ (distance between them)
=1212 ⨯ (7 + 13) ⨯ 4
= 40 cm2
​Hence, DC = 13 cm and area of trapezium = 40 cm2

Page No 388:

Question 8:

BD is one of the diagonals of a quad. ABCD. If AL ⊥ BD and CM ⊥ BD, show that ar(quad. ABCD)=12×BD×(AL + CM)ar(quad. ABCD)=12×BD×(AL + CM).

ANSWER:

ar(quad. ABCD) = ar(∆​ABD) + ar (∆DBC)
ar(∆ABD) = 1212⨯ base ⨯ height = 1212⨯ BD ⨯ AL             …(i) 
ar(∆DBC) = 1212 ⨯ BD ⨯ CL                  ...(ii)
From (i) and (ii), we get:
​ar(quad ABCD) = 1212 ⨯BD ⨯​ AL + 1212 ⨯ BD ⨯​ CL
​ar(quad ABCD) = 1212 ⨯ BD ⨯ ​(AL + CL)
Hence, proved.

Page No 388:

Question 9:

is the midpoint of the side AB of a parallelogram ABCD. If ar(AMCD) = 24 cm2, find ar(∆ABC).

ANSWER:


Join AC. 
AC divides parallelogram ABCD into two congruent triangles of equal areas. 
ar(ABC)=ar(ACD)=12ar(ABCD)ar△ABC=ar△ACD=12arABCD
M is the midpoint of AB. So, CM is the median. 
CM divides ABC in two triangles with equal area. 
ar(AMC)=ar(BMC)=12ar(ABC)ar△AMC=ar△BMC=12ar△ABC
ar(AMCD) = ar(ACD) + ar(AMC) = ar(ABC) + ar(AMC) = ​ar(ABC) + 1212​ar(ABC)
24=32ar(ABC)ar(ABC)=16 cm2⇒24=32ar△ABC⇒ar△ABC=16 cm2

Page No 388:

Question 10:

In the adjoining figure, ABCD is a quadrilateral in which diag. BD = 14 cm. If AL ⊥ BD and CM ⊥ BD such that AL = 8 cm and CM = 6 cm, find the area of quad. ABCD.

ANSWER:

​ar(quad ABCD) = ar(ABD) + ar(BDC)
1212 ⨯BD ⨯​ AL  +1212 ⨯BD ⨯​ CM
1212 ⨯BD ⨯​ ( AL + CM)
By substituting the values, we have;
ar(quad ABCD) = ​1212 ⨯ 14 ⨯ ( 8 + 6)
= 7 ​⨯14
= 98 cm2

Page No 388:

Question 11:

If P and Q are any two points lying respectively on the sides DC and AD of a parallelogram ABCD then show that ar(∆APB) = ar(∆BQC).

ANSWER:


We know
ar(∆APB) = 12ar(ABCD)12arABCD           …..(1)                     [If a triangle and a parallelogram are on the same base and between the same parallels then the area of the triangle is equal to half the area of the parallelogram]
Similarly, 
ar(∆BQC) = 12ar(ABCD)12arABCD           …..(2)
From (1) and (2)
ar(∆APB) = ar(∆BQC)
Hence Proved



Page No 389:

Question 12:

In the adjoining figure, MNPQ and ABPQ are parallelograms and T is any point on the side BP. Prove that
(i) ar(MNPQ) = are(ABPQ)
(ii) ar(∆ATQ) = 1212ar(MNPQ).

ANSWER:

(i) We know that parallelograms on the same base and between the same parallels are equal in area.
So, ar(MNPQ) = are(ABPQ)                   (Same base PQ and MB || PQ)                      …..(1)

(ii) If a parallelogram and a triangle are on the same base and between the same parallels then the area of the triangle is equal to half the area of the parallelogram. 
So, ar(∆ATQ) = 1212ar(ABPQ)                 (Same base AQ and AQ || BP)                       …..(2)
From (1) and (2)
ar(∆ATQ) = 1212ar(MNPQ)


 

Page No 389:

Question 13:

In the adjoining figure, ABCD is a trapezium in which AB || DC and its diagonals AC and BD intersect at O. Prove that ar(∆AOD) = ar(∆BOC).

ANSWER:

CDA and ​∆CBD lies on the same base and between the same parallel lines.
So, ar(​∆CDA) = ar(CDB)            …(i)
Subtracting ar(​∆OCD) from both sides of equation (i), we get:
ar(​∆CDA ar(​∆OCD) = ar(​​∆CDB ar (​​∆OCD)
⇒ ar(​​∆AOD) = ar(​​∆BOC)

Page No 389:

Question 14:

In the adjoining figure, DE || BC. Prove that
(i) ar(∆ACD) = ar(∆ABE),
(ii) ar(∆OCE) = ar(∆OBD),

ANSWER:

DEC and ​∆DEB lies on the same base and between the same parallel lines.
So, ar(​∆DEC) = ar(∆DEB)                      …(1)

(i) On adding​ ar(∆ADE)​ in both sides of equation (1), we get:
  ar(​∆DEC) + ar(∆ADE)​ = ar(∆DEB) + ar(∆ADE)​ ​                 
⇒ ar(​​∆ACD) = ar(​​∆ABE

  (ii) On subtracting​ ar(ODE)​ from both sides of equation (1), we get:​
   ar(​∆DEC ar(∆ODE)​ = ar(∆DEB ar(∆ODE)​ ​      ​
⇒ ar(​​∆OCE) = ar(​∆OBD)

Page No 389:

Question 15:

Prove that a median divides a triangle into two triangles of equal area.

ANSWER:



Let AD is a median of 
ABC and D is the midpoint of BC. AD divides ∆ABC in two triangles: ∆ABD and ADC.
To prove: ar(∆ABD) = ar(∆ADC)
Construction: Draw AL ⊥ BC.
Proof:
Since D is the midpoint of BC, we have:
BD = DC
Multiplying with 1212AL on both sides, we get:
1212 × BD × AL = 1212 × DC × AL 
⇒ ar(∆ABD) = ar(∆ADC)

Page No 389:

Question 16:

Show that a diagonal divides a parallelogram into two triangles of equal area.

ANSWER:



Let ABCD be a parallelogram and BD be its diagonal.

To prove: ar(∆ABD) = ar(∆CDB)

Proof: 
In ∆ABD and ∆CDBwe have:
AB = CD                    [Opposite sides of a parallelogram]
AD = CB                   [Opposite sides of a parallelogram]​

 BD  = DB                  [Common]
i.e., ∆ABD  CDB           [ SSS criteria]
∴ ar(∆ABD) = ar(∆CDB)

Page No 389:

Question 17:

In the adjoining figure, ABC and ABD are two triangles on the same base AB. If line segment CD is bisected by AB at O, show that ar(ΔABC) = ar(ΔABD)

ANSWER:

Line segment CD is bisected by AB at O                   (Given)
CO = OD                                …..(1)
In ΔCAO, 
AO is the median.                (From (1))
So, arΔCAO = arΔDAO          …..(2)
Similarly, 
In ΔCBD, 
BO is the median                 (From (1))
So, arΔCBO = arΔDBO          …..(3)
From (2) and (3) we have
arΔCAO + arΔCBO = arΔDBO + arΔDAO
ar(ΔABC) = ar(ΔABD)



 

Page No 389:

Question 18:

D and are points on sides AB and AC respectively of ∆ABC such that ar(∆BCD) = ar(∆BCE). Prove that DE || BC.

ANSWER:


ar(∆BCD) = ar(∆BCE)                     (Given)
We know, triangles on the same base and having equal areas lie between the same parallels.
Thus, DE || BC. 

Page No 389:

Question 19:

P is any point on the diagonal AC of a parallelogram ABCD. Prove that ar(∆ADP) = ar(∆ABP).

ANSWER:


Join BD. 
Let BD and AC intersect at point O. 
O is thus the midpoint of DB and AC. 
PO is the median of DPB, 
So, 
ar(DPO)=ar(BPO)                     .....(1)ar(ADO)=ar(ABO)                     .... (2)Case 1:(2)(1)ar(ADO)ar(DPO)=ar(ABO)ar(BPO)ar△DPO=ar△BPO                     …..1

ar△ADO=ar△ABO                     …..2Case 1:2-1⇒ar△ADO-ar△DPO=ar△ABO-ar△BPO
Thus, ar(∆ADP) = ar(∆ABP)

Case II: 

ar(ADO)+ar(DPO)=ar(ABO)+ar(BPO)ar△ADO+ar△DPO=ar△ABO+ar△BPO
Thus, ar(∆ADP) = ar(∆ABP)

Page No 389:

Question 20:

In the adjoining figure, the diagonals AC and BD of a quadrilateral ABCD intersect at O.
If BO = OD, prove that
ar(∆ABC) = ar(∆ADC),

ANSWER:

Given:  BO = OD
To prove: ar(∆ABC) = ar(∆ADC)
Proof
Since BO = ODO is the mid point of BD.
We know that a median of a triangle divides it into two triangles of equal areas.
CO is a median of ∆BCD.
i.e., ar(∆COD) = ar (∆COB)            …(i)

AO is a median of ∆ABD.
i.e., ar(∆AOD) = ar(∆AOB)              …(ii)

From (i) and (ii), we have:
ar(∆COD) + ar(∆AODar(∆COB) + ar(∆AOB)
∴ ar(∆ADC )​ = ar(∆ABC)

Page No 389:

Question 21:

The vertex A of ∆ABC is joined to a point D on the side BC. The midpoint of AD is E.
Prove that ar(ΔBEC)=12ar(ΔABC)ar(∆BEC)=12ar(∆ABC).

ANSWER:

Given:  D is the midpoint of BC and E is the midpoint of AD.
To prove: ar(ΔBEC)=12ar(ΔABC)ar(∆BEC)=12ar(∆ABC)
Proof: 
Since E is the midpoint of AD, BE is the median of ∆ABD.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., ar(∆BED ) = 1212ar(∆ABD)                 …(i)
Also, ar(∆CDE ) =1212 ar(∆ADC)             …(ii)

From (i) and (ii), we have:
ar(∆BED) + ar(∆CDE)​ 1212 ⨯​ ar(∆ABD)​ + 1212 ⨯​ ar(∆ADC)   
⇒ ar(∆BEC )​ = 1212⨯ [ar(∆ABD) + ar(∆ADC)] 
⇒ ​ar(∆BEC )​ =​ 1212 ⨯​ ar(∆ABC)



Page No 390:

Question 22:

is the midpoint of side BC of ∆ABC and E is the midpoint of BD. If is the midpoint of AE, prove that ar(∆BOE) = 1818ar(∆ABC).

ANSWER:

D is the midpoint of side BC of ∆ABC. 
AD is the median of ∆ABC. 
ar(ABD)=ar(ACD)=12ar(ABC)ar△ABD=ar△ACD=12ar△ABC
E is the midpoint of BD of ∆ABD, 
AE is the median of ∆ABD
ar(ABE)=ar(AED)=12ar(ABD)=14ar(ABC)ar△ABE=ar△AED=12ar△ABD=14ar△ABC
Also, O is the midpoint of AE, 
BO is the median of ∆ABE, 
ar(ABO)=ar(BOE)=12ar(ABE)=14ar(ABD)=18ar(ABC)ar△ABO=ar△BOE=12ar△ABE=14ar△ABD=18ar△ABC
Thus, ar(∆BOE) = 1818ar(∆ABC)

 

Page No 390:

Question 23:

In a trapezium ABCDAB || DC and is the midpoint of BC. Through M, a line PQ || AD has been drawn which meets AB in P and DC produced in Q, as shown in the adjoining figure. Prove that ar(ABCD) = ar(APQD).

ANSWER:

In MQC and MPB, 
MC = MB                            (M is the midpoint of BC)
CMQ = BMP                (Vertically opposite angles)
MCQ = MBP                (Alternate interior angles on the parallel lines AB and DQ)
Thus, MQC  MPB   (ASA congruency)
ar(MQC) = ar(MPB)
ar(MQC) + ar(APMCD) = ar(MPB) + ar(APMCD)
ar(APQD) = ar(ABCD)

Page No 390:

Question 24:

In the adjoining figure, ABCD is a quadrilateral. A line through D, parallel to AC, meets BC produced in P. Prove that ar(∆ABP) = ar(quad. ABCD).

ANSWER:

We have:
​ar(quad. ABCD) = ar(∆ACD) + ar(∆ABC)
ar(∆ABP) = ar(∆ACP)​​ + ar(∆ABC) 

ACD and ∆ACP are on the same base and between the same parallels AC and DP.
∴ ar(∆ACD) = ar(∆ ACP)​
By adding ar(∆ABC) on both sides, we get:
ar(∆ACDar(∆ABC) = ar(∆ACP)​​ + ar(∆ABC)                
⇒​ ar (quad. ABCD) = ar(∆ABP)
Hence, proved.

Page No 390:

Question 25:

In the adjoining figure, ∆ABC and ∆DBC are on the same base BC with A and D on opposite sides of BC such that ar(∆ABC) = ar(∆DBC). Show that BC bisects AD.

ANSWER:



Given: ∆ABC and ∆DBC are on the same base BC.
ar(∆ABC) = ar(∆DBC)​
To prove: BC bisects AD
Construction: Draw AL ⊥ BC and DM ⊥ BC.
Proof: 
Since ∆ABC and ∆DBC are on the same base BC and they have equal areas, their altitudes must be equal.
i.e., AL = DM
Let AD and BC intersect at O.
Now, in ∆ALO and ∆DMOwe have:
AL = DM
ALO = ∠DMO =  90o
∠​AOL = ∠DO​M                  (Vertically opposite angles)
i.e., ∆ ALO ≅ ∆ DMO
​​
∴​ OA = OD
Hence, BC bisects AD.

Page No 390:

Question 26:

ABCD is a parallelogram in which BC is produced to P such that CP BC, as shown in the adjoining figure. AP intersects CD at M. If ar(DMB) = 7 cm2, find the area of parallelogram ABCD.

ANSWER:

In MDA and MCP, 
DMA = CMP                  (Vertically opposite angles)
MDA = MCP                  (Alternate interior angles)
AD = CP                                 (Since AD = CB and CB = CP)
So, MDA  MCP         (ASA congruency)
DM = MC                         (CPCT)
M is the midpoint of DC
BM is the median of BDC
ar(BMC) = ar(DMB) = 7 cm2 
ar(BMC) + ar(DMB) = ar(DBC) = 7 + 7 = 14 cm2cm2
Area of parallelogram ABCD = ×× ar(DBC) = 2 ×× 14 = 28 cm2cm2 
 

Page No 390:

Question 27:

In a parallelogram ABCD, any point E is taken on the side BCAE and DC when produced meet at a point M. Prove that
ar(∆ADM) = ar(ABMC)

ANSWER:


Join BM and AC. 
ar(∆ADC) = 12bh12bh = 12×DC×h12×DC×h
ar(∆ABM) = 12×AB×h12×AB×h
AB = DC                               (Since ABCD is a parallelogram)
So, ar(∆ADC) = ar(∆ABM)
ar(∆ADC) + ar(∆AMC) = ar(∆ABM) + ar(∆AMC)
ar(∆ADM) = ar(ABMC)
Hence Proved

Page No 390:

Question 28:

PQRS are respectively the midpoints of the sides ABBCCD and DA of || gm ABCD. Show that PQRS is a parallelogram and also show that
ar(|| gm PQRS)=12×ar(∣∣∣∣gm ABCD)ar|| gm PQRS=12×ar(||gm ABCD).

ANSWER:

Given:  ABCD is a parallelogram and P, Q, R and S are the midpoints of sides AB, BC, CD and DA, respectively.
To prove: ar(parallelogram PQRS ) = 1212 × ar(parallelogram ABCD )
Proof: 
In ∆ABC, PQ || AC and PQ = 1212 × AC              [ By midpoint theorem] 
Again, in ∆DAC, the points S and R are the mid points of AD and DC, respectively.
∴ SR || AC and SR = 1212 × AC                                   [ By midpoint theorem] 
Now, PQ || AC and SR || AC  
⇒ ​PQ || SR
Also, PQ = SR = 
1212 × AC
∴ PQ || SR and PQ = SR
Hence, PQRS is a parallelogram.

Now, ar(parallelogram​ PQRS) = ar(∆PSQ) + ar(∆SRQ)                       …(i)
also, ar(parallelogram ABCD) = ar(parallelogram ABQS) + ar(parallelogram SQCD)            …(ii)

PSQ and parallelogram ABQS are on the same base and between the same parallel lines.
So, ar(∆PSQ ) =1212 × ar(parallelogram ABQS)                         …(iii)
Similarly, ∆SRQ and parallelogram SQCD are on the same base and between the same parallel lines.
So, ar(∆SRQ ) = 1212 × ar(parallelogram SQCD)                        …(iv)
Putting the values from (iii) and (iv) in (i), we get:
  ar(parallelogram​ PQRS) = 1212 × ar(parallelogram ABQS)​ + 1212 × ar(parallelogram SQCD)
From (ii), we get:
ar(parallelogram​ PQRS) = 1212 × ar(parallelogram ABCD)

This is the complete blog on RS Aggarwal Solutions Class 9 Maths Chapter 11 Areas of Parallelograms and Triangles. To know more about the CBSE Class 9 Maths exam, ask in the comments.

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