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R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 11 - Areas of Parallelograms and Triangles

Our RS Aggarwal & V Agarwal Textbook Solutions are considered extremely helpful for solving the tough questions which are asked in the CBSE Class 9 board exam. TopperLearning Textbook Solutions are compiled by subject experts. Herein, you can find all the answers to the textbook questions for Chapter 11 - Areas of Parallelograms and Triangles.

All our solutions are created in accordance with the latest CBSE syllabus, and they are amended from time to time to provide the most relevant answers. Our free RS Aggarwal & V Agarwal Textbook Solutions for CBSE Class 9 Mathematics will strengthen your fundamentals in this chapter and will help you in your attempts to score more marks in the final examination. CBSE Class 9 students can refer to our solutions any time — while doing their homework and while preparing for the exam.

Exercise/Page

R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 11 - Areas of Parallelograms and Triangles Page/Excercise MCQ

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Correct option: (c)

Area of gm ABCD = Base × Height = DC × DL

Solution 10

Correct option: (b)

Parallelograms on equal bases and between the same parallels are equal in area.

Solution 11

Correct option: (a)

ΔBMP and parallelogram ABPQ are on the same base BP and between the same parallels AQ and BP.

Parallelograms ABPQ and ABCD are on the same base AB and between the same parallels AB and PD.

Solution 12

Correct option: (a)

ΔABC is divided into four triangles of equal area.

A(parallelogram AFDE) = A(ΔAFE) + A(DFE)

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

Solution 32

Solution 33

Solution 34

Solution 35

Solution 36

Solution 37

Solution 2

Solution 3

Solution 5

Solution 6(ii)

Solution 6(i)

Solution 8

Solution 10

Solution 13

Solution 14

Solution 15

Solution 16

Solution 20

Solution 21

Solution 22

Solution 24

Solution 25

Solution 28

Solution 30

Solution 32

Solution 34

R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 11 - Areas of Parallelograms and Triangles Page/Excercise 11

Solution 1

Following figures lie on the same base and between the same parallels:

Figure (i): No

Figure (ii): No

Figure (iii): Yes, common base - AB, parallel lines - AB and DE

Figure (iv): No

Figure (v): Yes, common base - BC, parallel lines - BC and AD

Figure (vi): Yes, common base - CD, parallel lines - CD and BP

Solution 4

Solution 7

Solution 9

Construction: Join AC

Diagonal AC divides the parallelogram ABCD into two triangles of equal area.

ΔADC and parallelogram ABCD are on the same base CD and between the same parallel lines DC and AM.

Since M is the mid-point of AB,

Solution 11

Since ΔAPB and parallelogram ABCD are on the same base AB and between the same parallels AB and DC, we have

Similarly, ΔBQC and parallelogram ABCD are on the same base BC and between the same parallels BC and AD, we have

From (i) and (ii),

A(ΔAPB) = A(ΔBQC)

Solution 12

(i) Parallelograms MNPQ and ABPQ are on the same base PQ and between the same parallels PQ and MB.

(ii) ΔATQ and parallelogram ABPQ are on the same base AQ and between the same parallels AQ and BP.

Solution 17

We know that median of a triangle divides it into two triangles of equal area.

Now, AO is the median of ΔACD.

A(ΔCOA) = A(ΔDOA) ….(i)

And, BO is the median of ΔBCD.

A(ΔCOB) = A(ΔDOB) ….(ii)

Adding (i) and (ii), we get

A(ΔCOA) + A(ΔCOB) = A(ΔDOA) + A(ΔDOB)

A(ΔABC) = A(ΔABD)

Solution 18

Since ΔBCD and ΔBCE are equal in area and have a same base BC.

Therefore,

Altitude from D of ΔBCD = Altitude from E of ΔBCE

ΔBCD and ΔBCE are between the same parallel lines.

DE BC

Solution 19

Construction: Join BD.

Let the diagonals AC and BD intersect at point O.

Diagonals of a parallelogram bisect each other.

Hence, O is the mid-point of both AC and BD.

We know that the median of a triangle divides it into two triangles of equal area.

In ΔABD, OA is the median.

A(ΔAOD) = A(ΔAOB) ….(i)

In ΔBPD, OP is the median.

A(ΔOPD) = A(ΔOPB) ….(ii)

Adding (i) and (ii), we get

A(ΔAOD) + A(ΔOPD) = A(ΔAOB) + A(ΔOPB)

Solution 23

In ΔMCQ and ΔMPB,

QCM = PBM (alternate angles)

CM = BM (M is the mid-point of BC)

CMQ = PMB (vertically opposite angles)

ΔMCQ ΔMPB

A(ΔMCQ) = A(ΔMPB)

Now,

A(ABCD) = A(APQD) + A(DMPB) - A(ΔMCQ)

A(ABCD) = A(APQD)

Solution 26

AMD = PMC (vertically opposite angles)

And, DM = CM (c.p.c.t.)

BM is the median of ΔBDC.

A(ΔDMB) = A(ΔCMB)

A(ΔBDC) = 2 × A(ΔDMB) = 2 × 7 = 14 cm2

Now,

A(parallelogram ABCD) = 2 × A(ΔBDC) = 2 × 14 = 28 cm2

Solution 27

Construction: Join AC and BM

Let h be the distance between AB and CD.

Solution 29

Construction: Join EF

Since the line segment joining the mid-points of two sides of a triangle is parallel to the third side,

FE  BC

Clearly, ΔBEF and ΔCEF are on the same base EF and between the same parallel lines.

A(ΔBEF) = A(ΔCEF)

A(ΔBEF) - A(ΔGEF) = A(ΔCEF) - A(ΔGEF)

A(ΔBFG) = A(ΔCEG) …(i)

We know that a median of a triangle divides it into two triangles of equal area.

A(ΔBEC) = A(ΔABE)

A(ΔBGC) + A(ΔCEG) = A(quad. AFGE) + A(ΔBFG)

A(ΔBGC) + A(ΔBFG) = A(quad. AFGE) + A(ΔBFG) [Using (i)]

Solution 31

ΔDBC and ΔEBC are on the same base and between the same parallels.

A(ΔDBC) = A(ΔEBC) ….(i)

BE is the median of ΔABC.

Solution 33

ΔBCF and ΔACF are on the same base CF and between the same parallel lines CF and BA.

A(ΔBCF) = A(ΔACF)

A(ΔBCF) - A(ΔCGF) = A(ΔACF) - A(ΔCGF)

A(ΔCBG) = A(ΔAFG)

Solution 35

Construction: Join DB. Let DB cut MN at point Y.

M and N are the mid-points of AD and BC respectively.

MN AB CD

Y is the mid-point of DB.

Similarly, in ΔBDC,

Now, MN = MY + YN

Construction: Draw DQ AB. Let DQ cut MN at point P.

Then, P is the mid-point of DQ.

i.e. DP = PQ = h (say)

Solution 36

Construction: Join AC. Let AC cut EF at point Y.

E and F are the mid-points of AD and BC respectively.

EF AB CD

Y is the mid-point of AC.

Similarly, in ΔABC,

Now, EF = EY + YF

Construction: Draw AQ DC. Let AQ cut EF at point P.

Then, P is the mid-point of AQ.

i.e. AP = PQ = h (say)

Solution 37

Since D and E are the mid-points of AB and AC respectively,

DE BC PQ

In ΔACP, AP DE and E is the mid-point of AC.

D is the mid-point of PC (converse of mid-point theorem)

In ΔABQ, AQ DE and D is the mid-point of AB.

E is the mid-point of BQ (converse of mid-point theorem)

From (i) and (ii),

AP = AQ

Now, ΔACP and ΔABQ are on the equal bases AP and AQ and between the same parallels BC and PQ.

A(ΔACP) = A(ΔABQ)

Solution 38

In ΔRSC and ΔPQB,

CRS = BPQ (RC PB, corresponding angles)

RSC = PQB (RC PB, corresponding angles)

SC = QB (opposite sides of a parallelogram BQSC)

ΔRSC ΔPQB (by AAS congruence criterion)

A(ΔRSC) = A(ΔPQB)

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TopperLearning provides step-by-step solutions for each question in each chapter in the RS Aggarwal & V Aggarwal Textbook. Access the CBSE Class 9 Mathematics Chapter 11 - Areas of Parallelograms and Triangles for free. The questions have been solved by our subject matter experts to help you understand how to answer them. Our RS Aggarwal Solutions for class 9 will help you to study and revise the whole chapter, and you can easily clear your fundamentals in Chapter 11 - Areas of Parallelograms and Triangles now.

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