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# Frank Modern Certificate Solution for Class 9 Mathematics Chapter 21 - Areas Theorems on Parallelograms

## Frank Textbook Solutions Chapter 21 - Areas Theorems on Parallelograms

Frank Textbook Solutions are considered extremely helpful for solving difficult questions in the ICSE Class 9 Mathematics exam. TopperLearning Textbook Solutions are compiled by our subject experts. Herein, you can find all the answers to the questions of   Chapter 21 - Areas Theorems on Parallelograms for the Frank textbook.

Frank Textbook Solutions for class 9  are in accordance with the latest ICSE syllabus, and they are amended from time to time to be most relevant. Our free Frank Textbook Solutions for ICSE Class 9 Mathematics will give you deeper insight on the chapters and will help you to score more marks in the final examination. ICSE Class 9 students can refer to our solutions while doing their homework and while preparing for the exam.

Exercise/Page

## Frank Modern Certificate Solution for Class 9 Mathematics Chapter 21 - Areas Theorems on Parallelograms Page/Excercise 21.1

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

ΔPQT and parallelogram PQRS are on the same base PQ and between the same parallel lines PQ and SR.

ΔPSN and parallelogram PQRS are on the same base PS and between the same parallel lines PS and QN.

Adding equations (i) and (ii), we get

Subtracting A(ΔRTN) from both the sides, we get

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Joining AC and FE, we get

ΔAFC and ΔAFE are on the same base AF and between the same parallels AF and CE.

A(ΔAFC) = A(ΔAFE)

A(ΔABF) + A(ΔABC) = A(ΔABF) + A(ΔBFE)

A(ΔABC) = A(ΔBFE)

A(parallelogram ABCD) = A(parallelogram BFGE)  A(parallelogram ABCD) = A(parallelogram BFGE)

Solution 11

Solution 12

Joining TR, we get

ΔPQR and ΔQTR are on the same base QR and between the same parallel lines QR and PT.

ΔQTR and ΔTQS are on the same base QT and between the same parallel lines QT and RS.

From (i) and (ii), we get

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

SM PN

SM TN

Also, SR MN

ST MN

Hence, SMNT is a parallelogram.

SM PN

SM PO

Also, PS QM

PS OM

Hence, SMOP is a parallelogram.

Now, parallelograms SMNT and SMOP are on the same base SM and between the same parallels SM and PN.

A(parallelogram SMNT) = A(parallelogram SMOP) ….(i)

Similarly, we can show that quadrilaterals PQRS is a parallelogram.

Now, parallelograms PQRS and SMOP are on the same base PS and between the same parallels PS and QM.

A(parallelogram PQRS) = A(parallelogram SMOP) ….(ii)

From (i) and (ii), we have

A(parallelogram SMNT) = A(parallelogram PQRS)

Solution 31

Solution 32

Solution 33

Solution 34(a)

AD is the median of ΔABC.

Therefore it will divide ΔABC into two triangles of equal areas.

Area(ΔABD) = Area(ΔACD) ….(i)

Similarly, ED is the median of ΔEBC.

Area(DEBD) = Area(DECD) ….(ii)

Subtracting equation (ii) from (i), we have

Area(ΔABD) - Area(ΔEBD) = Area(ΔACD) - Area(ΔECD)

Area(ΔABE) = Area(ΔACE)

Solution 34(b)

AD is the median of ΔABC.

Therefore it will divide ΔABC into two triangles of equal areas.

Area(ΔABD) = Area(ΔACD) ….(i)

Similarly, ED is the median of ΔEBC.

Area(ΔEBD) = Area(ΔECD) ….(ii)

Subtracting equation (ii) from (i), we have

Area(ΔABD) - Area(ΔEBD) = Area(ΔACD) - Area(ΔECD)

Area(ΔABE) = Area(ΔACE) ….(iii)

Since E is the mid-point of median AD,

AE = ED

Now,

ΔABE and ΔBED have equal bases and a common vertex B.

Area(ΔABE) = Area(ΔBED) ….(iv)

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

Area(ΔABE) = A(ΔBED) = Area(ΔACE) = Area(ΔEDC) ….(v)

Now,

Area(ΔABC) = Area(ΔABE) + A(ΔBED) + Area(ΔACE) + Area(ΔEDC)

= 4 × Area(ΔABE) [From (v)]

Solution 35

Construction: Join SM and SQ.

In a parallelogram PQRS, SQ is the diagonal.

So, it bisects the parallelogram.

Area(DPSQ)

SM is the median of ΔPSQ.

Area(ΔPSM)

Again, MN is the median of ΔPSM.

Area(ΔPMN)

Solution 36

Construction: Join QR. Let the diagonals PR and QS intersect each other at point O.

Since diagonals of a parallelogram bisect each other, therefore O is the mid-point of both PR and QS.

Now, median of a triangle divides it into two triangles of equal area.

In ΔPSQ, OP is the median.

Area(ΔPOS) = Area(ΔPOQ) ….(i)

Similarly, OT is the median of ΔTSQ.

Area(ΔTOS) = Area(ΔTOQ) ….(ii)

Subtracting equation (ii) from (i), we have

Area(ΔPOS) - Area(ΔTOS) = Area(ΔPOQ) - Area(ΔTOQ)

Area(ΔPTQ) = Area(ΔPTS)

Area(ΔPTS) = 18 square units

Solution 37

## Browse Study Material

TopperLearning provides step-by-step solutions for each question in each chapter in the Frank textbook recommended by ICSE schools. Access Chapter 21 - Areas Theorems on Parallelograms here. Our Frank Textbook Solutions for ICSE Class 9 Mathematics are designed by our subject matter experts. These solutions will help you to revise the whole chapter, so you can clear your fundamentals before the examination.

# Text Book Solutions

ICSE IX - Mathematics

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