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# Frank Modern Certificate Solution for Class 9 Mathematics Chapter 19 - Quadrilaterals

## Frank Textbook Solutions Chapter 19 - Quadrilaterals

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 19 - Quadrilaterals 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 19 - Quadrilaterals Page/Excercise 19.1

Solution 1(a)

Solution 1(b)

Solution 1(c)

Solution 1(d)

Solution 1(e)

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Construction: Join PR.

Solution 14

Solution 15

## Frank Modern Certificate Solution for Class 9 Mathematics Chapter 19 - Quadrilaterals Page/Excercise 19.2

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13(a)

Solution 13(b)

Solution 13(c)

Solution 14

Solution 15(a)

Construction:

Join BS and AQ.

Join diagonal QS.

Solution 15(b)

Construction:

Join BS and AQ.

Join diagonal QS.

Solution 16(a)

Construction:

Draw SM PQ and RN PQ

Solution 17

Solution 18

Let ABCD be a quadrilateral, whose diagonals AC and BD bisect each other at right angle.

i.e. OA = OC, OB = OD

And, AOB = BOC = COD = AOD = 90°

To prove ABCD a rhombus, we need to prove ABCD is a parallelogram and all sides of ABCD are equal.

Now, in ΔAOD and DCOD

OA = OC  (Diagonal bisects each other)

AOD = COD (Each 90°)

OD = OD   (common)

∴ΔAOD ≅ΔCOD  (By SAS congruence rule)

Similarly, we can prove that

AD = AB and CD = BC ….(ii)

From equations (i) and (ii), we can say that

AB = BC = CD = AD

Since opposite sides of quadrilateral ABCD are equal, so, we can say that ABCD is a parallelogram.

Since all sides of a parallelogram ABCD are equal, so, we can say that ABCD is a rhombus.

Solution 19

Consider ABCD is a kite.

Then, AB = AD and BC = DC

Solution 20

## 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 19 - Quadrilaterals 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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