Chapter 19 : Quadrilaterals - Frank Solutions for Class 9 Maths ICSE

Mathematics in ICSE Class 9 is one of the most challenging and trickiest subjects of all. It includes complex topics such as logarithms, expansions, indices and Pythagoras Theorem which are difficult to understand for an average student. TopperLearning provides study materials for ICSE Class 9 Mathematics to make the subject easy and help students to clear all their concepts. Our study materials comprise numerous video lessons, question banks, revision notes and sample papers which help achieve success in the examination.

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Chapter 19 - Quadrilaterals Excercise Ex. 19.1

Question 1

In the following figures, find the remaining angles of the  parallelogram

Solution 1

Question 2

In the following figures, find the remaining angles of the  parallelogram

Solution 2

Question 3

In the following figures, find the remaining angles of the  parallelogram

Solution 3

Question 4

In the following figures, find the remaining angles of the  parallelogram

Solution 4

Question 5

In the following figures, find the remaining angles of the  parallelogram

Solution 5

Question 6
Solution 6
Question 7

The consecutive angles of a parallelogram are in the ratio 3:6. Calculate the measures of all the angles of the parallelogram.

Solution 7

 

 

Question 8

Solution 8

Question 9

In the given figure, ABCD is a parallelogram, find the values of x and y.

 

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12
Solution 12
Question 13
Solution 13
Question 14
Solution 14
Question 15

Solution 15

Question 16

Solution 16

Question 17

In the given figure, MP is the bisector of P and RN is the bisector of R of parallelogram PQRS. Prove that PMRN is a parallelogram.

Solution 17

 

Construction: Join PR.

 

 

Question 18

Solution 18

Question 19

Solution 19

Chapter 19 - Quadrilaterals Excercise Ex. 19.2

Question 1
Solution 1
Question 2
Solution 2
Question 3
Solution 3
Question 4
Solution 4
Question 5
Solution 5
Question 6
Solution 6
Question 7
Solution 7
Question 8
Solution 8
Question 9
Solution 9
Question 10
Solution 10
Question 11
Solution 11
Question 12
Solution 12
Question 13

In a parallelogram PQRS, M and N are the midpoints of the opposite sides PQ and RS respectively. Prove that

RN and RM trisect QS.

Solution 13

 

 

Question 14

In a parallelogram PQRS, M and N are the midpoints of the opposite sides PQ and RS respectively. Prove that

PMRN is a parallelogram.

Solution 14

 

 

Question 15

In a parallelogram PQRS, M and N are the midpoints of the opposite sides PQ and RS respectively. Prove that

MN bisects QS.

Solution 15

 

 

Question 16
Solution 16
Question 17

In the given figure, PQRS is a parallelogram in which PA = AB = Prove that:

SA QB and SA = QB.

 

Solution 17

Construction:

Join BS and AQ.

Join diagonal QS.

 

 

Question 18

In the given figure, PQRS is a parallelogram in which PA = AB = Prove that:

SAQB is a parallelogram.

Solution 18

Construction:

Join BS and AQ.

Join diagonal QS.

 

 

Question 19

In the given figure, PQRS is a trapezium in which PQ SR and PS = QR. Prove that:

PSR = QRS and SPQ = RQP

 

Solution 19

Construction:

Draw SM PQ and RN PQ

 

Question 20

In a parallelogram ABCD, E is the midpoint of AB and DE bisects angle D. Prove that:

  1. BC = BE.
  2. CE is the bisector of angle C and angle DEC is a right angle
Solution 20

 

 

Question 21

Prove that if the diagonals of a quadrilateral bisect each other at right angles then it is a rhombus.

Solution 21

 

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)

AD = CD  ….(i)

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.

Question 22

Prove that the diagonals of a kite intersect each other at right angles.

Solution 22

Consider ABCD is a kite.

Then, AB = AD and BC = DC

 

 

 

Question 23

Prove that the diagonals of a square are equal and perpendicular to each other.

Solution 23

 

 

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