# FRANK Solutions for Class 9 Maths Chapter 19 - Quadrilaterals

If you are given the measure of one angle of a parallelogram, can you find the measure of the remaining angles? You can by referring to our Frank Solutions for ICSE Class 9 Mathematics Chapter 19 Quadrilaterals during revision. Further, you can practise Maths questions and answers on how to prove that the given figure is a parallelogram or a rectangle.

The Frank textbook solutions at the TopperLearning portal even includes problems on providing proofs to show that the given quadrilateral is a rhombus. If you are looking for more such ICSE Class 9 Maths chapter solutions, check the Selina solutions and solved sample question papers.

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## Chapter 19 - Quadrilaterals Exercise Ex. 19.1

Question 1(a)

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

Solution 1(a)

Question 1(b)

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

Solution 1(b)

Question 1(c)

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

Solution 1(c)

Question 1(d)

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

Solution 1(d)

Question 1(e)

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

Solution 1(e)

Question 2
Solution 2
Question 3

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

Solution 3

Question 4

Solution 4

Question 5

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

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 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 13

Construction: Join PR.

Question 14

Solution 14

Question 15

Solution 15

## Chapter 19 - Quadrilaterals Exercise 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(a)

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(a)

Question 13(b)

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 13(b)

Question 13(c)

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

MN bisects QS.

Solution 13(c)

Question 14
Solution 14
Question 15(a)

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

SA QB and SA = QB.

Solution 15(a)

Construction:

Join BS and AQ.

Join diagonal QS.

Question 15(b)

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

SAQB is a parallelogram.

Solution 15(b)

Construction:

Join BS and AQ.

Join diagonal QS.

Question 16(a)

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

PSR = QRS and SPQ = RQP

Solution 16(a)

Construction:

Draw SM PQ and RN PQ

Question 17

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 17

Question 18

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

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.

Question 19

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

Solution 19

Consider ABCD is a kite.

Then, AB = AD and BC = DC

Question 20

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

Solution 20

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