# Class 9 FRANK Solutions 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.

## Quadrilaterals Exercise Ex. 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

## Quadrilaterals Exercise Ex. 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 13(a)

### Solution 13(b)

### Solution 13(c)

### Solution 12

### 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)

∴ 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.

### Solution 19

Consider ABCD is a kite.

Then, AB = AD and BC = DC

### Solution 20