# Class 9 SELINA Solutions Maths Chapter 14 - Rectilinear Figures [Quadrilaterals: Parallelogram, Rectangle, Rhombus, Square and Trapezium]

## Rectilinear Figures [Quadrilaterals: Parallelogram, Rectangle, Rhombus, Square and Trapezium] Exercise Ex. 14(A)

### Solution 1

The sum of the interior angle=4 times the sum of the exterior angles.

Therefore the sum of the interior angles = 4×360° =1440°.

Now we have

_{}

Thus the number of sides in the polygon is 10.

### Solution 2

Let the angles of the pentagon are 4x, 8x, 6x, 4x and 5x.

Thus we can write

_{}

Hence the angles of the pentagon are:

4×20= 80, 8×20= 160, 6×20= 120, 4×20= 80, 5×20= 100

### Solution 3

Let the measure of each equal angles are x.

Then we can write

_{}

Therefore the measure of each equal angles are 116

### Solution 4

Let the number of sides of the polygon is *n* and there are k angles with measure 195^{o}.

Therefore we can write:

_{}

In this linear equation n and k must be integer. Therefore to satisfy this equation the minimum value of k must be 6 to get n as integer.

Hence the number of sides are: 5 + 6 = 11.

### Solution 5

Let the measure of each equal angles are x.

Then we can write:

_{}

Thus the measure of each equal angles are 126^{o}.

### Solution 6

Let the measure of each equal sides of the polygon is x.

Then we can write:

_{}

Thus the measure of each equal angles are 127^{o}.

### Solution 7

Let the measure of the angles are 3x, 4x and 5x.

Thus

_{}

Thus the measure of angle E will be 4×30=120

### Solution 8

(i)

Let each angle of measure x degree.

Therefore measure of each angle will be:

(ii)

Let each angle of measure x degree.

Therefore measure of each exterior angle will be:

(iii)

Let the number of each sides is n.

Now we can write

Thus the number of sides are 12.

### Solution 9

Let measure of each interior and exterior angles are 3k and 2k.

Let number of sides of the polygon is n.

Now we can write:

Again

From (1)

Thus the number of sides of the polygon is 5.

### Solution 10

For (n-1) sided regular polygon:

Let measure of each angle is x.

Therefore

For (n+1) sided regular polygon:

Let measure of each angle is y.

Therefore

Now we have

Thus the value of n is 13.

### Solution 11

(i)

Let the measure of each exterior angle is x and the number of sides is n.

Therefore we can write:

Now we have

(ii)

Thus the number of sides in the polygon is:

## Rectilinear Figures [Quadrilaterals: Parallelogram, Rectangle, Rhombus, Square and Trapezium] Exercise Ex. 14(B)

### Solution 1

(i)True.

This is true, because we know that a rectangle is a parallelogram. So, all the properties of a parallelogram are true for a rectangle. Since the diagonals of a parallelogram bisect each other, the same holds true for a rectangle.

(ii)False

This is not true for any random quadrilateral. Observe the quadrilateral shown below.

Clearly the diagonals of the given quadrilateral do not bisect each other. However, if the quadrilateral was a special quadrilateral like a parallelogram, this would hold true.

(iii)False

Consider a rectangle as shown below.

It is a parallelogram. However, the diagonals of a rectangle do not intersect at right angles, even though they bisect each other.

(iv)True

Since a rhombus is a parallelogram, and we know that the diagonals of a parallelogram bisect each other, hence the diagonals of a rhombus too, bisect other.

(v)False

This need not be true, since if the angles of the quadrilateral are not right angles, the quadrilateral would be a rhombus rather than a square.

(vi)True

A parallelogram is a quadrilateral with opposite sides parallel and equal.

Since opposite sides of a rhombus are parallel, and all the sides of the rhombus are equal, a rhombus is a parallelogram.

(vii)False

This is false, since a parallelogram in general does not have all its sides equal. Only opposite sides of a parallelogram are equal. However, a rhombus has all its sides equal. So, every parallelogram cannot be a rhombus, except those parallelograms that have all equal sides.

(viii)False

This is a property of a rhombus. The diagonals of a rhombus need not be equal.

(ix)True

A parallelogram is a quadrilateral with opposite sides parallel and equal.

A rhombus is a quadrilateral with opposite sides parallel, and all sides equal.

If in a parallelogram the adjacent sides are equal, it means all the sides of the parallelogram are equal, thus forming a rhombus.

(x)False

Observe the above figure. The diagonals of the quadrilateral shown above bisect each other at right angles, however the quadrilateral need not be a square, since the angles of the quadrilateral are clearly not right angles.

### Solution 2

From the given figure we conclude that

_{}

Again from the _{}

_{}

Hence _{}

### Solution 3

In the given figure

### Solution 4

### Solution 5

In the given figure _{}is an equilateral triangle

Therefore all its angles are _{}

Again in the

_{}

_{}

Again

_{}

Now

Therefore

_{}

(i)_{}

(ii)

(iii)

(iv)Reflex

### Solution 6

Given that the figure ABCD is a rhombus with angle A = 67^{o}

### Solution 7

(i)ABCD is a parallelogram

Therefore

_{}

Thus

_{}

Solving equations (i) and (ii) we have

x=5

y=3

(ii)

In the figure ABCD is a parallelogram

_{}

Therefore

_{}

Solving (i), (ii) we have

### Solution 8

Given that the angles of a quadrilateral are in the ratio _{}Let the angles be _{}

_{}

Therefore the angles are

_{}

Since all the angles are of different degrees thus forms a trapezium

### Solution 9

Given AB = 20 cm and AD = 12 cm.

From the above figure, it's evident that ABF is an isosceles triangle with angle BAF = angle BFA = x

So AB = BF = 20

BF = 20

BC + CF = 20

CF = 20 - 12 = 8 cm

### Solution 10

We know that AQCP is a quadrilateral. So sum of all angles must be 360.

∴ x + y + 90 + 90 = 360

x + y = 180

Given x:y = 2:1

So substitute x = 2y

3y = 180

y = 60

x = 120

We know that angle C = angle A = x = 120

Angle D = Angle B = 180 - x = 180 - 120 = 60

Hence, angles of parallelogram are 120, 60, 120 and 60.

## Rectilinear Figures [Quadrilaterals: Parallelogram, Rectangle, Rhombus, Square and Trapezium] Exercise Ex. 14(C)

### Solution 1

Let us draw a parallelogram _{}Where F is the midpoint

Of side DC of parallelogram _{}

To prove:_{} is a parallelogram

Proof:

Therefore _{}

_{}

Also AD|| EF

therefore AEFD is a parallelogram.

### Solution 2

GIVEN:_{} is a parallelogram where the diagonal _{} bisects

parallelogram _{} at angle _{}

TO PROVE:_{ } is a rhombus

Proof
: Let us draw a parallelogram _{} where the
diagonal _{} bisects the
parallelogram at angle _{}.

Consruction :Let us join AC as a diagonal of the parallelogram

_{}

Since _{} is a
parallelogram

Therefore

_{}

Diagonal _{} bisects angle _{}

So _{}

Again _{} also bisects
at _{}

Therefore _{}

Thus _{} is a rhombus.

Hence proved

### Solution 3

### Solution 4

Let us join PQ.

### Solution 5

Given _{}is a parallelogram

To prove:_{}

Proof: _{}is a parallelogram

_{}

Again,

_{}

NOW

_{}

Hence proved

### Solution 6

Given ABCD is a parallelogram. The bisectors of _{}and_{} meet at E. The bisectors of and_{} meet at F

From the parallelogram _{}we have

_{}

In triangle ECD sum of angles_{}

_{}

Similarly taking triangle _{}it can be prove that _{}

Now since

Therefore the lines _{}and BF are parallel

Hence proved

### Solution 7

Given:_{} is a parallelogram

_{}

TO PROVE:_{} is a rectangle

Proof :

_{}

_{}is a right triangle because its acute interior angles are complementary.

Similarly

_{}

since 3 angles of quadrilateral _{}are right angles,si is the 4^{th} one and so _{}is a rectangle ,since its interior angles are all right angles

Hence proved

### Solution 8

Given: A parallelogram _{}in which _{}

Are the bisects of _{}respectively forming quadrilaterals_{}.

To prove: _{}is a rectangle

Proof :

_{}Also in

_{}

From the above equation we get

_{}

Hence PQRS is a rectangle

### Solution 9

(i)Let_{}

_{}

Also _{}is the bisector _{}

_{}

Now ,

_{}

Therefore _{}

Now

_{}

Therefore

_{}

Also ,_{}

In _{}

Therefore _{}

Hence _{}bisect _{}

(ii)

Opposite angles are supplementary

Therefore

_{}

_{}

Hence proved

### Solution 10

Points _{}are taken on the diagonal _{}of a parallelogram _{}such that _{}.

Prove that _{}is a parallelogram

CONSTRUCTION: Join _{}to meet _{}in _{}.

PROOF: We know that the diagonals of parallelogram bisect each other.

Now,_{}bisect each other at _{}.

_{}

Thus in a quadrilateral _{},diagonal _{}are such that _{}and _{}

Therefore the diagonals _{}bisect each other.

Hence _{}is a parallelogram

### Solution 11

ABCD is a Parallelogram.

Hence opposite sides are parallel and equal.

AD = BC

DC = AB

Also,

DC = AB = 2DP = 2PC.

Now,

DP || AB and AP transversal.

Hence alternate angles are equal

∠DPA = ∠PAB …(1)

Also,

DP || AB and PB transversal.

Hence alternate angles are equal

Hence ∠CPB = ∠PBA … (2)

In ∆ADP

AD = DP

Hence ∆ADP is isosceles.

As a result,

∠DAP = ∠DPA … (3)

Also, in ∆PCB

PC = CB

Hence ∆PCB is isosceles.

As a result,

∠CPB = ∠CBP … (4)

So from (1) and (3),

∠PAB = ∠DAP

Hence AP bisects angle A

And (2) and (4)

∠PBA = ∠CBP

Hence BP bisects angle B

Now in ∆APB, using the angle sum property

∠PAB +∠PBA + ∠APB = 180 …(5)

Also

AD || BC and AB transversal,

Hence interior angles are supplementary

∠DAB + ∠CBA = 180^{o}

Now as AP bisects angle A and BP bisects angle B

2∠DAP + 2∠CBP = 180^{o} … (6)

From (5) and (6)

∠PAB +∠PBA + ∠APB = 2∠DAP + 2∠CBP

Also we have

∠PAB = ∠DAP

∠PBA = ∠CBP

Hence we get

∠DAP + ∠CBP + ∠APB = 2∠DAP + 2∠CBP

∠APB = ∠DAP + ∠CBP

### Solution 12

ABCD is a square and _{}

### Solution 13

Given: _{}is quadrilateral,

_{}

To prove: (i) AC bisects angle BAD.

(ii) AC is perpendicular bisector of BD.

Proof :

_{}

Therefore _{}bisects _{}

_{}

Thus _{}is perpendicular bisector of _{}

Hence proved

### Solution 14

Given _{} is a
trapezium,_{}

To prove(i)_{}DAB = _{}CBA

(ii) _{}ADC = _{}BCD

(iii) AC = BD

(iv) OA = OB and OC = OD

Proof :(i) Since _{} and
transversal _{}cuts them at _{}respectively.

Therefore, _{}

Since _{}

Therefore _{} is a
parallelogram

_{}

_{}

In _{},we have

_{}

Since _{}

Again _{}

Hence proved

### Solution 15

### Solution 16

(i)

▭ABCD is a parallelogram

Hence CD || AB, also if we consider AD as transversal, we get interior angles supplementary.

Hence,

∠CDA + ∠BAD = 180

Now multiply ½ throughout.

½∠CDA + ½∠BAD = 90 …(1)

Now we have DO and AO as bisectors of ∠CDA and ∠BAD respectively.

Hence,

½∠CDA = ∠ODA

½∠BAD = ∠OAD

So our equation (1) becomes

∠ODA + ∠OAD = 90

Hence,
the bisectors of any two adjacent angles intersect at 90^{o}.

(ii)

▭ABCD is a parallelogram

Hence ∠DAB = ∠BCD

Now multiply ½ throughout.

½∠DAB = ½∠DCB …(1)

Now we have AX and CY as bisectors of ∠DAB and ∠DCB respectively.

Hence,

½∠DAB = ∠XAY

½∠BCD = ∠XCY

So our equation (1) becomes

∠XAY = ∠XCY …(2)

Also, XC || YB and YC is transversal, hence alternate angles are equal.

∠XCY = ∠CYB …(3)

From (2) and (3) we have

∠XAY = ∠CYB

Now, these are pair of corresponding angles formed by two lines AX and CY and the transversal AB.

Hence by converse of corresponding angle test, we can say

AX||CY.

Thus, the bisectors of opposite angles are parallel to each other.

### Solution 17

### Solution 18

### Solution 19