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

Study from Selina Solutions for ICSE Class 9 Mathematics Chapter 14 Rectilinear figures [Quadrilaterals: Parallelogram, Rectangle, Rhombus, Square and Trapezium]. These are reference solutions freely accessible at TopperLearning to help you with your Maths revision. In this chapter, learn how to compute answers of the interior and exterior angles of various polygons.

Rhombus, parallelograms and rectangles are types of quadrilaterals that you can easily understand using our Selina ICSE Class 9 Maths textbook solutions. If you require more resources to ace your Maths exam, you can find online concept videos, practice tests, mock exam papers with solutions and more on TopperLearning.com.

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

The sum of the inteior angles of a polygon is four times the sum of its exterior angles. Find the number of sides in the polygon.

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.

The angles of a pentagon are in the ratio 4 : 8 : 6 : 4 : 5. Find each angle of the pentagon.

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

One angle of a six-sided polygon is 140^{o} and the other angles are equal. Find the measure of each equal angle.

Let the measure of each equal angles are x.

Then we can write

_{}

Therefore the measure of each equal angles are 116

In a polygon there are 5 right angles and the remaining angles are equal to 195^{o} each. Find the number of sides in the polygon.

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.

Three angles of a seven sided polygon are 132^{o} each and the remaining four angles are equal. Find the value of each equal angle.

Let the measure of each equal angles are x.

Then we can write:

_{}

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

Two angles of an eight sided polygon are 142^{o} and 176^{o}. If the remaining angles are equal to each other; find the magnitude of each of the equal angles.

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}.

In a pentagon ABCDE, AB is parallel to DC and A : E : D = 3 : 4 : 5. Find angle E.

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

Thus

_{}

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

AB, BC and CD are the three consecutive sides of a regular polygon. If BAC = 15^{o}; find,

(i) Each interior angle of the polygon.

(ii) Each exterior angle of the polygon.

(iii) Number of sides of the polygon.

(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.

The ratio between an exterior angle and an interior angle of a regular polygon is 2 : 3. Find the number of sides in the polygon.

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.

The difference between an exterior angle of (n - 1) sided regular polygon and an exterior angle of (n + 2) sided regular polygon is 6^{o} find the value of n.

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.

Two alternate sides of a regular polygon, when produced, meet at right angle. Find:

(i) The value of each exterior angle of the polygon;

(ii) The number of sides in the polygon.

(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:

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

State, 'true' or 'false'

(i) The diagonals of a rectangle bisect each other.

(ii) The diagonals of a quadrilateral bisect each other.

(iii) The diagonals of a parallelogram bisect each other at right angle.

(iv) Each diagonal of a rhombus bisects it.

(v) The quadrilateral, whose four sides are equal, is a square.

(vi) Every rhombus is a parallelogram.

(vii) Every parallelogram is a rhombus.

(viii) Diagonals of a rhombus are equal.

(ix) If two adjacent sides of a parallelogram are equal, it is a rhombus.

(x) If the diagonals of a quadrilateral bisect each other at right angle, the quadrilateral is a square.

(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.

In the figure, given below, AM bisects angle A and DM bisects angle D of parallelogram ABCD. Prove that : _{}AMD = 90^{o.}

From the given figure we conclude that

_{}

Again from the _{}

_{}

Hence _{}

In the following figure, AE and BC are equal and parallel and the three sides AB, CD and DE are equal to one another. If angle A is 102^{o}. Find angles AEC and BCD.

In the given figure

In a square ABCD, diagonals meet at O. P is a point on BC such that OB = BP.

Show that:

(i) _{}POC = _{}

(ii) _{}BDC = 2 _{}POC

(iii) _{}BOP = 3 _{}CPO

The given figure shows a square ABCD and an equilateral triangle ABP. Calculate:

(i) _{}AOB

(ii) _{}BPC

(iii) _{}PCD

(iv) Reflex _{}APC

In the given figure _{}is an equilateral triangle

Therefore all its angles are _{}

Again in the

_{}

_{}

Again

_{}

Now

Therefore

_{}

(i)_{}

(ii)

(iii)

(iv)Reflex

In the given figure ABCD is a rhombus with angle A = 67^{o}

If DEC is an equilateral triangle, calculate:

(i) _{}CBE(ii) _{}DBE.

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

In each of the following figures, ABCD is a parallelogram.

(i)

(ii)

In each case, given above, find the values of x and y.

(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

The angles of a quadrilateral are in the ratio 3 : 4 : 5 : 6. Show that the quadrilateral is a trapezium.

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

In a parallelogram ABCD, AB = 20 cm and AD = 12 cm. The bisector of angle A meets DC at E and BC produced at F. Find the length of CF.

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

In parallelogram ABCD, AP and AQ are perpendiculars from vertex of obtuse angle A as shown. If _{} find angles of the parallelogram.

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.

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

E is the mid-point of side AB and F is the mid point of side DC of parallelogram ABCD. Prove that AEFD is a parallelogram.

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 AEFC is a parallelogram.

The diagonal BD of a parallelogram ABCD bisects angles B and D. Prove that ABCD is a rhombus.

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

The alongside figure shows a parallelogram ABCD in which AE = EF = FC. Prove that:

(i) DE is parallel to FB

(ii) DE = FB

(iii) DEBF is a parallelogram.

In the alongside diagram, ABCD is a parallelogram in which AP bisects angle AP bisects angle A and BQ bisects angle B. Prove that :

(i) AQ = BP

(ii) PQ = CD.

(iii) ABPQ is a parallelogram.

Let us join PQ.

In the given figure, ABCD is a parallelogram. Prove that: AB = 2 BC.

Given _{}is a parallelogram

To prove:_{}

Proof: _{}is a parallelogram

_{}

Again,

_{}

NOW

_{}

Hence proved

Prove that the bisectors of opposite angles of a parallelogram are parallel.

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

Prove that the bisectors of interior angles of a parallelogram form a rectangle.

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

Prove that the bisectors of the interior angles of a rectangle form a square.

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

In parallelogram ABCD, the bisector of angle A meets DC at P and AB = 2 AD.

Prove that:

(i) BP bisects angle B.(ii) Angle APB = 90^{o}.

(i)Let_{}

_{}

Also _{}is the bisector _{}

_{}

Now ,

_{}

Therefore _{}

Now

_{}

Therefore

_{}

Also ,_{}

In _{}

Therefore _{}

Hence _{}bisect _{}

(ii)

Opposite angles are supplementary

Therefore

_{}

_{}

Hence proved

Points M and N are taken on the diagonal AC of a parallelogram ABCD such that AM = CN. Prove that BMDN is a parallelogram.

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

In the following figure, ABCD is a parallelogram. Prove that:

(i) AP bisects angle A

(ii) BP bisects angle B

(iii) _{}DAP + _{}BCP = _{}APB

Consider _{}

_{}

Therefore _{}

_{}

_{}

Hence proved

ABCD is a square. A is joined to a point P on BC and D is joined to a point Q on AB. If AP = DQ; prove that AP and DQ are perpendicular to each other.

ABCD is a square and _{}

In a quadrilateral ABCD, AB = AD and CB = CD. Prove that :

(i) AC bisects angle BAD.

(ii) AC is perpendicular bisector of BD.

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

The following figure shows a trapezium ABCD in which AB is parallel to DC and AD = BC. Prove that:

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

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

(iii) AC = BD

(iv) OA = OB and OC = OD

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

In the given figure, AP is bisector of _{}A and CQ is bisector of _{}C of parallelogram ABCD. Prove that APCQ is a parallelogram.

In case of a parallelogram prove that:

(i) The bisectors of any two adjacent angles intersect at 90^{o}.

(ii) The bisectors of opposite angles are parallel to each other.

_{}is a parallelogram ,the bisectors of _{}and

_{}meet at a point _{}and the bisectors of _{}AND _{}meet at F.

We have to prove that the _{}and _{}

Proof :In the parallelogram _{}

_{}

Similarly taking triangle BCF it can be proved that_{}

Also _{}

Now since _{}

Hence proved

The diagonals of a rectangle intersect each other at right angles. Prove that the rectangle is a square.

In the following figure, ABCD and PQRS are two parallelograms such that ∠D = 120° and ∠Q = 70°. Find the value of x.

In the following figure, ABCD is a rhombus and DCFE is a square.

If ∠ABC =56°, find:

(i) ∠DAE

(ii) ∠FEA

(iii) ∠EAC

(iv) ∠AEC

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