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

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

Question 1

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.

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.

Question 2

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

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:

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

Question 3

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

Solution 3

Let the measure of each equal angles are x.

Then we can write

Therefore the measure of each equal angles are 116degree

Question 4

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

Solution 4

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

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.

Question 5

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

Solution 5

Let the measure of each equal angles are x.

Then we can write:

Thus the measure of each equal angles are 126o.

Question 6

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

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 127o.

Question 7

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

Solution 7

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

Thus

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

Question 8

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

(i) Each interior angle of the polygon.

(ii) Each exterior angle of the polygon.

(iii) Number of sides of the polygon.

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.

Question 9

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.

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.

Question 10

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

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.

Question 11

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.

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:

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

Question 1

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.

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.

Question 2

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

Solution 2

From the given figure we conclude that

Again from the

Hence

Question 3

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 102o. Find angles AEC and BCD.

Solution 3

In the given figure

 

 

Given space that space space space AE space equals BC
space We space have space to space find space angle AEC space space angle BCD space
space Let space us space join space space EC space and space BD.
In space the space quadrilateral space AECB
AE space equals BC space and space AB equals EC
also space AE space vertical line vertical line space space BC
rightwards double arrow AB space vertical line vertical line space space EC
space So space quadrilateral space is space straight a space parallelogram.

In space parallelogram space consecutive space angles space are space supplementary
rightwards double arrow angle straight A plus angle straight B equals 180 to the power of degree
rightwards double arrow 102 to the power of degree space plus angle straight B equals 180 to the power of degree
rightwards double arrow angle straight B equals 78 to the power of degree

In space parallelogram space opposite space angles space are space equal
rightwards double arrow angle straight A equals angle BEC space and space angle straight B equals angle AEC space
rightwards double arrow angle BEC equals 102 to the power of degree space space and space space angle AEC equals space 78 to the power of degree space

Now space consider space increment ECD
EC space equals ED equals CD space space space space space space left square bracket Since space AB equals EC right square bracket
Therefore space increment ECD space is space an space equilateral space triangle.
rightwards double arrow angle ECD equals 60 to the power of degree space

angle BCD space equals angle BEC plus angle ECD
rightwards double arrow angle BCD space equals 102 to the power of degree plus 60 to the power of degree space
rightwards double arrow angle BCD space equals 162 to the power of degree

Therefore space space angle AEC equals space 78 to the power of degree space space and space angle BCD space equals 162 to the power of degree

Question 4

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

Solution 4

Question 5

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

(i)  AOB

(ii)  BPC

(iii)  PCD

(iv) Reflex  APC

Solution 5

 

In the given figure  is an equilateral triangle

Therefore all its angles are

Again in the

Again

triangle B P C
rightwards double arrow angle B P C equals 75 to the power of ring operator space left square bracket Since space BP space equals CB right square bracket

Now

angle C equals angle B C P plus angle P C D
rightwards double arrow angle P C D equals 90 to the power of ring operator minus 75 to the power of ring operator
rightwards double arrow angle P C D equals 15 to the power of ring operator

Therefore

angle A P C equals 60 to the power of ring operator plus 75 to the power of ring operator
rightwards double arrow angle A P C equals 135 to the power of ring operator
rightwards double arrow R e f l e x space angle A P D equals 360 to the power of ring operator minus 135 to the power of ring operator equals 225 to the power of ring operator


(i)

(ii)angle B P C equals 75 to the power of ring operator space

(iii)angle P C D equals 15 to the power of ring operator

(iv)Reflex angle A P D equals 225 to the power of ring operator

Question 6

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

If DEC is an equilateral triangle, calculate:

(i)  CBE(ii)  DBE.

Solution 6

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



text In   the   rhombus    We   have end text
angle A equals 67 to the power of degree equals angle C space space space left square bracket text Opposite   angles end text right square bracket
angle A plus angle D equals 180 to the power of degree left square bracket text Consecutive   angles   are   supplementary end text right square bracket
rightwards double arrow angle D equals 113 degree
rightwards double arrow angle A B C equals 113 degree

text Consider end text space triangle D B C comma
D C equals C B space left square bracket text Sides   of   rhombous ] end text
S o triangle D B C space text is   an   isoscales   triangle end text
rightwards double arrow angle C D B equals angle C B D
A l s o comma
angle C D B plus angle C D B plus angle B C D equals 180 to the power of degree
rightwards double arrow 2 angle C B D equals 113 to the power of degree
rightwards double arrow angle C D B equals angle C B D equals 56.5 to the power of degree............ left parenthesis i right parenthesis

text Consider end text space triangle D C E comma
E C equals C B
S o triangle D C E space text is   an   isoscales   triangle end text
rightwards double arrow angle C B E equals angle C E B
A l s o comma
angle C B E plus angle C E B plus angle B C E equals 180 to the power of degree
rightwards double arrow 2 angle C B E equals 53 to the power of degree
rightwards double arrow angle C D E equals 26.5 to the power of degree

F r o m space left parenthesis i right parenthesis
angle C B D equals 56.5 to the power of degree
rightwards double arrow angle C B E plus angle D B E equals 56.5 to the power of degree
rightwards double arrow 26.5 to the power of degree plus angle D B E equals 56.5 to the power of degree
rightwards double arrow angle D B E equals 30.5 to the power of degree

 

Question 7

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

(i)

(ii)


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

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

7 y equals 6 y plus 3 y minus 8 to the power of degree space space space space space space space end exponent left parenthesis i right parenthesis space space space left square bracket S i n c e space angle A equals angle C right square bracket
4 x plus 20 to the power of degree equals 0 space space space space space space space left parenthesis i i right parenthesis

Solving (i), (ii) we have

X equals 12 to the power of degree
Y equals 16 to the power of degree space space space space

Question 8

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

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

Question 9

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.

 

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

 

Question 10

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

 

 

 

 

 

 

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. 

 

 

 

 

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

Question 1

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.

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 AEFC is a parallelogram.

Question 2

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

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

Question 3

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.

Solution 3

Question 4

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.

Solution 4


Let us join PQ. 


text Consider   the end text space triangle A O Q space text and end text space triangle B O P
angle A O Q equals angle B O P space space space left square bracket text opposite   angles end text right square bracket
angle O A Q equals angle B P O space space space space left square bracket text alternate   angles end text right square bracket
rightwards double arrow triangle A O Q space text ≅ end text space triangle B O P space space left square bracket text AA   test end text right square bracket

H e n c e space A Q space equals B P

text Consider   the end text space triangle Q O P space text and end text space triangle A O B
angle A O B equals angle Q O P space space space left square bracket text opposite   angles end text right square bracket
angle O A B equals angle A P Q space space space space left square bracket text alternate   angles end text right square bracket
rightwards double arrow triangle Q O P space text ≅ end text space triangle A O B space space left square bracket text AA   test end text right square bracket

H e n c e space P Q space equals A B space equals C D

text Consider   the end text space text quadrilateral end text space Q P C D
D Q equals C P space a n d space D Q space vertical line vertical line space C P space left square bracket text Since end text space A D space equals B C space a n d space A D space vertical line vertical line B C space right square bracket
text Also end text space Q P equals D C space a n d space A B vertical line vertical line Q P vertical line vertical line D C

H e n c e space q u a d r i l a t e r a l space Q P C D space text is   a   parallelogram. end text

 

Question 5

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

Solution 5

Given  is a parallelogram

To prove:

 

Proof:  is a parallelogram

Again,

NOW

Hence proved

Question 6

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

Solution 6

Given ABCD is a parallelogram. The bisectors of  and meet at EThe bisectors of angle A B Cand  meet at F

From the parallelogram  we have

In triangle ECD sum of angles

Similarly taking triangle  it can be prove that


Now since

angle B F C equals angle C E D equals 90 to the power of degree

Therefore the lines  and BF are parallel

Hence proved

Question 7

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

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 4th one and so  is a rectangle ,since its interior angles are all right angles

Hence proved

Question 8

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

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

Question 9

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 = 90o.

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

Question 10

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

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

Question 11

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

(i) AP bisects angle A

(ii) BP bisects angle B

(iii)  DAP +  BCP =  APB

Solution 11

Consider  

Therefore

Hence proved

Question 12

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.

Solution 12

ABCD is a square and

C o n s i d e r triangle D A Q space a n d space triangle A B P
angle D A Q equals angle A B P equals 90 to the power of degree
space D Q space equals A P
A D equals A B
triangle D A Q space approximately equal to space triangle A B P
rightwards double arrow angle P A B equals angle Q D A

N o w comma
angle P A B plus angle A P B equals 90 to the power of degree
a l s o space angle Q D A plus angle A P B equals 90 to the power of degree space space left square bracket angle P A B equals angle Q D A right square bracket

C o n s i d e r space triangle A O Q space B y space A S P
angle Q D A plus angle A P B plus angle A O D equals 180 to the power of degree
rightwards double arrow 90 to the power of degree plus angle A O D equals 180 to the power of degree
rightwards double arrow angle A O D equals 90 to the power of degree

H e n c e space A P space a n d space D Q space a r e space p e r p e n d i c u l a r.

Question 13

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

(i) AC bisects angle BAD.

(ii) AC is perpendicular bisector of BD.

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

Question 14

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

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

Question 15

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

Solution 15

 

Question 16

In case of a parallelogram prove that:

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

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

Solution 16

 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

Question 17

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

Solution 17

  

Question 18

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

  

Solution 18

Question 19

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

Solution 19

  

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