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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:

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

Solution 3

Let the measure of each equal angles are x.

Then we can write

Therefore the measure of each equal angles are 116degree

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.

Solution 5

Let the measure of each equal angles are x.

Then we can write:

Thus the measure of each equal angles are 126o.

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.

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

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

 

 

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

Solution 4

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 C equals 225 to the power of ring operator

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 B 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

 

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 x 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 plus 7 y equals 180 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

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

begin mathsize 11px style Thus comma space in space quadrilateral space DEBF space we space have
OB space equals space OD space and space OE space equals space OF
rightwards double arrow Diagonals space of space straight a space quadrilateral space bisect space each space other.
rightwards double arrow DEBF space is space straight a space parallelogram.
rightwards double arrow DE space is space parallel space to space FB
rightwards double arrow DE space minus space FB space left parenthesis Opposite space sides space are space equal right parenthesis end style

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 A B P Q
A Q equals B P space a n d space A Q space vertical line vertical line space B 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 A B 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 A B P Q space text is a parallelogram. end text

 

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 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

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

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 = 180o

Now as AP bisects angle A and BP bisects angle B

2∠DAP + 2∠CBP = 180o … (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

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.

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 90o.

 

(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

  

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