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

Ex. 14(A)

Ex. 14(B)

Ex. 14(C)

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

Solution 3

Let the measure of each equal angles are x.

Then we can write

Therefore the measure of each equal angles are 116 $degree$

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 $degree$ =120 $degree$

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 = 67^o

$text In the rhombus We have end text angle A equals 67 to the power of degree equals angle C space space space left square bracket text Opposite angles end text right square bracket angle A plus angle D equals 180 to the power of degree left square bracket text Consecutive angles are supplementary end text right square bracket rightwards double arrow angle D equals 113 degree rightwards double arrow angle A B C equals 113 degree text Consider end text space triangle D B C comma D C equals C B space left square bracket text Sides of rhombous] end text S o triangle D B C space text is an isoscales triangle end text rightwards double arrow angle C D B equals angle C B D A l s o comma angle C D B plus angle C D B plus angle B C D equals 180 to the power of degree rightwards double arrow 2 angle C B D equals 113 to the power of degree rightwards double arrow angle C D B equals angle C B D equals 56.5 to the power of degree............ left parenthesis i right parenthesis text Consider end text space triangle D C E comma E C equals C B S o triangle D C E space text is an isoscales triangle end text rightwards double arrow angle C B E equals angle C E B A l s o comma angle C B E plus angle C E B plus angle B C E equals 180 to the power of degree rightwards double arrow 2 angle C B E equals 53 to the power of degree rightwards double arrow angle C B E equals 26.5 to the power of degree F r o m space left parenthesis i right parenthesis angle C B D equals 56.5 to the power of degree rightwards double arrow angle C B E plus angle D B E equals 56.5 to the power of degree rightwards double arrow 26.5 to the power of degree plus angle D B E equals 56.5 to the power of degree rightwards double arrow angle D B E equals 30.5 to the power of degree$

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

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 E. The bisectors of $angle A B C$ 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

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

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 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 ,

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

$C o n s i d e r triangle D A Q space a n d space triangle A B P angle D A Q equals angle A B P equals 90 to the power of degree space D Q space equals A P A D equals A B triangle D A Q space approximately equal to space triangle A B P rightwards double arrow angle P A B equals angle Q D A N o w comma angle P A B plus angle A P B equals 90 to the power of degree a l s o space angle Q D A plus angle A P B equals 90 to the power of degree space space left square bracket angle P A B equals angle Q D A right square bracket C o n s i d e r space triangle A O Q space B y space A S P angle Q D A plus angle A P B plus angle A O D equals 180 to the power of degree rightwards double arrow 90 to the power of degree plus angle A O D equals 180 to the power of degree rightwards double arrow angle A O D equals 90 to the power of degree H e n c e space A P space a n d space D Q space a r e space p e r p e n d i c u l a r.$