SELINA Solution Class 9 Chapter 14 Rectilinear Figures (Quadrilaterals: Parallelogram , Rectangle , Rhombus, Square and Trapezium) Exercise 14B

Question 0.1

State, 'true' or 'false'
The diagonals of a rectangle bisect each other.

Sol: True

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Question 0.2

State, 'true' or 'false'
The diagonals of a quadrilateral bisect each other.

Sol: False

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Question 0.3

State, 'true' or 'false'
The diagonals of a parallelogram bisect each other at right angle.

Sol: False

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Question 0.4

State, 'true' or 'false'
Each diagonal of a rhombus bisects it.

Sol: True

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Question 0.5

State, 'true' or 'false'
The quadrilateral, whose four sides are equal, is a square.

Sol: False

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Question 0.6

State, 'true' or 'false'
Every rhombus is a parallelogram.

Sol: True

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Question 0.7

State, 'true' or 'false' 
Every parallelogram is a rhombus.

Sol: False

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Question 0.8

State, 'true' or 'false'
 Diagonals of a rhombus are equal.

Sol: False 

 Question 0.9

State, 'true' or 'false' 

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

Sol: True

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Question 1.1

State, 'true' or 'false'
 If the diagonals of a quadrilateral bisect each other at right angle, the quadrilateral is a square.

Sol: False

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Question 2

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

Sol:

From the given figure we conclude that
∠A + ∠D = 180°  ...[Since consecutive angles are supplementary ]

${\displaystyle \frac{\angle A}{2}+\frac{\angle D}{2}}$= 90°

Again from the ΔADM

${\displaystyle \frac{\angle A}{2}+\frac{\angle D}{2}}$ + ∠M = 180°   
⇒ 90° + ∠M = 180°     ...${\displaystyle [\sin ce\frac{\angle A}{2}+\frac{\angle D}{2}=90°]}$
⇒ ∠M = 90°
Hence ∠AMD = 90°

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 102^o. Find angles AEC and BCD.

Sol:

In the given figure

Given that AE =BC
We have to find ∠AEC  ∠BCD
Let us Join EC and BD.

In the quadrilateral AECB
AE = BC and AB = EC
also AE || BC
⇒ AB || EC
So quadrilateral is a parallelogram

In parallelogram consecutive angles are supplementary
⇒  ∠A + ∠B = 180°
⇒  102° + ∠B = 180°
⇒  ∠B = 78°
In parallelogram  opposite angles are equal
⇒ ∠A = ∠BEC and ∠B = ∠AEC
⇒ ∠BEC = 102° and ∠AEC = 78°

Now consider ΔECD
EC = ED = CD [SInce AB = EC ]

Therefore ΔEDC is an equilateral triangle.
⇒ ∠ECD = 60°
∠BCD = ∠BEC + ∠ECD
⇒ ∠BCD = 102° + 60°
⇒ ∠BCD = 162°
Therefore ∠AEC = 78° and ∠BCD = 162°

Question 4

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

Show that: 

(i) ∠POC =  ${\displaystyle \left[22\frac{1°}{2}\right]}$

(ii) ∠BDC = 2 ∠POC

(iii) ∠BOP = 3 ∠CPO

SOl:

Let ∠POC = x°
Diagonals of a square bisect the angles.
∴  ∠OCP = ∠OBP = ${\displaystyle \frac{90°}{2}}$ = 45°
Using exterior angle property for ΔOPC,
∠OPB = ∠OCP + ∠POC
⇒ ∠OPB = 45° + x°           ....(i)

In ΔOBP,
OB = BP                ...(Given)
∴ ∠OPB = ∠BOP    ..( angles opposite to equal sides are equal )
⇒ ∠BOP = 45° + x  ...(ii) ...[ From (i) ]

Diagonals of a Square are perpendicular to  each other.
∴ ∠BOP + ∠POC = 90°
⇒ 45 + x + x = 90°
⇒ X = 22.5°
⇒ ∠POC = x° = ${\displaystyle \left[22\frac{1°}{2}\right]}$

Now ⇒ ∠BDC = 45°   .....( Diagonals of a square bisects the angles )
⇒ ∠BCD = 2 x 22.5° = 2∠POC

And ,∠BOP = 45° + x°      ...[ From (ii) ]
⇒ ∠BOP = 45° + 22.5° = 3 x 22.5° = 3∠POC

Question 5

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

Calculate: (i) ∠AOB
                (ii) ∠BPC
                (iii) ∠PCD
                (iv) Reflex ∠APC

Sol:

In the given figure ΔAPB is an equilateral triangle.
Therefore all its angles are 60°
Again in the
ΔADB,
∠ABD = 45°
∠AOB = 180° - 60° - 45° = 75°

Again
ΔBPC
⇒ ∠BPC = 75°              ....[ Since BP = CB ]
Now,
∠C = ∠BCP + ∠PCD 
⇒ ∠PCD = 90° - 75°
⇒ ∠PCD = 15°
Therefore,
∠APC = 60° + 75°
⇒ ∠APC = 135°
⇒ Reflex ∠APD = 360° - 135° = 225°

(i) ∠AOB = 75°
(ii) ∠BPC = 75°
(iii) ∠PCD = 15°
(iv) Reflex ∠APC = 135°.  Reflex ∠APD = 225°

Question 6

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

If DEC is an equilateral triangle, calculate:
(i) ∠CBE
(ii) ∠DBE.

Sol:

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

In the rhombus we have
∠A = 67° = ∠C                 ....[ Opposite angles ]
∠A + ∠D = 180°               ....[ Consecutive angles are supplementary. ]
⇒ ∠D = 113°
⇒ ∠ABC = 113°

Consider ΔDBC,
DC = CB                          ....[ Sides of rhombous ]
So, ΔDBC is an isoscales triangle,
⇒ ∠CDB = ∠CBD
Also,
∠CDB + ∠CBD + ∠BCD = 180°
⇒ 2∠CBD = 113°
⇒ ∠CDB = ∠CBD = 56.5°    ....(i)

Consider ΔDCE,
EC = CB
So ΔDCE is an isoscales triangle
⇒  ∠CBE = ∠CEB
Also,
∠CBE + ∠CEB + ∠BCE = 180°
⇒ 2∠CBE = 53°
⇒ ∠CDE = 26.5°

From (i)
∠CBD = 56.5°
⇒ ∠CBE + ∠DBE = 56.5°
⇒ 26.5° + ∠DBE = 56.5°
⇒ ∠DBE = 56.5° - 26.5° = 30°

Question 7.1

In the following figures, ABCD is a parallelogram.

find the values of x and y.

SoL:

ABCD is a parallelogram.
Therefore
AD = BC
AB = DC
Thus
4y = 3x - 3                ....[ Since AD = BC ]
⇒ 3x - 4y = 3            ....(i)
6y + 2 = 4x               ....[ Since AB = DC ]
⇒  4x - 6y = 2           ....(ii)

Solving equations (i) and (ii) we have
x = 5
y = 3.

Question 7.2

In the following figures, ABCD is a parallelogram.

find the values of x and y.

SOl:

In the figure, ABCD is a parallelogram
∠A = ∠C 
∠B = ∠D              .....[ Since opposite angles are equal. ]
Therefore,
7y = 6y + 3y - 8°       ...(i)( Since ∠A = ∠C )
4x + 20° = 0              ...(ii)

Solving (i), (ii) we have
x = 12° 
Y = 16°

Question 8

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

Sol:

Given that the angles of a quadrilateral are in the ratio 3:4:5:6 
Let the angles be 3x, 4x, 5x, 6x.
3x + 4x + 5x + 6x = 360°
⇒ x = ${\displaystyle \frac{360°}{18}}$

⇒ x =  20°
Therefore the angles are
3 x 20 = 60°
4 x 20 = 80°
5 x 20 = 100°
6 x 20 = 120°
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.

Sol:

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 the vertex of obtuse angle A as shown.
If  ∠x: ∠y = 2: 1.
find angles of the parallelogram.

Sol:

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 a parallelogram are 120, 60, 120 and 60.