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

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.

Sol:

Let us draw a parallelogram ABCD Where F is the midpoint Of side DC and E is the mid-point of side AB of a parallelogram  ABCD.

To prove:  AEFD is a parallelogram

Proof: 
In parallelogram ABCD
AB || DC
BC || AD
AB = DC
${\displaystyle \frac{1}{2}\text{AB}=\frac{1}{2}}$DC
AE = DF
Also AD || EF
Therefore, AEFD is a parallelogram.

Question 2

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

Sol:

Given: ABCD is a parallelogram where the diagonal BD bisects
parallelogram  ABCD at angle B and D

To Prove:  ABCD is a rhombus

Proof: Let us draw a parallelogram  ABCD where the diagonal BD bisects the parallelogram at an angle B and D.

Construction: Let us join AC as a diagonal of the parallelogram ABCD

Since ABCD  is a parallelogram
Therefore
AB = DC
AD =BC
Diagonal  BD bisects angle B and D
So ∠COD = ∠DOA
Again AC also bisects at A and C   

Therefore ∠AOB =∠ BOC
Thus ABCD 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.

Sol:

Construction : 
Join DF and EB
Join diagonal BD

Since diagonals of a parallelogram bisect each other.
∴  OA = OC and OB = OD
Also, AE = EF = FC

Now, OA = OC and AE = FC
⇒  OA - AE = OC - FC
⇒  OE = OF

Thus, in quadrilateral DEFB, bisect each other.
OB = OD and OE = OF
⇒  Diagonals of a quadrilateral DEFB bisect each other.
⇒ DEFB is a parallelogram.
⇒  DE is parallel to FB
⇒  DE = FB                      .....( Opposite sides are equal )

Question 4

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

(i) AQ = BP
(ii) PQ = CD.
(iii) ABPQ is a parallelogram.

Sol:

Let us join PQ. 
Consider the ΔAOQ and ΔBOP
∠AOQ = ∠BOP                  ..[ opposite angles ]
∠OAQ = ∠BPO                 ...[ alternate angles ]
⇒ ΔAOQ ≅ ΔBOP            ...[ AA test ]
Hence AQ = BP

Consider the ΔQOP and ΔAOB
∠AOB = ∠QOP                 ...[ opposite angles ]
∠OAB = ∠APQ                 ...[ alternate angles ]
⇒ ΔQOP ≅ ΔAOB              ...[ AA test ]
Hence PQ = AB = CD

Consider the quadrilateral QPCD
DQ = CP and DQ || CP ||       ...[ Since AAD = BC and AD || BC ]
Also QP = DC and AB || QP || DC

Hence Quadrilateral QPCD is a Parallelogram.

Question 5

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

Sol:

Given ABCD is a parallelogram
To prove: AB = 2BC

Proof:  ABCD is a parallelogram
A + D + B + C = 180°

From the AEB we have
⇒ ${\displaystyle \frac{\text{∠A}}{2}+\frac{\text{∠B}}{2}}$ + E = 180°
⇒ ∠A - ${\displaystyle \frac{\text{∠A}}{2}}$ + ∠D + ∠E1 = 180° ...[ taking E1 as new angle ]
⇒ ∠A + ∠D + ∠E1 = 180° + ${\displaystyle \frac{\text{∠A}}{2}}$ 
⇒ ∠E1 = ${\displaystyle \frac{\text{∠A}}{2}}$  ...[ Since ∠A + ∠D = 180°

Again,
similarly ,
∠E1 = ${\displaystyle \frac{\text{∠B}}{2}}$ 
Now
AB = DE + EC
= AD + BC
= 2BC                                ...[ since AD = BC]
Hence proved.

Question 6

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

Sol:

Given ABCD is a parallelogram. The bisectors of ∠ADC and ∠BCD meet at E. The bisectors of ∠ABC and ∠BCD meet at F

From the parallelogram ABCD we have

∠ADC + ∠BCD = 180° ...[ sum of adjacent angles of a parallalogram ]
⇒  ${\displaystyle \frac{\text{∠ADC}}{2}+\frac{\text{∠BCD}}{2}}$ = 90°
⇒ ∠EDC + ∠EDC = 90°

In triangle ECD sum of angles = 180°
⇒ ∠EDC + ∠ECD + ∠CED = 180°
= ∠CED = 90°

Similarly taking triangle BCF it can prove that ∠BFC = 90°
Now since
∠BFC = ∠CED = 90°

Therefore the lines DE and BF are parallel
Hence proved

Question 7

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

Sol:

Given:
ABCD is a parallelogram
AE bisects ∠BAD
BF bisects ∠ABC
CG bisects ∠BAD
DH bisects ∠ADC

To prove: LKJI is a rectangle

Proof :
∠BAD + ∠ABC = 180° ...[ adjacent angles of a parallelogram are supplementary ] 
∠BAJ = ${\displaystyle \frac{1}{2}}$ ∠BAD ...[AE bisects  BAD ]
∠ABJ = ${\displaystyle \frac{1}{2}}$ ∠ABC ... [DH  bisect ABC ]
∠BAJ + ∠ABJ = 90° ...[ halves of supplementary angles are complementary ]

ΔABJ is a right triangle because its acute interior angles are complementary.
Similarly
∠DLC = 90°
∠AID = 90°

Then ∠JIL = 90° because ∠AID and ∠JIL are vertical angles

since 3 angles of a quadrilateral, LKJI are right angles, si is the 4^th one and so is LKJI 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.

Sol:

Given: A parallelogram ABCD in which AR, BR, CP, DP 
Are the bisects of ∠A, ∠B, ∠C, ∠D, respectively forming quadrilaterals PQRS.

To prove: PQRS is a rectangle

Proof :
∠DCB + ∠ ABC =180° ...[ co - interior angles of parallelogram are supplementary ]

⇒ ${\displaystyle \frac{1}{2}}$ ∠DCB + ${\displaystyle \frac{1}{2}}$∠ABC = 90° 
⇒ ∠1 + ∠2 = 90° 
ΔCQB, ∠1 + ∠2 + ∠CQB = 180° 

From the above equation we get
∠CQB = 180° - 90° = 90° 
∠ RQP = 90°         ...[ ∠CQB = ∠ RQP , vertically opposite angles ]
∠QRP = ∠RSP = ∠SPQ = 90° 
So, PQRS is a square.
Hence Proved.

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 = 90^o.

Sol:

(i) Let AD = x
AB = 2AD = 2x
Also AP is the bisector ∠A
∠1 = ∠2
Now,
∠2 = ∠5                   ...[ alternate angles ]
Therefore ∠1 = ∠5
Now
AP = DP = x ...[ sides opposite to equal angles are also equal ]
Therefore
AB = CD ...[ opposite sides of  parallelogram are equal ]
CD = 2x
⇒ DP + PC = 2x
⇒ x + PC = 2x
⇒ PC = x
Also, BC = x
ΔBPC
⇒ ∠6 = ∠4 ...[ angles opposite to equal sides are equal ]
⇒ In ∠6 = ∠3
Therefore ∠3 =∠ 4
Hence BP bisect ∠B

(ii)
Opposite angles are supplementary
Therefore

∠1 + ∠2 + ∠3 + ∠4 = 180°
⇒ 2 ∠2 + 2 ∠3 =180°     .....[ ∠1 = ∠2 , ∠3 = ∠4 ]
⇒ ∠2 + ∠3 = 90°
ΔAPB
∠2 + ∠3 ∠APB = 180°
⇒ ∠APB = 180° - 90° ...[ by angle sum property ] 
⇒ ∠APB = 90° 
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.

Sol:

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

construction: Join B to D to meet AC in O.

Proof: We know that the diagonals of a parallelogram bisect each other.
Now, AC and BD bisect each other at O.
OC = OA
AM = CN
OA - AM = OC - CN
OM = ON

Thus in a quadrilateral BMDN, diagonal BD and MN are such that OM = ON and OD = OB

Therefore the diagonals AC and PQ bisect each other.
Hence  BMDN 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

Sol:

Consider ΔADP and ΔBCP,
AB = BC               ....[ Since ABCD is a parallelogram. ]
DC = AB               ....[ Since ABCD is a parallelogram. ]
∠A ≅ ∠C              ....[ Opposite angles ]
ΔADP ≅ ΔBCP     .....[ SAS ]

Therefore, AP = BP
AP bisects ∠A
BP bisects ∠B

In ΔAPB, AP = BP
AP bisects ∠A
BP bisects ∠B

In ΔAPB,
AP = PB
∠APB = ∠DAP + ∠BCP
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.

Sol:

ABCD is a square and AP = PQ.

Consider ΔDAQ and ΔABP,
∠DAQ = ∠ABP = 90°
DQ = AP 
AD = AB
ΔDAQ ≅ ΔABP
⇒  ∠PAB = ∠QDA

Now,
∠PAB + ∠APB = 90°
also ∠QDA + ∠APB = 90°      ....[ ∠PAB = ∠QDA ]

Consider ΔAOQ by ASP
∠QDA + ∠APB + ∠AOD = 180° 
⇒ 90° + ∠AOD = 180° 
⇒ ∠AOD = 90° 
Hence AP and DQ are perpendicular.

Question 13

In a quadrilateral ABCD, AB = AD and CB = CD.
Prove that :
(i) AC bisects angle BAD.
(ii) AC is the perpendicular bisector of BD.

Sol:

Given: ABCD is quadrilateral,
AB = AD
CB = CD

To prove:
(i) AC bisects angle BAD.
(ii) AC is the perpendicular bisector of BD.

Proof:
In ΔABC and ΔADC,
AB = AD                    ....(given)
CB = CD                   .....(given)
AC = AC                   ......(Common side)
ΔABC ≅ ΔADC         .......(SSS)

∠BAD = ∠DAO       .......(AC bisects A)

Therefore AC bisects ∠BAD
OD = OB
OA = OC                 ......( diagonals bisect each other at O )
Thus AC is perpendicular bisector of BD.
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.

Sol:

Given ABCD is a trapezium, AB || DC and AD = BC.

Prove that:
(i) ∠DAB = ∠CBA
(ii) ∠ADC = ∠BCD
(iii) AC = BD
(iv) OA = OB and OC = OD.

Proof: (i) Since AD || CE and transversal AE cuts them at A and E respectively.
Therefore, ∠A + ∠B = 180°
Since, AB || CD and AD || BC
Therefore, ABCD is a parallelogram.
∠A = ∠C
∠B = ∠D              ....[ Since ABCD is a parallelogram ]
Therefore,
∠DAB = ∠CBA
∠ADC = ∠BCD

In ΔABC and ΔBAD, we have
BC = AD                 ....( given )
AB = BA                 ....( Common )
∠A = ∠B                 ....( proved )
ΔABC ≅ ΔBAD        ....( SAS )  
ΔABC ≅ ΔBAD
Since, Therefore AC = BD....( Corresponding parts of congruent triangles are equal. )
OA = OB
Again OC = OD      ....( Since diagonals bisect each other at O )
Hence proved.

Question 15

In the given figure, AP is the bisector of ∠A and CQ is the bisector of ∠C of parallelogram ABCD.

Prove that APCQ is a parallelogram.

Sol:

Construction: Join AC

Proof:
∠BAP = ${\displaystyle \frac{1}{2}}$∠A             ...( AP is the bisector  of ∠A )

∠DCQ = ${\displaystyle \frac{1}{2}}$∠C            ...( CQ is the bisector of ∠C ) 
⇒ ∠BAP = ∠DCQ           ....(i)....[ ∠A = ∠R ( Opposite angles of a parallelogram.) ]
Now,
∠BAC = ∠DCA                ....(ii)....[ Alternate angles since AB || DC ]

Subtracting (ii) from (i), We get
∠BAP - ∠BAC = ∠DCQ - ∠DCA 
⇒ ∠CAP = ∠ACQ
⇒ AP || QC                    .....( Alternate angles are equal )
Similarly, PC || AQ.
Hence, APCQ is a parallelogram.

Question 16

In case of a parallelogram
prove that:
(i) The bisectors of any two adjacent angles intersect at 90^o.
(ii) The bisectors of the opposite angles are parallel to each other.

Sol:

ABCD is a parallelogram, the bisectors of ∠ADC and ∠BCD meet at a point E and the bisectors of ∠BCD and ∠ABC meet at F.

We have to prove that the ∠CED = 90° and ∠CFG = 90°

Proof: In the parallelogram ABCD
∠ADC + ∠BCD = 180°       ....[ sum of adjacent angles of a parallelogram ]

⇒ ${\displaystyle \frac{\text{∠ADC}}{2}+\frac{\text{∠BCD}}{2}}$ = 90°

⇒ ∠EDC + ∠ECD + ∠CED = 180°
⇒ ∠CED = 90°

Similarly taking triangle BCF it can be proved that ∠BFC = 90°
∠BFC + ∠CFG = 180°               ....[ adjacent angles on a line ]
Also ⇒ ∠CFG = 90°
Now since ∠CFG = ∠CED = 90° ....[ It means that the lines DE and BG are parallel ]
Hence proved.

Question 17

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

Sol:

To prove: ABCD is a square,
that is, to prove that sides of the quadrilateral are equal
and each angle of the quadrilateral is 90°,
ABCD is a rectangle,
⇒ ∠A = ∠B = ∠c = ∠D = 90° and diagonals bisect each other.

that is, MD = BM                     ..(i)
Consider ΔAMD and ΔAMB,
MD =  BM                               ....( from(i) )
∠AMD = ∠AMB = 90°             .....(given)
AM = AM                                ......( common side )
ΔAMD ≅ ΔAMB                      ....(SAS congruence criterion)
⇒ AD = AB                                 ...( c.p.c.t.c. )
Since ABCD is a rectangle, AD = BC and AB = CD
Thus, AB = BC = CD = AD and ∠A = ∠B = ∠C = ∠D = 90°
⇒ ABCD is a square.

Question 18

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

Sol:

ABCD is a parallelogram.
⇒ Opposite angles of a parallelogram are congruent.
⇒ ∠DAB = ∠BCD and ∠ABC = ∠ADC = 120°
In ABCD,
∠DAB + ∠BCD + ∠ABC + ∠ADC = 360°   ....( sum of the measures of angles of a quadrilateral )
⇒ ∠BCD + ∠BCD + 120° + 120° = 360°
⇒  2∠BCD = 360° - 240°
⇒  2∠BCD = 120°
⇒  ∠BCD = 60°
PQRS is parallelogram.
⇒ ∠PQR = ∠PSR = 70°
In ΔCMS,
∠CMS + ∠CSM + ∠MCS = 180°      ....( angle sum property )
⇒ x + 70° + 60°  = 180° 
⇒ x  = 50°

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

Sol:

ABCD is a rhombus.
⇒ AD = CD and ∠ADC = ∠ABC = 56°
DCFE is a square.
⇒ ED = CD and ∠FED = ∠EDC = ∠DCF = ∠CFE = 90°
⇒ AD = CD = ED
In ΔADE,
AD = ED
⇒ ∠DAE = ∠AED                ...(i)
∠DAE + ∠AED + ∠ADE = 180°
⇒ 2∠DAE + 146° = 180°            ....( Since ∠ADE = ∠EDC + ∠ADC = 90° + 56° = 146° )
⇒ 2∠DAE = 34°
⇒ ∠DAE = 17°
⇒ ∠DEA = 17°                   ....(ii)

In ABCD,
∠ABC + ∠BCD + ∠ADC + ∠DAB = 360°
⇒ 56° + 56° + 2 ∠DAB = 360°   ....( ∵ Opposite angles of a rhombus are equal.)
⇒ 2∠DAB = 248°
⇒ ∠DAB = 124°
We know that diagonals of a rhombus, bisect its angles.

⇒ ∠DAC = ${\displaystyle \frac{124°}{2}}$ = 62°

⇒ ∠EAC = ∠DAC - ∠DAE = 62° - 17° = 45°
Now,
∠FEA = ∠FED - ∠DEA  
          = 90° - 17°             ....( From(ii) and each angle of a square is 90° )
        = 73°       
We know that diagonals of a square bisect its angles.
⇒ ∠CED = ${\displaystyle \frac{90°}{2}}$ = 45°
So,
∠AEC = ∠CED - ∠DEA
          = 45° - 17°
          = 28°
Hence, ∠DAE = 17°, ∠FEA = 73°, ∠EAC = 45° and ∠AEC = 28°.           

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

---

Question 0.3

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

Sol: False

---

Question 0.4

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

Sol: True

---

Question 0.5

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

Sol: False

---

Question 0.6

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

Sol: True

---

Question 0.7

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

Sol: False

---

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

---

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.

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

Question 1

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

Sol:

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
( 2n - 4 ) x 90° = 1440°
2n - 4 = 16
2n = 20
n = 10
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.

Sol:

Let the angles of the pentagon are 4x, 8x, 6x, 4x and 5x.
Thus we can write 
4x + 8x+ 6x + 4x +5x = 540°
                             27x =540°
                                 x = 20°

Hence the angles of the pentagon are:
4 × 20° = 80° ,
8 × 20°= 160° ,
6 × 20°= 120° ,
4 × 20°= 80° ,
5 × 20°= 100° 

Question 3

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

Sol:

Let the measure of each equal angles are x.
Then we can write
140° + 5x = ( 2 x 6 - 4 ) x 90°
140° + 5x = 720°
           5x = 580°
             x = 116°
Therefore the measure of each equal angles are 116°

Question 4

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.

Sol:

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

Therefore we can write :
5 x 90° + k  x 195° = ( 2n - 4 )90°
180°n - 195° k = 450 - 360°
180°n - 195° k = 90°
   12n - 13k = 6

In this linear equation n and k must be an integer. Therefore to satisfy this equation the minimum value of k must be 6 to get n as an integer.
Hence the number of sides are: 5 + 6 = 11.

Question 5

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.

Sol:

Let the measure of each equal angles are x.

Then we can write:
3 x 132° + 4 x = ( 2 x 7 - 4 ) 90°
4 x = 900° - 369°
4 x = 504
x = 126°
Thus the measure of each equal angles are 126°.

Question 6

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. 

Sol:

Let the measure of each equal sides of the polygon is x.
Then we can write:
142° + 176° + 6 x = ( 2 x 8 - 4 ) 90°
6 x = 1080° - 318°
6 x = 762°
x = 127°
Thus the measure of each equal angles are 127°.

Question 7

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

Sol:

Let the measure of the angles are 3x, 4x and 5x.
Thus
∠A + ∠B + ∠C + ∠D + ∠E =540°
3 x + ( ∠B + ∠C ) + 4 x + 5 x = 540°
                           12x + 180° = 540°
                                      12x = 360°
                                          x = 30°
Thus the measure of angle E will be 4 × 30° = 120°

Question 8

AB, BC, and CD are the three consecutive sides of a regular polygon. If BAC = 15°;
find,
(i) Each interior angle of the polygon.
(ii) Each exterior angle of the polygon.
(iii) The number of sides of the polygon.

Sol:

(i) Let each angle of measure x degree.
Therefore the measure of each angle will be :
x = 180° - 2 x 15° = 150°

(ii) Let each angle of measure x degree.
Therefore the measure of each exterior angle will be :
x = 180° - 150° = 30°

(iii) Let the number of each side is n.
Now we can write
n.150° = ( 2n - 4 ) x 90°
180°n - 150°n = 360°
30°n = 360°
n = 12.
Thus the number of sides is 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.

Sol:

Let the measure each interior and exterior angles are 3k and 2k.
Let the number of sides of the polygon is n.
Now we can write:
n.3k = ( 2n - 4 ) x 90°
3nk = ( 2n - 4 ) 90°                  ....(1)
Again
n.2k = 360°
nk = 180°
From (1)
3.180° = ( 2n - 4 ) 90°
3 = n - 2 
n = 5
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 6° find the value of n.

Sol:

For (n-1) sided regular polygon:
Let the measure of each angle is x.
Therefore,
( n - 1 )x = [ 2( n - 1 ) - 4 ] 90°

x = ${\displaystyle \frac{n-3}{n-1}}$ 180°

For (n+1) sided regular polygon:
Let measure of each angle is y.
Therefore
( n + 2 )y = [ 2( n + 2 ) - 4 ] 90°

y = ${\displaystyle \frac{n}{n+2}}$ 180°

Now we have
y - x = 6°

${\displaystyle \frac{n}{n+2}180°-\frac{n-3}{n-1}180°}$ = 6°

${\displaystyle \frac{n}{n+2}-\frac{n-3}{n-1}=\frac{1}{30}}$

30n( n - 1 ) - 30( n - 3 )( n + 2 ) = ( n + 2 )( n - 1 )
- 30n + 30n + 180 = n² + n - 2
n² + n - 182 = 0
( n - 13 )( n + 14 ) = 0
n = 13, - 14
Thus the value of n is 13.

Question 11

Two alternate sides of a regular polygon, when produced, meet at the right angle.
Find:
(i)The value of each exterior angle of the polygon;
(ii) The number of sides in the polygon.

Sol:

(i) Let the measure of each exterior angle is x and the number of sides is n.
Therefore we can write :
n = ${\displaystyle \frac{360°}{x}}$
Now We have
x + x + 90° = 180°
2x = 90°
x = 45°

(ii) Thus the number of sides in the polygon is :
n = ${\displaystyle \frac{360°}{45°}}$ 
n = 8.