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.

SChand CLASS 9 Chapter 11 Rectilinear Figures TEST

  TEST

 Question 1

Sol : 

The given figure is trapezium if. A pair of two opposite side are parallel (b) option (b) is right 

 Question 2

Sol: If The given Figure is of a parallelogram. so Their opposite sides are equal and parallel

$\begin{aligned} \Rightarrow & x=3 \\ & 2 x+1=3 x-2 \Rightarrow 3 x-2 x=1+2 \\ \Rightarrow & x=3 \\ & \text { so } x=3 \end{aligned}$

option(e) is right 

 Question 3

Sol: In figure , ABCD is a rectangle whose diagonals bisect each other At O.

$\triangle A B D$ is a rignt triangle as $\angle A=90^{\circ}$. 

option (b) is rigut

 Question 4

(i) In figure ABCD is a parallelogram 

So Their opposite side are equal and parallel and opposite angle are equal 

$x+6=5 x-8 \Rightarrow 5 x-x=6+8$

$\Rightarrow 4 x=14 \Rightarrow x=\frac{14}{4}=3.5$

$14 y+6 y=180$

$20 y=180^{\circ} \Rightarrow y=\frac{180^{\circ}}{20^{\circ}}=9$

$6 y=a \Rightarrow 6 \times 9=a \Rightarrow 9=54^{\circ}$

$b=14 y=14 \times 9=126^{\circ}$

So $a=54, b=126^{\circ}$

(ii) P Q R S is a rectangle

(IMAGE TO BE ADDED)

opposite sides are equal and parallel 

Diagonals bisect Each other.

$c=330^{\circ}$

Similarly a=e 

e$=<1$

and $\angle L+33^{\circ}=90^{\circ}$

$\Rightarrow \angle L=90^{\circ}-33^{\circ}=57^{\circ}$

So $\begin{aligned} e=\angle 1 &=57^{\circ} \text { and } a=e=57^{\circ} \\ \text { if } O P &=O Q \\ \text { so } c &=\angle 2=33^{\circ} \\ \text { so } d &=180^{\circ}-c-\angle 2=180^{\circ}-30^{\circ}-35^{\circ} \\ &=180-66^{\circ}=114^{\circ} \\ & \text { and } b+d=180^{\circ} \\ \Rightarrow b &=180^{\circ}-d=180^{\circ}-114^{\circ}=66^{\circ} \end{aligned}$

Hence 

$a=57^{\circ}, b=66^{\circ}, c=33^{\circ}, d=194^{\circ}, e=57^{\circ}$

(iii) MNPQ is a rhombus in which all sides are equal and diagonals bisect each other at 90 and diagonals bisects opposite angles 

$b=53^{\circ}$

$a=d$

$e=53^{\circ}$

but $e+d=90^{\circ}$

$53^{\circ}+d=90^{\circ} \Rightarrow d=90^{\circ}-53^{\circ}=37^{\circ}$

$a=d=37^{\circ}$

$L c=90^{\circ}$

$a=37^{\circ}, b=53^{\circ}, \angle c=90^{\circ}, d=37^{\circ}, e=53^{\circ}$

 Question 5

Sol: In the figure ABCD is a kite in which BC = CD , AB = AD and diagonals AC and BC intersect at right angle at O and AC bisects the opposite angles 

$\angle O B C=58^{\circ}, \angle D A B=50^{\circ}$

(IMAGE TO BE ADDED)

(i) In $\triangle B C D \cdot B C=C O$

so $\angle O B C=\angle O D C=58^{\circ}$

So $\angle B C U=180^{\circ}-\angle O B C-\angle O D C$

$=180^{\circ}-58^{\circ}-58^{\circ}=180^{\circ}-116^{\circ}=64^{\circ}$

(ii) $\angle D A O=\frac{1}{2} \angle B A D=\frac{1}{2} \times 50^{\circ}=25^{\circ}$

(iii) $\angle O D A=90^{\circ}-\angle O A O=90^{\circ}-25^{\circ}=65^{\circ}$

(iv) $\angle A D C=\angle O O A+\angle O D C$

$=65^{\circ}+58^{\circ}=123^{\circ}$

 Question 6

Sol: In $A B C, P C$ and $Q R$ are mid segments $A B=B C$

So $P Q \| B C$ and $P Q=\frac{1}{2} \quad B C=B R$.............(i)

But AB =BC and PR Their mid points 

So PB = BR ...........(ii)

And QR||AB and QR =$\frac{1}{2} A B=P B$..........(ii)

and $Q R \| A B$ and $A R=\frac{1}{2} A B=P B$..........(iii)

From (i), (ii) and (iii)

$P B=B R=Q R=P Q$

SQ BPQR is a rhombus or a square 

But $\angle B \neq 90^{\circ}$

So $B P Q R$ is a rhombus.

SChand CLASS 9 Chapter 11 Rectilinear Figures Exercise 11(D)

 Exercise 11 D

Question 1

Sol: (i) Draw a line segment BC=5.2cm

 (ii) At B draw a ray BX making an angle of 45 and cut off BA=6.5cm

(iii) With center C and radius 6.5cm and with center A and radius 5.2 cm, draw arcs intersecting each other at D.

 (iv) Join AD and CD 

ABCD is the required ||gm

(IMAGE TO BE ADDED)

Question 2

Sol:(i)Draw diagonal BD =6.8 cm and bisect it at O.

(IMAGE TO BE ADDED)

(ii) From O, Draw a ray OX making an angle of 60 and  product 

(iii) Cut of OA = OC =4.4\2= 2.2cm

(iv) Join AB, BC , CD and DA 

ABCD is the required ||gm

Question 3

 Sol:(i) Draw a line segment AB=6cm

(ii) At A , draw a perpendicular AX and Cut off AP =2.9cm

(iii) From P, and EPF ||AB

(iv) With center B and A and radius 4cm draw arcs cutting EF at D and C.

(v) Join AD and BC.

Then ABCD is the required ||gm

 (IMAGE TO BE ADDED)

Question 4

Sol:(i) Draw a line segment AB=4.3cm

(ii) At B draw a ray BY. 

(iii) At A, draw perpendicular AX and Cut off AE =3cm

(iv) From E draw a parallel line to AB which Intersects By at C.

(v)With center A and radius BC, draw an arc cutting EF at D.

(vi) Join AD.

Then ABCD is the required ||gm

 (IMAGE TO BE ADDED)

 

Question 5

Sol: (i)Draw a line segment $A B=6 \mathrm{~cm}$

(ii) At A draw a ray AX making an angle of $45^{\circ}$ and cut off $A D=3 \mathrm{~cm}$.

(iii)With center B and Radius 3cm and with center D and Radius 6cm, Draw arcs intersecting each other at C.

(iv) Join BC and DC 

ABCD is the required ||gm

(v) Draw the angle bisector of A meeting CD at P and join BP

 (IMAGE TO BE ADDED)

To prove: $\angle A P B=90^{\circ}$

proof: In $\triangle A D B$

$\angle D P A=\angle P A B$

But $\angle D A P=\angle P A B$

$\angle D A P=\angle B P A=\frac{45^{\circ}}{2}=22 \frac{1}{2}^{\circ}$

$A D=D O=3 \mathrm{~cm}$

Similarly in $\triangle B C P$

${ }_{P C}=D C-D P=6-3=3 \mathrm{~cm}$

So $P C=C B=3 \mathrm{~cm}$

So $\angle C P B=\angle C B P$

But $\angle C P B+\angle C B P=180^{\circ}-\angle B C P$

$=180^{\circ}-45^{\circ}=135^{\circ}$

$\angle C P B=\angle C B P=\frac{139^{\circ}}{2}=67 \frac{1}{2}^{\circ}$

But $\angle D P A+\angle C P B+\angle A P B=180^{\circ}$

$\Rightarrow 22 \frac{1}{2}+67 \frac{1}{2}^{\circ}+\angle A P B=180^{\circ}$

$=90^{\circ}+\angle A P B=180^{\circ}$

$\Rightarrow \angle A P B=180^{\circ}-90^{\circ}=90^{\circ}$

Hence $\angle A P Q=90^{\circ}$

Question 6

Sol: The diagonal AC and BD of a ||gm ABCD bisect each other 

$A O=O C=\frac{10}{2}=5 \mathrm{~cm}$

and $B O=O D=\frac{6}{2}=3 \mathrm{~cm}$

(i) Draw a line segment AB =4cm 

(ii) With center a and radius 5cm and with center B and radius 3cm draw which intersect each other at O. 

(iii) Join OA and OC.

(iv) Produce AO to C such that Ao=OC  and produce BO to B such that BO=OD 

(v) Join AD, DC ,CB

Then ABCD is the required ||gm

Question 7

Sol: (i) Draw a line segment AC =10cm and bisect it at O.

(ii) AT O, Draw a ray OX making an angle of 60 and produce XO to Y. 

(iii) From O, cut off OD = OB = $\frac{6}{2}=3 \mathrm{~cm}$

(iv) Join AB,BC , CD and DA. 

ABCD is the required ||gm

Whose longer side is BC or AD on measuring it , it is 6cm long. 

(IMAGE TO BE ADDED)

Question 8

Sol;(i) Draw a line AX.

(ii) AT A, Draw a perpendicular and cut off AL = 3cm 

(iii) Through L draw LY parallel to AX.

(iv) From A, Draw a ray making an angle of 60 which meets LY at D. 

(v) AT AD. draw a perpendicular and cut off AM = 4cm

(vi) Through M, draw a line parallel to AD which Meets LY at C. 

Then ABCD is the required parallelogram 

(IMAGE TO BE ADDED)

Question 9

Sol: If diagonal of a ||gm bisect each other 

AO=OC and OB =OD = 2.5cm

Steps of construction :

(i) Draw a line segment AC=7cm and bisect it at O. 

(ii) AT O. draw a line making an angle of 75 and produce it to both sides

(iii) From O, Cut off OB =OD $=\frac{5}{2}=2.5 \mathrm{~cm} .$ 

(iv) Join AB, BC, CD and DA 

ABCD is the required ||gm

Measuring its shorter side AD=4cm 

(IMAGE TO BE ADDED)

Question 10

Sol: (i) Draw a line segment AB = 3.6cm

(ii) AT, A draw a ray AX making an angle of 45 and cut off AB=3.6cm

(iii)With center B and D and radius 3.6cm draw arcs intersecting each other at C.

(iv) Join BC and CD 

Then ABCD is the required rhombus 

(IMAGE TO BE ADDED)

Question 11

Sol: (i) Draw a line segment AB =3.2cm

(ii) AT A and B draw perpendiculars AX and BY and Cut off AD = BC =3.2cm

(iii) Join DC 

ABCD is the required Square 

(iv) Join its diagonals AC and BD 

On Measuring each =4.5 cm and each angles between these diagonals is 90 $\angle A O B=90^{\circ}$

(IMAGE TO BE ADDED)

Question 12

Sol: We know that diagonals of a square bisect each other at right angles and diagonals are also equal 

Steps of construction: 

(i) Draw a line segment PR =5cm and draw its perpendicular bisects XY intersecting it at O.

(ii)Cut off OQ = OS =$\frac{5}{2}=2.5 \mathrm{~cm}$

(iii) Join PQ, QR, RS and SP respectively. 

Then PQRS is the required square on measuring its side PQ which is = 3.6cm

(IMAGE TO BE ADDED)

Question 13

Sol: Steps of construction: 

(i) Draw a line segment AB = 4.2 cm 

(ii)AT A and B, Draw perpendicular AX and BY 

(iii) Join CD

Then ABCD is the required rectangle 

(IMAGE TO BE ADDED)

Question 14

Sol: Diagonals of a rectangle bisect each other and arc equal in length 

Steps of Construction: 

(i) Draw a line segment AC = 6cm

(ii) Bisect AC and O. 

(iii) From O, draw a line XY 

Making an angle of 37 and produce it to both side of O.

(iv) From XY Cut off OC =OD = 3cm

 (v) Join AB, BC , CD and AD,

Then ABCD is the required rectangle.

(IMAGE TO BE ADDED)

Question 15

Sol: The diagonals of a square bisect each other at right angles and arc also equal in length. 

Steps of Construction: 

(i) Draw a line segment AC =5cm

(ii) Draw its perpendicular bisector XY intersecting at O. 

(iii) Cut off OB =OD =2.5CM

(iv) Join AB, BC, CD and DA respectively 

Then ABCD is the required square

On measuring its side it is 3.6 cm each 

(IMAGE TO BE ADDED)

Question 16

Sol: Diagonals of rectangle bisect each other and are equal to each other in length.

Steps of constructions:

(i) draw a line segment $A C=6 \mathrm{~cm}$ and bisect it at O .

(ii)At O , draw a line XY making an angle of $45^{\circ}$ and produce it to both of O .

(iii) cut off $O B=O D=\frac{6}{2}=3 \mathrm{~cm}$.

(iv)Join AB, BC, CD and DA respectively

Then ABCD is the required rectangle. 

(IMAGE TO BE ADDED)

Question 17

SoL: Diagonals of a rhombus bisect each other at right angles.

Steps of constructions:

(i) Draw a line segment $P R=8 \mathrm{~cm}$

(ii) Draw its perpendicular bisector XY which intersects $P R$ at 0 .

(iii) from XY act off $O Q=O S=\frac{6}{2}=3 \mathrm{~cm}$.

(iv) Join $P Q, Q R, R S$ and SP

Then PQRS is the required rhombus 

(IMAGE TO BE ADDED)

(v) if the required point X is equidistant rom $p$ and PS.

So it lies on the angle bisectors of LQPS which is the diagonal PR

it is equidistant from $R$ and S

 Then it will be on the perpendicular bisector of ks Now draw perpendicular bisector of RS which intersects PR at X

Then X is the required point 

Construction the trapezium PQRS in which PQ is parallel to RS from the given measurement in problem 18 to 20 without using set square and protractor as far as possible

Question 18

Sol:  Steps of constructions:

(i) Draw a line segment $P Q=7 \mathrm{~cm}$.

(ii) Take a point E such that AE =SR = 2.5 cm and then EQ = 7 -2.5 =4.5cm

(iii) With center E and radius 3cm and with center Q and radius 3.5cm Draw arcs which intersect each other at R 

(iv)Join $E R$ and $Q R$.

(v)with conter $R$ and radius $2.5 \mathrm{~cm}$ and with center $P$ and radius $3 \mathrm{~cm}$ draw arcs intersecting each other at S .

 (vi) Join RS and PS.

Then PQRS is the required trapezium 

(IMAGE TO BE ADDED)

Question 19

Sol: Steps of constructions:

(i) draw a line segment $P Q=8 \mathrm{~cm}$.

(ii) cut off $P E=R S=3 \mathrm{~cm}$.

(iii) With center E and Radius 2cm and with center    Q and radius 4cm draw intersecting each other at R. 

(IV) Join $E R$ and $Q R$

(v) with center $P$ and radius $2 \mathrm{~cm}$ and with conter $R$ and radius $3 \mathrm{~cm}$ draw arcs intersecting each other at S.

(vi) Join PS and SR.

Then PQRS is the required trapezium On measuring the angle QPS it is 50

(IMAGE TO BE ADDED)

Question 20

Sol:if $\angle A B C=120^{\circ}$ and $A B \| D C$

 So $\angle B C D=180^{\circ}-120^{\circ}=60^{\circ}$

(i) draw a line segment $A B=4 \mathrm{~cm}$

(ii) AT B draw a ray BX making an angle of 120 and cut off BC = 2.5cm

(iii) At c draw and the ray CY making an angle of $60^{\circ}$

(iv) with centre $A$ and radius $2.4 \mathrm{~cm}$ draw an are which intersects $\mathrm{CY}$ at $\mathrm{D}^{\prime}$ and O

If $\angle A$ is obtuse and by joining AD then angle 

So Join AD 

Then ABCD is the required trapezium on measuring AC it is 5.5cm

(IMAGE TO BE ADDED)

Question 21

(i) draw a line segment $A B=6.5 \mathrm{~cm}$

(ii)produce it to $P$ such that- BP $=3 \mathrm{~cm}$

(iii)with conter p and radius $5 \mathrm{~cm}$ and with center $A$ and radius $7 \mathrm{~cm}$ draw ares which intersect each other at C

(iv) Join $A C$ and $P C$ and $B C$.

(v)With conter B and recdius $5 \mathrm{~cm}$ and with conter c and radius $3 \mathrm{~cm}$, draw ares intersecting each other at $D$.

(vi) Join AD , BD and CD 

Then $A B C D$ is the required trapezium.

(IMAGE TO BE ADDED)

Question 22

Sol: The radius of the circumcircle al a regular hexagon is the length of the side of the regular hexagon.

(i) Draw a circle with center O and radius 4cm

(ii) Take a point $A$ on it and from $A$, draw ars of $4 \mathrm{~cm}$ cutting the circle at $B, C, D$, F and $F$.

(iii) Join $A D, B C, C D, D E, E F$ and $F A$.

(IMAGE TO BE ADDED)

Then $A B C D E F$ is the required regular $h$ exagon The area ag regular hexagon $=6 \times \frac{\sqrt{3}}{4}(\text { side })^{2}$

$=\frac{6 \sqrt{3}}{4}(4)^{2}$

$=\frac{6 \times \sqrt{3} \times 16}{4}=24 \sqrt{3} \mathrm{~cm}^{2}$

$=24(1.732) \mathrm{cm}^{2}$

$=41.568 \mathrm{~cm}^{2}$

Question 23

Sol; The radius of the circumcircle of a regular hexagon is the length of the side of this hexagon. 

(i) Draw a circle with center O and radius 3.2cm

(ii)Take a point A on the circle 

(iii) Start from A and Radius 3.2cm the arcs on the circle at B,C,D E and F

(IV) Join AB, BC , CD , DE , EF and FA 

Then ABCDEF is required regular hexagon 

(IMAGE TO BE ADDED)