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Question-1

The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.

Solution:
Let ABCD be a quadrilateral in which    A : B : C : D = 3 : 5 : 9 : 13

Sum of the ratios = 3 + 5 + 9 + 13 = 30

Also, A + B + C + D = 360°
   

Sum of all the angles of a quadrilateral is 360°  

A = × 360° = 36°  
   
    B = × 360° = 60°

    C = × 360° = 108°

and D = × 360° = 156°

Question-2

 If the diagonals of a parallelogram are equal, then show that it is a rectangle.
 

 


Solution:
Given: In parallelogram ABCD, AC = BD

To Prove: Parallelogram ABCD is a rectangle.

Proof: In Δ ACB and Δ BDA,

AC = BD | Given

AB = BA  | Common

BC = AD |Opposite sides of the parallelogram ABCD


D ACB Δ BDA | SSS Rule

ABC = BAD ...(1) CPCT

Again AD || BC | Opposite sides of parallelogram ABCD


AD || BC and the transversal AB intersects them.


BAD + ABC = 180° ...(2) |Sum of consecutive interior angles on |the same side of the transversal is 180° 

From (1) and (2),

BAD = ABC = 90°


A = 90° and C = 90°


Similarly we can prove B = 90° and D = 90°

Parallelogram ABCD is a rectangle.

Question-3

Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
 

 


Solution:
Given: ABCD is a quadrilateral where diagonals AC and BD intersect each other at right angles at O.

To Prove: Quadrilateral ABCD is a rhombus.

Proof: In Δ AOB and Δ AOD,

AO = AO  | Common

OB = OD | Given AOB = AOD | Each = 90°          
Δ AOB
Δ AOD   | SAS Rule

AB = AD ...(1)       | CPCT

Similarly, we can prove that

AB = BC ...(2)

BC = CD ...(3)

CD = AD ...(4)

From (1), (2), (3) and (4), we obtain

AB = BC = CD = DA Quadrilateral ABCD is a rhombus.

Question-4

Show that the diagonals of a square are equal and bisect each other at right angles.
 

 

Solution:
Given: ABCD is a square.

To Prove: (i) AC = BD

               (ii) AC and BD bisect each other at right angles.


Proof: (i) In Δ ABC and Δ BAD,

AB = BA  | Common

BC = AD | Opposite sides of square ABCD


ABC = BAD | Each = 90° (ABCD is a square)

Δ ABC
Δ BAD | SAS Rule

AC = BD | CPCT


(ii) In Δ OAD and Δ OCB,

AD = CB | Opposite sides of square ABCD

OAD = OCB   | AD || BC and transversal AC intersects them ODA = OBC   | AD || BC and transversal BD intersects them Δ OAD   Δ OCB   | ASA Rule

OA = OC …..(1)

Similarly, we can prove that

OB = OD …..(2)

From (1) and (2),

AC and BD bisect each other.

Again, in Δ OBA and Δ ODA,

OB = OD | From (2) above

BA = DA | Sides of square are equal

OA = OA | Common
Δ OBA Δ ODA  | SSS Rule
AOB = AOD   …..(3) | CPCT

But AOB + AOD = 180° | Linear Pair Axiom   
AOB = AOD = 90° | From (3)
AC and BD bisect each other at right angles.

Question-5

Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Solution:
Given: The diagonals AC and BD of a quadrilateral ABCD and equal and bisect each other at right angles.

To Prove: Quadrilateral ABCD is a square.

Proof: In Δ OAD and Δ OCB,

OA = OC  | Given
   
OD = OB | Given AOD = COB   | Vertically Opposite Angles  
Δ OAD Δ OCB   | SAS Rule   AD = CB      | c.p.c.t.   ODA = OBC | c.p.c.t. ∴∠ ODA = OBC
AD || BC

Now, AD = CB and AD || CB

Quadrilateral ABCD is a || gm.

Similarly we can prove that AB = DC and AB || DC

In Δ AOB and Δ AOD,

AO = AO | Common

OB = OD | Given
  AOB = AOD | Each = 90° (Given) Δ AOB Δ AOD | SAS Rule AB = AD

Now, ABCD is a parallelogram and AB = AD
ABCD is a rhombus.

Again, in Δ ABC and Δ BAD,

AC = BD | Given

BC = AD |
ABCD is a rhombus

AB = BA | Common
Δ ABC Δ BAD | SSS Rule ABC = BAD | c.p.c.t. AD || BC | Opp. sides of || gm ABCD and transversal AB intersects them.

ABC + BAD = 180°   | Sum of consecutive interior angles on the same side of the transversal is 180°

ABC = BAD = 90°

Similarly, BCD = ADC = 90°

ABCD is a square.

Question-6

Diagonal AC of a parallelogram ABCD bisects  A (see figure). Show that: 
(i) it bisects  C also
(ii) ABCD is a rhombus.
 

 


Solution:
Given: Diagonal AC of a parallelogram ABCD bisects A.
To Prove: (i) it bisects C also.
                (ii) ABCD is a rhombus.

Proof: (i) In Δ ADC and Δ CBA,

          AD = CB              | Opp. sides of
parallelogram  ABCD

          CA = CA              | Common

          DC = BA              | Opp. sides of
parallelogram  ABCD
   Δ ADC Δ CBA         | SSS Rule   
ACD = CAB        | c.p.c.t.

and DAC = BCA        | c.p.c.t.

But CAB = DAC        | Given

   \ ACD = BCA

AC bisects
 C also.
(ii) From above, ACD = CAD

  \ AD = CD                   | Opposite sides of equal angles of a triangle are equal

  \ AB = BC = CD = DA   |
ABCD is a parallelogram

ABCD is a rhombus.

Question-7


ABCD is a rhombus. Show that diagonal AC bisects  A as well as  C and diagonal BD bisects  B as well as D.
 

 


Solution:
Given : ABCD is a rhombus.
To Prove : (i) Diagonal AC bisects A as well as C.
                (ii) Diagonal BD bisects B as well as D.

Proof: (i) ABCD is a rhombus

AD = CD
DAC = DCA ...(1)
  | Angles opposite to equal sides of a triangle are equal

Also, CD || AB  and transversal AC intersects them
DAC = BCA …(2) |Alternate Interior angles

From (1) and (2) DCA = BCA

AC bisects C

Similarly AC bisects A.


(ii) Proceeding similarly as in (i) above, we can prove that BD bisects B as well as D.

Question-8

ABCD is a rectangle in which diagonal AC bisects  A as well as  C. Show that (i) ABCD is a square (ii) diagonal BD bisects  B as well as  D.
 

 


Solution:
Given: ABCD is a rectangle in which diagonal AC bisects A as well as C.

To Prove: (i) ABCD is a square.
               (ii) diagonal BD bisects B as well as D.


Proof: (i) AB || DC and transversal AC intersects them.

ACD = CAB |Alternate interior angles

But CAB = CAD  
ACD = CAD      
AD = CD       |Sides opposite to equal angles of a triangle are equal
ABCD is a square.


(ii) In Δ BDA and Δ DBC,

BD = DB | Common

DA = BC | Sides of a square ABCD

AB = DC  | Sides of a square ABCD
Δ BDA Δ DBC | SSS Rule
ABD = CDB  | CPCT

But CDB = CBD    | CB = CD (Sides of a square ABCD)    
ABD = CBD .......... (1)
BD bisects B.


Now, ABD = CBD | from (1)
       ABD = ADB | AB = AD
       CBD = CDB | CB = CD

        CDB = CBD = ABD = ADB

       CDB = ADB BD bisects D.

Question-9

In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see figure). Show that:
(i) Δ APD Δ CQB
(ii) AP = CQ
(iii) Δ AQB Δ CPD
(iv) AQ= CP
(v) APCQ is a parallelogram

Solution:
style=""> Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ.



To Prove: (i) Δ APD Δ CQB

(ii) AP = CQ                         

(iii) Δ AQB = Δ CPD              

(iv) AQ = CP                       

(v) APCQ is a parallelogram


Construction: Join AC to intersect BD at O.
Proof:
First we need to prove APCQ is a parallelogram
(v) The diagonals of a parallelogram bisect each other.
OB = OD
OB - BQ = OD - DP | BQ = DP given
OQ = OP      .......... (1)

Also, OA = OC       .......... (2) | Diagonals of a parallelogram bisect each other

From (1) and (2), APCQ is a parallelogram.

(ii) APCQ is a parallelogram | Proved in (v) above AP = CQ | Opposite sides of a parallelogram are equal

(iv) APCQ is a parallelogram| Proved in (i) above
AQ = CP| Opposite sides of a parallelogram are equal


(i) In Δ APD and Δ CQB,

AP = CQ | Opposite sides of a parallelogram are equal | from (ii)

PD = QB | Given

AD = CB| Opposite sides of a parallelogram are equal
Δ APD Δ CQB.| SSS rule


(iii) In Δ AQB and Δ CPD,
AQ = CP | APCQ is a parallelogram

QB = PD | Given

AB = CD | Opposite sides of parallelogram ABCD

AQB
CPD | SSS Rule

Question-10


ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD respectively (see figure). 
Show that: 
(i) 
D APB  Δ CQD

(ii) AP = CQ.

 

 


Solution:
Given: ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD respectively.

To Prove: (i) Δ APB Δ CQD
               (ii) AP = CQ.


Proof: (i) In Δ APB and Δ CQD,

AB = CD | Opposite sides of parallelogram ABCD
ABP = CDQ |
AB || DC and transversal BD intersects them APB = CQD | Each = 90°
APB
CQD | AAS Rule

(ii) Δ APB Δ CQD | Proved above in
(i) AP = CQ. | CPCT

Question-11

In D ABC and Δ DEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see figure). Show that:
 
 

(i) Quadrilateral ABED is a parallelogram
(ii) Quadrilateral BEFC is a parallelogram
(iii) AD || CF and AD = CF
(iv) Quadrilateral ACFD is a parallelogram
(v) AC = DF
(vi) Δ ABC Δ DEF

 


Solution:
Given: In Δ ABC and Δ DEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively.
To Prove: (i) Quadrilateral ABED is a parallelogram

               (ii) Quadrilateral BECF is a parallelogram

               (iii) AD || CF and AD = CF

               (iv) Quadrilateral ACFD is a parallelogram

               (v) AC = DF

               (vi) Δ ABC Δ DEF

Proof: (i) In quadrilateral ABED,

 AB = DE and AB || DE | Given
 Quadrilateral ABED is a parallelogram

(ii) In quadrilateral BEFC,

BC = EF and BC || EF  | Given
Quadrilateral BECF is a parallelogram


(iii) ABED is a parallelogram | From (i)
AD || BE and AD = BE ...(1) | 

∵ Opposite sides of a parallelogram are parallel and equal

BEFC is a parallelogram | From (ii)
BE || CF and BE = CF ...(2) | 

∵ Opposite sides of a parallelogram are parallel and equal

From (1) and (2), we obtains AD || CF and AD = CF

(iv) In quadrilateral ACFD,

AD || CF and AD = CF | Proved in (iii)  
Quadrilateral ACFD is a parallelogram


(v) ACFD is a parallelogram | Proved in (iv)

AC || DF and AC = DF |

∵ In a parallelogram opposite sides are parallel and of equal length

(vi) In Δ ABC and Δ DEF

AB = DE  |

∵ ABED is a parallelogram
BC = EF  | 
  BEFC is a parallelogram

AC = DF | Proved in (v)
Δ ABC Δ DEF | SSS Rule

Question-12

ABCD is a trapezium in which AB || CD and AD = BC (see figure). Show that:  
(i)  A =  B
(ii)  C =  D
(iii) Δ ABC  Δ BAD
(iv) Diagonal AC = Diagonal BD
[Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E]
 

 
Solution:
Given: ABCD is a trapezium in which AB || CD and AD = BC
To Prove: (i) A = B

               (ii)C = D

              (iii) Δ ABC Δ BAD

              (iv) Diagonal AC = Diagonal BD

Construction: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.
Proof:
(i) AB || CD  | Given and AD || EC | By construction
AECD is a parallelogram
   
AD = EC | Opposite sides of a parallelogram are equal

But AD = BC | Given

EC = BC
CBE = CEB ...(1) | Angles of opposite to equal sides of a triangle are equalB + CBE = 180° ...(2)  | Linear Pair AxiomA +CEB = 180°....(3)  | The sum of consecutive interior angles on the same side of the transversal is 180°         


From (2) and (3),B + CBE = A + CEB

But CBE = CEB
B = A or A = B


(ii) AB || CD A + D = 180°

and B + C = 180°

A + D = B + C     | Proved in (i)

But A = B  

D = C or C = D


(iii) In Δ ABC and Δ BAD, AB = BA | Common

BC = AD | Given ABC = BAD | From (i)

∴ Δ ABC Δ BAD | SAS Rule         

(iv)  Δ ABC = Δ BAD | From (iii) above     
AC = BD  | CPCT

Question-13

ABCD is a quadrilateral in which p, Q, R and S are mid-points of the sides AB, BC, CD and DA, AC is a diagonal. Show that: 

(i) SR || AC and SR = AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.
 

 


Solution:
Given: ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA. AC is a diagonal.

To Prove: (i) SR || AC and SR = AC
               (ii) PQ = SR
               (iii) PQRS is a parallelogram.


Proof: (i) In Δ DAC,
S is the mid-point of DA and R is the mid-point of DC
SR
|| AC and SR = AC |Mid-point theorem

(ii) In Δ BAC,
P is the mid-point of AB and Q is the mid-point of BC
PQ
|| AC and PQ = AC | Mid-point theorem
But from (i), SR = AC   PQ = SR

(iii) PQ
|| AC  | From (ii)

SR
|| AC | From(i) PQ || SR | Two lines parallel to the same line are parallel to each other

Also, PQ = SR | From (ii)

PQRS is a parallelogram.

Question-14

 ABCD is a rhombus and P, Q, R and S are the mid - points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.
 

 


Solution:
Given: ABCD is a rhombus. P, Q, R, S are the mid - points of AB, BC, CD, DA respectively. PQ, QR, RS and SP are joined.

To prove: PQRS is a rectangle.

Construction: Join AC and BD.

Proof: In triangles RDS and PBQ,

DS = QB  |
∵ Halves of opposite sides of parallelogram ABCD which are equal

DR = PB | Halves of opposite sides of parallelogram ABCD which are equal

SDR = QBP |∵ Opposite angles of parallelogram ABCD which are equal

Δ RDS    Δ PBQ |∵ SAS Axiom

SR = PQ | c.p.c.t.

In triangles RCQ and PAS,

RC = AP  |
 Halves of opposite sides of parallelogram ABCD which are equal

CQ = AS | Halves of opposite sides of parallelogram ABCD which are equal

RCQ = PAS | ∵ Opposite angles of parallelogram ABCD which are equal

Δ RCQ = Δ PAS

RQ = SP

In PQRS,

SR = PQ and RQ = SP PQRS is a parallelogram,

In Δ CDB, R and Q are the mid-points of DC and CB respectively.

RQ || DB RF || EO.

Similarly, RE || FO

OFRE is a
parallelogram  

R = EOF = 90°
   |

 Opposite angles of a parallelogram are equal and diagonals of a rhombus intersect at 90°

Thus PQRS is a rectangle.

Question-15

ABCD is a rectangle and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

Solution:
Given: ABCD is a rectangle. P, Q, R and S are mid-points of AB, BC, CD and DA respectively. PQ, QR, RS and SP are joined.

To prove: Quadrilateral PQRS is a rhombus.

Construction: Join AC.



Proof: In Δ ABC,
P and Q are the mid-points of AB and BC respectively.
PQ || AC and PQ = AC …..(1)

In Δ ADC,
S and R are the mid-points of AD and DC respectively.
SR || AC and SR = AC …..(2)

From (1) and (2),

PQ || SR and PQ = SR
Quadrilateral PQRS is a parallelogram ….(3)


In rectangle ABCD,

AD = BC | Opposite sides AD = BC |


 Halves of equals are equal
 

AS = BQ

In Δ APS and Δ BPQ,

AP= BP |

 P is the mid-point of AB

AS = BQ | Proved above PAS = PBQ
        
| Each = 90° Δ APS
≅ Δ BPQ | SAS Axiom

PS = PQ ...(4) | c.p.c.t.


In view of (3) and (4), PQRS is a rhombus.

Question-16

 ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see figure). Show that F is the mid-point of BC.
 

 

Solution:
Given: ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at R


To Prove: F is the mid-point of BC.


Proof: Let DB intersect EF at G.

In Δ DAB, E is the mid-point of DA and EG || AB

G is the mid-point of DB | By converse of mid-point theorem

Again, in Δ BDC, G is the mid-point of BD and GF ||AB || DC

F is the mid-point of BC. | By converse of mid-point theorem

Question-17

In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively, (see figure). Show that the line segments AF and EC trisect the diagonal BD.
 

 


Solution:
Given: In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively.

To Prove: Line segments AF and EC trisect the diagonal BD.

Proof: AB || DC (  Opposite sides of parallelogram ABCD )
AE || FC ...............(1)
AB = DC (
 Opposite sides of parallelogram ABCD )

AB = DC ( Halves of equals are equal ) AE = CF ...............(2)

In view of (1) and (2),

AECF is a parallelogram
EC || AF ...........(3) ( Opposite sides of parallelogram AECF )

In Δ DBC, F is the mid-point of DC

and FP || CQ ( 
∵  EC || AF)

P is the mid-point of DQ ( By converse of mid-point theorem) DP = PQ ............(4)

Similarly, in Δ BAP,

BQ = PQ ...........(5)

From (4) and (5), we obtain

DP = PQ = BQ

Line segments AF and EC trisect the diagonal BD.

Question-18

Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.
 
Solution:
Given: ABCD is a quadrilateral. P, Q, R and S are the mid-points of the sides DC, CB, BA and AD respectively.

To prove: PR and QS bisect each other.

Construction: Join PQ, QR, RS, SP, AC and BD.

Proof: In Δ ABC,
R and Q are the mid-points of AB and BC respectively.

RQ || AC and RQ = AC.

Similarly, we can show that

PS || AC and PS = AC

RQ || PS and RQ = PS.

Thus a pair of opposite sides of a quadrilateral PQRS are parallel and equal.

PQRS is a parallelogram.

Since the diagonals of a parallelogram bisect each other.
PR and QS bisect each other.

Question-19

ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that: 
(i) D is the mid-point of AC
(ii) MD  AC
(iii) CM = MA=  AB
 

 


Solution:
Given: ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D.

To Prove: (i) D is the mid-point of AC.

(ii) MD AC

(iii) CM = MA = AB.


Proof: (i) In Δ ACB, M is the mid-point of AB and MD || BC D is the mid-point of AC               | By converse of mid-point theorem

(ii) MD || BC and AC intersects them
ADM = ACB   | Corresponding angles

But ACB = 90°   | Given

ADM = 90° MD AC           

(iii) Now, ADM + CDM = 180° | Linear Pair Axiom
ADM = CDM = 90°

In Δ ADM and Δ CDM,

AD = CD |
D is the mid-point of AC ADM = CDM | Each = 90°

DM = DM | Common
Δ ADM
Δ CDM   | SAS Rule
MA = MC            | CPCT


But M is the mid-point of AB
MA = MB = MA = MC = CM = MA =




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