Rs Aggarwal 2020 2021 Solutions for Class 9 Maths Chapter 10 Quadrilaterals are provided here with simple step-by-step explanations. These solutions for Quadrilaterals are extremely popular among Class 9 students for Maths Quadrilaterals Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 9 Maths Chapter 10 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

Page No 312:

Answer:

Given: Three angles of a quadrilateral are 75°, 90° and 75°.

Let the fourth angle be x.

Using angle sum property of quadrilateral,

75°+90°+75°+x=360°240°+x=360°

x=360°-240°x=120°

So, the measure of  the fourth angle is 120°.

Page No 312:

Answer:

Let A = 2x​.
Then B = (4x)C = (5x) and D = (7x)
Since the sum of the angles of a quadrilateral is 360o, we have:
2x + 4x + 5x + 7x = 360   
⇒ 18 x = 360 ​
x = 20
 A = 40; B = 80C = 100; D = 140

Page No 312:

Answer:

  We have AB || DC.

  A  and   D are the interior angles on the same side of transversal line AD, whereas B and   C are the interior angles on the same side of transversal line BC. 
Now,  A + D = 180
⇒  D = 180A  
 D = 180 55 = 125

Again ,  B + C = 180
⇒  C  = 180 B  
C = 180 −  70 = 110
 

Page No 312:

Answer:

Given:  ABCD is a square in which AB = BC = CD = DA. ∆EDC is an equilateral triangle in which ED = EC = DC and  
 EDC  =    DEC = ​DCE =  60.
To prove:  AE = BE and 
DAE = 15
Proof: In ADE and ∆BCE, we have:
AD = BC              [Sides of a square]

DE = EC​             [Sides of an equilateral triangle]
ADE BCE = 90 +  60 = 150​ 
∴ ADE ≅  ∆BCE
i.e., AE =  BE  

Now, 
ADE = 150
DA = DC     [Sides of a square]
DC = DE      [Sides of an equilateral triangle]
So, DA = DE
ADE and BCE are isosceles triangles.
i.e., DAE = DEA = 12180° - 150° = 30°2 =15°
 

Page No 312:

Answer:

Given: A quadrilateral ABCD, in which BM ​⊥ AC and DN ⊥ AC and BM = DN.
To prove: AC bisects BD; or DO = BO

Proof:
Let AC and BD intersect at O.
Now, in ∆OND and ∆OMB, we have:
∠OND = ∠OMB                 (90o each)
∠DON = ∠ BOM                  (Vertically opposite angles)
Also, DN = BM                         (Given)
i.e.,
∆OND ≅ ∆OMB             (AAS congurence rule)
∴ OD = OB                          (CPCT)
​Hence, AC bisects BD.

Page No 312:

Answer:

Given:  ABCD is a quadrilateral in which AB = AD and BC = DC  
(i)
In ∆ABC and ∆ADCwe have:
AB = AD                                                  (Given)

BC = DC                                                 (Given)
AC is common.
i.e., ∆ABC ≅ ∆ADC                                    (SSS congruence rule)
∴ ∠BAC = ∠DAC and ∠BCA = ∠D​CA        (By CPCT)
Thus, AC bisects ∠A and ∠ C.

(ii)
Now, in 
∆ABE and ∆ADE, we have:
  AB = AD                                      (Given)​
∠BAE = ∠DAE​                               (Proven above)
 AE is common.
∴ ∆ABE ≅  ∆ADE                          (SAS congruence rule)
⇒ BE = DE                                                 (By CPCT)
 
(iii)  
∆ABC ≅  ∆ADC                  (Proven above)
∠ABC = ∠AD​C                           (By CPCT)​
 

Page No 312:

Answer:

Given: ABCD is a square and ∠PQR = 90°.
Also, PB = QC = DR

(i) We have:
  BC = CD                (Sides of square)
  CQ = DR                   (Given)
  BC = BQ + CQ
⇒ CQ = BC − BQ
∴ DR = BC − BQ           ...(i)
   
Also, CD = RC+ DR
∴ DR = CD −  RC = BC − RC            ...(ii)
From (i) and (ii), we have:
BC − BQ = ​BC − RC
∴ BQ = RC

(ii) In ∆RCQ and ∆QBP, we have:
PB = QC   (Given)
BQ = RC  (Proven above)
∠RCQ = ∠QBP   (90o each)
i.e., ∆RCQ ≅ ∆QBP       (SAS congruence rule)
∴ QR =  PQ                (By CPCT)

(iii) ∆RCQ ≅ ∆QBP and QR = PQ (Proven above)
∴ In ∆RPQ, ∠QPR = ∠QRP = 12180° - 90° = 90°2 = 45°

Page No 312:

Answer:


Let ABCD be a quadrilateral whose diagonals are AC and BD and O is any point within the quadrilateral. 
Join O with A, B, C, and D.
We know that the sum of any two sides of a triangle is greater than the third side.
So, in ∆AOC, OA + OC > AC

Also, in ∆ BOD, OB + OD > BD
Adding these inequalities, we get:
(OA + OC) + (OB + OD) > (AC + BD)
OA + OB + OC + OD > AC + BD


 



Page No 313:

Answer:

Given: ABCD is a quadrilateral and AC is one of its diagonal.

(i)  We know that the sum of any two sides of a triangle is greater than the third side.
In ∆ABC, AB + BC > AC            ...(1)

In ∆ACD, CD + DA > AC            ...(2)
​Adding inequalities (1) and (2), we get:
AB + BC + CD + DA > 2AC 

(ii) In ∆ABC, we have :
 ​  AB + BC > AC            ...(1)
  We also know that the length of each side of a triangle is greater than the positive difference of the length of the other two sides.
   In ∆ACD, we have:​
  AC > |DA − CD|​        ...(2)
   From (1) and (2), we have:
   AB + BC > |DA − CD|​
  ⇒ AB + BC + CD > DA

(iii) In ∆ABC, AB + BC > AC
    In ∆ACD, CD + DA > AC
    In ∆ BCD, BC CD > BD
   In ∆ ABD, DAAB > BD
   ​Adding these inequalities, we get:
   2(AB + BC + CD + DA) > 2(AC + BD)
   ⇒ (AB + BC + CD + DA) > (AC + BD)
 

Page No 313:

Answer:


Let ABCD be a quadrilateral and ∠1, ∠2, ∠3 and ∠4  are its four angles as shown in the figure.
Join BD which divides ABCD in two triangles, ∆ABD and ∆BCD.
In ∆ABD, we have:
∠1 + ∠2 + ​∠A = 180o      ...(i)
In ​∆BCD, we have:
∠3 + ∠4 + ∠C = 180o     ...(ii)
On adding (i) and (ii), we get:

(∠1 + ∠3) + ∠A + ∠C + (∠4 + ∠2) ​= 360o  
⇒ ∠A + ∠C  + ∠B + ∠D = 360o                         [ ∵ ∠1 + ∠3 = ∠B; ∠4 + ∠2 = ∠D]
∴ ∠A + ∠C  + ∠B + ∠D = 360o
 



Page No 328:

Answer:

ABCD is parallelogram and ∠A = 72°​.
We know that opposite angles of a parallelogram are equal.
∴∠A ​= ∠C and B ​= ∠D ​ ​
∴ ∠C = 72o
A and ∠B are adajcent angles.
i.e., ∠A ​+ ∠B​ = 180o
⇒ ∠B = 180o   ∠A
⇒ ∠B​ = 180o 72o = 108o
∴​ ∠B​ =​ ∠D108o
Hence, ∠B​ =​ ∠D = 108o​ and ∠C​ = 72o

Page No 328:

Answer:

Given:  ABCD is parallelogram and ∠DAB = 80°​ and ∠DBC = 60°
To find: Measure of ∠CDB and ∠ADB

In parallelogram ABCD, AD ||​ BC
∴ ∠DBC = ∠ ADB = 60o     (Alternate interior angles)     ...(i)

As ∠DAB and ∠ADC are adajcent angles, ∠DAB ​+ ∠ADC​ = 180o
∴ ∠ADC = 180o  − ∠DAB
​    ⇒∠ADC​ = 180o − 80o = 100o

Also, ∠ADC​ = ∠ADB + ∠C​​DB
∴​ ∠ADC​ =​ 100o 
⇒ ∠ADB + ∠C​​DB = 100o              ...(ii)
From (i) and (ii), we get:
60o + ∠C​​DB = 100o 
⇒ ∠C​​DB = 100o − 60o = 40o
Hence, ∠CDB​ =​ 40o and ∠ADB​ = 60o



Page No 329:

Answer:



Given: parallelogram ABCD, M is the midpoint of side BC and ∠BAM = ∠DAM.

To prove: AD = 2CD
 
Proof:

Since, ADBC and AM is the transversal.

So, DAM=AMB      (Alternate interior angles)

But, DAM=BAM (Given)

Therefore, AMB=BAM

AB=BM             (Angles opposite to equal sides are equal.)    ...(1)

Now, AB = CD       (Opposite sides of a parallelogram are equal.)

2AB=2CD(AB+AB)=2CD

BM+MC=2CD      (AB BM and MC = BM)

BC=2CDAD=2CD       AD=BC, Opposite sides of a parallelogram are equal.

Page No 329:

Answer:

ABCD is a parallelogram.
∴ ​∠A = ​∠C and B =​ ∠D (Opposite angles)
And ​∠A + ​∠B = 180o                (Adjacent angles are supplementary)
∴ ​∠B = 180o − ∠
180o − 60o = 120o             ( ∵∠A = 60o)
∴ ​∠A = ​∠C = 60o and B =​ ∠D​ = 120o

(i) In ∆ APB, ∠​PAB = 60°2=30° and ∠PBA = 120°2=60°
   ∴​ ∠​APB​ = 180o − (30o + 60o) = 90o

(ii) In ∆ ADP, ∠​PAD = 30o and ∠ADP = 120o
    ∴ ∠APB = 180o − (30o + 120o) = 30o
   Thus, ∠​PAD = ​∠APB = ​30o
   Hence, ∆ADP is an isosceles triangle and AD = DP.

   In ∆ PBC, ∠​ PBC = 60o ∠​ BPC = 180o − (90o +30o) = 60o and ∠​ BCP  = 60o (Opposite angle of ∠A)
   ∴ ∠ PBC = ∠​ BPC = ∠​ BCP
   Hence, ∆PBC is an equilateral triangle and, therefore, PB = PC = BC.​

(iii) DC = DP + PC
     From (ii), we have:
    DC = AD + BC                     [AD = BC, opposite sides of a parallelogram]
    ⇒ DC = AD + AD

    ⇒ DC = 2 AD

Page No 329:

Answer:

ABCD is a parallelogram.
∴ AB ∣∣​ DC and BC ​∣∣​ AD

(i) In ∆AOB, ∠BAO = 35°, ​∠AOB = ∠COD = 105°  (Vertically opposite angels)
∴ ​∠ABO = 180o − (35o + 105o) = 40o

(ii)∠ODC and ∠ABO are alternate interior angles.
∴ ∠ODC = ∠ABO = 40o

(iii) ∠ACB = ∠​CAD = 40o                             (Alternate interior angles)

(iv) ∠CBDABC  ABD            ...(i)

     ∠ABC = 180o − ∠BAD                        (Adjacent angles are supplementary)   ​
⇒∠ABC = 180o − 75o = 105o   
⇒∠CBD = 105o ABD                         (∠ABDABO)
⇒∠CBD = 105o 40o =  65o

Page No 329:

Answer:

ABCD is a parallelogram.
i.e., ∠A = C and B∠D                  (Opposite angles)
Also, ∠A + ∠B = 180o                            (Adjacent angles are supplementary)   ​
∴​ (2x + 25)°​ + (3x − 5)°​ = 180
⇒ ​5x +20 = 180
⇒​ 5x = 160
⇒​ x = 32o
∴​∠A = 2 ⨯ 32 + 25 = 89o and ∠B = 3 32 − 5 = 91o
Hence, x = 32o, ∠AC89o and ∠BD = 91o 

Page No 329:

Answer:

 Let ABCD be a parallelogram. 
∴ ∠​A = ∠C and B = ∠D           (Opposite angles)
Let A = xo and ∠B4x5°
Now, ∠​A + ∠B = 180o                 (Adjacent angles are supplementary)
  x + 4x5 = 180o9x5 = 180ox = 100oNow, A = 100o and B = 45×100° = 80o
 Hence, AC = 100o; B = ∠D = 80o

Page No 329:

Answer:

 Let ABCD be a parallelogram. 
 ∴ ∠​A = ∠​and B = ∠D          (Opposite angles)
 Let A be the smallest angle whose measure is xo.
 ∴​ ∠​B = (2x − 30)o
Now, ∠​A + ∠B = 180o                (Adjacent angles are supplementary) 
   ⇒ x + 2x − 30o = 180o
   ⇒ 3x = 210o
   ⇒ x = 70o
∴ ​∠​B = 2 ⨯ 70o − 30o = 110o
 HenceAC = 70o; ∠B = ∠D = 110o

Page No 329:

Answer:

ABCD is a parallelogram.
The opposite sides of a parallelogram are parallel and equal.
∴ AB = DC = 9.5 cm
Let BC = AD = x
∴​ Perimeter of ABCD = AB + BC + CD + DA = 30 cm
⇒ 9.5 + x + 9.5 + x = 30
⇒ 19 + 2x = 30
⇒ 2x = 11
x = 5.5 cm
Hence, AB = DC = 9.5 cm and BC = DA = 5.5 cm

Page No 329:

Answer:

ABCD is a rhombus and a rhombus is also a parallelogram. A rhombus has four equal sides.
(i)​ In ∆ABC, ∠​BAC = ∠BCA = 12180 - 110 = 35o
i.e., x = 35o   
Now, ∠​B + ∠C = 180o                 (Adjacent angles are supplementary)  ​
But ∠​C​ = x + y = 70o 
 
⇒​ y = 70o    x  
 
⇒​y =  70o − 35o = 35o
 Hence, x = 35o; y = 35o

(ii) The diagonals of a rhombus are perpendicular bisectors of each other.
   So, in ∆​AOB, ​∠OAB = 40o, ∠AOB = 90o and ∠ABO = 180o − (40o + 90o) = 50o
   ∴ ​x = 50o
   In ∆​AB​D, AB = AD
  
So, ∠ABD = ​∠ADB = ​50o
   Hence, x = 50o;  y = 50o


​(iii) ∠​BAC = ∠​DCA                   (Alternate interior angles)​
     i.e., x  = 62o  
     In ∆BOC∠​BCO =  62o              [In ∆​ ABC, AB = BC, so ∠​BAC = ∠​ACB]
    Also, ∠​BOC = 90o
  ∴ ∠​OBC = 180o − (90o + 62o) = 28o
  Hence, x = 62o; y = 28o

Page No 329:

Answer:


Let ABCD be a rhombus.
AB = BC = CD = DA
Here, AC and BD are the diagonals of ABCD, where AC = 24 cm and BD = 18 cm.
Let the diagonals intersect each other at O.
We know that the diagonals of a rhombus are perpendicular bisectors of each other.
∴​ ∆AOB is a right angle triangle in which OA = AC/2 = 24/2 = 12 cm and OB = BD/2 = 18/2 = 9 cm.
Now, AB2= OA2 + OB2              [Pythagoras theorem]
⇒​ AB2=​ (12)2 + (9)2
⇒​ AB2=​ 144 + 81 = 225
⇒​ AB=​ 15 cm

Hence, the side of the rhombus is 15 cm.

Page No 329:

Answer:


Let ABCD be a rhombus.
∴ AB = BC = CD = DA = 10 cm
Let AC and BD be the diagonals of ABCD. Let AC = x and BD = 16 cm and O be the intersection point of the diagonals.
We know that the diagonals of a rhombus are perpendicular bisectors of each other.
∴​ ∆AOB is a right angle triangle, in which OA = AC ÷ 2 = ÷ 2 and OB = BD ÷2 = 16 ÷ 2 = 8 cm.

Now, AB2= OA2 + OB2              [Pythagoras theorem]
102 = x22 + 82100 - 64 = x2436 ×4 = x2

x2 =144
 x = 12 cm

Hence, the other diagonal of the rhombus is 12 cm.
∴ Area of the rhombus =  12×12×16 = 96 cm2



Page No 330:

Answer:

(i) ABCD is a rectangle.
The diagonals of a rectangle are congruent and bisect each other. Therefore, in​ ∆ AOB, we have:
   OA = OB   
 ∴​ ∠​OAB = ∠​OBA = 35o
∴​ x = 90o − 35o = 55o
And ∠AOB = 180o − (35o + 35o) = 110o
∴​ y = ∠AOB​ = 110o​                     [Vertically opposite angles]
Hence, x = 55o and y = 110o​​

(ii) In ∆AOB, we have:
      OA = OB   
Now, ∠​OAB = ∠OBA = 12×180° - 110° = 35o
∴​ y = ∠BAC = 35o                 [Interior alternate angles]
Also, x = 90oy                          [ ​∵∠C = 90o =  x + y ]
⇒​ x = 90o − 35o = 55o                     
Hence, x = 55o and y = 35o​​

Page No 330:

Answer:



Given: ABCD is a rhombus, DE is altitude which bisects AB i.e. AE = EB

In AED and BED,

DE=DE                         (Common side)

DEA=DEB=90°     (Given)

AE=EB                          (Given)

 AEDBED         (By SAS congruence Criteria)

AD=BD                    (CPCT)

Also, AD=AB            (Sides of rhombus are equal)

AD=AB=BD

Thus, ABD is an equilateral triangle.

Therefore, A=60°

C=A=60°                         (Opposite angles of rhombus are equal)

And, ABC+BCD=180°           (Adjacent angles of rhombus are supplementary.)

ABC+60°=180°ABC=180°-60°ABC=120°ADC=ABC=120°

Hence, the angles of rhombus are 60°, 120°, 60° and 120°.

Page No 330:

Answer:

The angles of a square are bisected by the diagonals.
∴ ∠​OBX = 45o                        [∠​ABC = 90o and BD bisects ∠​ABC​]
And ∠​BOX = ∠COD = 80o           [Vertically opposite angles]
∴​ In ∆BOX, we have:
∠AXO = ∠OBX + ​∠BOX        [Exterior angle of ∆BOX]
⇒​ ∠AXO = 45o + 80o = 125o
∴ ​x =125o

Page No 330:

Answer:


Given: 
A rhombus ABCD.
 
To prove: Diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.
 
Proof:

In ABC,

AB = BC         (Sides of rhombus are equal.)

4=2          (Angles opposite to equal sides are equal.)     ...(1)

Now,

ADBC          (Opposite sides of rhombus are parallel.)

AC is transversal.

So, 1=4        (Alternate interior angles)          ...(2)

From (1) and (2), we get

1=2 

Thus, AC bisects A.

Similarly,

Since, ABDC and AC is transversal.

So, 2=3     (Alternate interior angles)     ...(3)

From (1) and (3), we get

4=3

Thus, AC bisects ∠C.

Hence, AC bisects C and A

In DAB,

AD = AB                   (Sides of rhombus are equal.)

ADB=ABD          (Angles opposite to equal sides are equal.)     ...(4)

Now,

DCAB                        (Opposite sides of rhombus are parallel.)

BD is transversal.

So, CDB=DBA        (Alternate interior angles)          ...(5)

From (4) and (5), we get

ADB=CDB 

Thus, DB bisects D.

Similarly,

Since, ADBC and BD is transversal.

So, CBD=ADB     (Alternate interior angles)     ...(6)

From (4) and (6), we get

CBD=ABD

Thus, BD bisects ∠B.

Hence, BD bisects D and B.

Page No 330:

Answer:


Given: 
In a parallelogram ABCD, AM CN.
 
To prove: AC and MN bisect each other.
 
Construction: Join AN and MC.

Proof:
Since, ABCD is a parallelogram.

ABDCAMNC

Also, AM = CN           (Given)

Thus, AMCN is a parallelogram.

Since, diagonals of a parallelogram bisect each other.

Hence, AC and MN bisect each other.

 

Page No 330:

Answer:

We have:
B = ∠D                        [Opposite angles of parallelogram ABCD]
AD = BC and AB = DC      [Opposite sides of parallelogram ABCD]
Also, AD || BC and AB|| DC
It is given that AP=13AD and CQ=13BC.
∴ ​AP = CQ                                   [∵ AD = BC]
In ∆​DPC and ∆​BQA, we have:
AB = CD, ∠B = ∠D and DP = QB                        [∵DP23AD and QB = 23BC 
i.e., ∆​DPC ≅ ∆​BQA
∴​ PC  = QA

Thus, in quadrilatreal AQCP, we have:
  AP = CQ                   ...(i)
 PC  = QA                   ...(ii)
∴ ​AQCP is a parallelogram.

Page No 330:

Answer:

In ∆​ODF and ∆​OBE, we have:
OD = OB                                  (Diagonals bisects each other)
DOF = ∠BOE                         (Vertically opposite angles)
∠FDO = ∠OBE                         (Alternate interior angles)
 i.e., ∆​ODF ≅ ∆​OBE
∴​ OF = OE                                 (CPCT)
Hence, proved.

Page No 330:

Answer:


Given: In parallelogram ABCDDP ABAQ  BC and ∠PDQ = 60°

In quadrilateral DPBQ, by angle sum property, we have

PDQ+DPB+B+BQD=360°60°+90°+B+90°=360°B=360°-240°B=120°

Therefore, B = 120°

Now,
B=D=120°           (Opposite angles of a parallelogram are equal.)

A+B=180°           (Adjacent angles of a parallelogram are supplementary.)

A+120°=180°A=180°-120°A=60°

Also, 
A=C=60°         (Opposite angles of a parallleogram are equal.)

So, the angles of a parallelogram are 60°, 120°, 60° and 120°.

Page No 330:

Answer:


Given: In rectangle ABCDAC bisects ∠A, i.e. ∠1 = ∠2 and AC bisects ∠C, i.e. ∠3 = ∠4.
 
To prove:
(i) ABCD is a square,
(ii) diagonal BD bisects ∠B as well as ∠D.

Proof:

(i)
Since, ADBC        (Opposite sides of a rectangle are parallel.)

So, 1=4         (Alternate interior angles)

But, 1=2        (Given)

So, 2 = 4

In ABC,

Since, 2=4

So, BC=AB                            (Sides opposite to equal angles are equal.)

But these are adjacent sides of the rectangle ABCD.

Hence, ABCD is a square.

(ii)
Since, the diagonals of a square bisects its angles.
So, diagonals BD bisects B as well as ∠D.



Page No 331:

Answer:

In ∆ODC and ∆​OEB, we have:
DC = BE                                   (∵ DC = AB)
∠COD = ∠BOE                        (Vertically opposite angles)
OCD = ∠OBE                        ( Alternate interior angles)
  
i.e., ∆​ODC ≅ ∆​OEB
⇒ OC = OB                                 (CPCT)
We know that BC = OC + OB.
∴ ED bisects BC.
 

Page No 331:

Answer:

Given: ABCD is a parallelogram.
BE = CE    (E is the mid point of BC)
DE and AB when produced meet at F.
To prove: AF = 2AB

Proof:
In parallelogram ABCD, we have:
          AB || DC
       ∠DCE = ∠EBF            (Alternate interior angles)
In ∆DCE and ∆BFE, we have:
 ∠DCE = ∠EBF              (Proved above)

 ∠DEC = ∠BEF              (Vertically opposite angles)
Also, BE = CE           (Given)
∴ ∆DCE ≅​ ∆BFE  (By ASA congruence rule)
∴ DC = BF         (CPCT)
But DC = AB, as ABCD is a parallelogram.
∴ DC = AB =  BF                   ...(i)

Now, AF = AB + BF                ...(ii)
From (i), we get:
AF = AB + AB = 2AB
Hence, proved. 

Page No 331:

Answer:



Given: l || m and the bisectors of interior angles intersect at and D.

To prove: ABCD is a rectangle.

Proof:


Since, 
l || m                       (Given)

So, PAC=ACR            (Alternate interior angles)

12PAC=12ACR

BAC=ACD

but, these are a pair of alternate interior angles for AB and DC.

ABDC

Similarly, BCAD

So, ABCD is a parallelogram.

Also,
 PAC+CAS=180°      (Linear pair)

12PAC+12CAS=90°BAC+CAD=90°BAD=90°

But, this an angle of the parallleogram
ABCD.

Hence, ABCD is a rectangle.

Page No 331:

Answer:


Given: In square ABCD, AK BL CM DN

To prove: KLMN is a square.
 
Proof:

In square 
ABCD,

AB = BC = CD = DA             (All sides of a square are equal.)

And, AK BL CM DN     (Given)

So, AB - AK = BC - BL = CD - CM = DA - DN

 KB = CL = DM = AN        ...(1)

In NAK and KBL,

NAK=KBL=90°    (Each angle of a square is a right angle.)

AK=BL                        (Given)

AN=KB                        [From (1)]

So, by SAS congruence criteria,

NAKKBL

NK=KL     (CPCT)           ...(2)

Similarly,


MDNNAK DNMCMLMCLLBK

MN = NK  and DNM=KNA     (CPCT)       ...(3)
 MN = JM
  and DNM=CML        (CPCT)        ...(4)
ML = LK and CML=BLK            (CPCT)        ...(5)

From (2), (3), (4) and (5), we get

NK = KL = MN = ML      ...(6)


And, DNM=AKN=KLB=LMC

Now,

In NAK,

NAK = 90°

Let AKN = x°

So, DNK=90°+x° (Exterior angles equals sum of interior opposite angles.)

DNM+MNK=90°+x°x°+MNK=90°+x°MNK=90°

Similarly,
NKL=KLM=LMN=90°       ...(7)

Using (6) and (7), we get
 
All sides of quadrikateral KLMN are equal and all angles are 90°.

So, KLMN is a square.

Page No 331:

Answer:

 BC || QA and CA || QB
i.e., BCQA is a parallelogram.
∴ BC = QA                           ...(i)
Similarly, BC || AR and AB || CR.
i.e., BCRA is a parallelogram.
BC = AR                         ...(ii)
But QR = QA + AR
From (i) and (ii), we get:
QR = BC + BC
QR = 2BC
∴ BC12QR

Page No 331:

Answer:

Perimeter of ∆​ABC = AB + BC + CA                         ...(i)
Perimeter of ∆PQR  =​ PQ + QR + PR                   ...(ii)

 BC || QA and CA || QB
 i.e., BCQA is a parallelogram.
 ∴ BC = QA                                  ...(iii)
Similarly, BC || AR and AB || CR 
i.e., BCRA is a parallelogram.
∴ BC = AR                                    ...(iv)
But, QR = QA + AR
From (iii) and (iv), we get:
QR = BC + BC
QR = 2BC
∴ BC = 12QR
Similarly, CA = ​12PQ and AB = ​12PR

From (i) and (ii), we have:
Perimeter of ∆​ABC  = 12QR + 12PQ + 12PR
                              = 12PR + QR + PQ
 
i.e., Perimeter of ∆​ABC  = 12 (Perimeter of ∆​PQR)
∴ Perimeter of ∆PQR = 2 ⨯ Perimeter of ∆ABC 



Page No 345:

Answer:



Given: In quadrilateral ABCD, P, Q, R and S are respectively the midpoints of the sides AB, BC, CD and DA.

To prove:

(i) PQ || AC and PQ 12AC

(ii) PQ || SR

(iii) PQRS is a parallelogram.

Proof:

(i)
In ABC,

Since, P and Q are the mid points of sides AB and BC, respectively.      (Given)

ACPQ and PQ=12AC            (Using mid-point theorem.)

(ii)
In ADC,

Since, S and R are the mid-points of AD and DC, respectively.        (Given)

SRAC and SR=12AC           (Using mid-point theorem.)            ...(1)

From (i) and (1), we get

PQ || SR

(iii)
From (i) and (ii), we get
 
PQ=SR=12AC

So, PQ and SR are parallel and equal.

Hence, PQRS is a parallelogram.

Page No 345:

Answer:



Given: In an isosceles right ∆ABC, CMPN is a square.
 
To prove: P bisects the hypotenuse AB i.e., AP = PB.
 
Proof:

In square CMPN,

 
∴ CM = MP = PN = CN            (All sides are equal.)
 
Also, ∆ABC is an isosceles with AC = BC.

⇒ AN + NC = CM + MB

⇒ AN = MB          (∵ CN = CM)       ...(i)

Now,

In ∆ANP and ∆PMB,

AN = MB                [From (i)]

ANP = ∠PMB = 90°

PN = PM             (Sides of square CMPN)

∴ By SAS congruence criteria,

∆ANP  BMP


Hence, AP = PB         (By CPCT)



Page No 346:

Answer:

In parallelogram ABCD, we have:
AD || BC and AB || DC

AD = BC and AB = DC 
AB = AE
+ BE and DC = DF + FC
AE = BE = DF = FC   
Now, DF = AE and DF || AE.
i.e., AEFD is a parallelogram​​.
∴​ AD || EF​

Similarly, ​BEFC is also a parallelogram. 
∴​ EF || BC
∴​ AD || EF || BC
Thus, AD, EF and BC are three parallel lines cut by the transversal line DC at D, F and C, respectively such that DF = FC.
These lines AD, EF and BC​ are also cut by the transversal AB at A, E and B, respectively such that​ AE =  BE. ​
Similarly, they are also cut by​ GH.
∴ GP = PH            (By intercept theorem) 

Page No 346:

Answer:

 
Given: A parallelogram ABCD
 
To prove: MN is bisected at O

Proof:

IOAM and OCN,

OA = OC                 (Diagonals of parallelogram bisect each other)

AOM = CON      (Vertically opposite angles)

MAO = OCN       (Alternate interior angles)

 By ASA congruence criteria,

OAM  OCN

 OM = ON  (CPCT)

Hence, MN is bisected at O.

Page No 346:

Answer:

 
Given: In trapezium PQRS, 
PQ || SR, is the midpoint of PS and MN || PQ.

To prove: N is the midpoint of QR.

Construction: Join QS.

Proof:

In SPQ,

Since, M is the mid-point of SP and MO || PQ.

Therefore, O is the mid-point of SQ.      (By Mid-point theorem)

Similarly, in SRQ,

Since, O is the mid-point of SQ and ON || SR           (SR || PQ and MN || PQ)

Therefore, N is the mid-point of QR.      (By Mid-point theorem)

Page No 346:

Answer:



Given: In parallelogram PQRS, PQ = 12 cm and PS = 9 cm. The bisector of ∠SPQ meets SR at M.

Let ∠SPQ = 2x.

⇒ ∠SRQ = 2x and ∠TPQ = x.

Also, PQ SR

⇒ ∠TMR = ∠TPQ = x.

In △TMR, SRQ is an exterior angle.

⇒ ∠SRQ = ∠TMR + ∠MTR

⇒ 2= + ∠MTR

⇒ ∠MTR = x

⇒ △TPQ is an isosceles triangle.

TQ = PQ = 12 cm

Now,

RT = TQ QR

      
= TQ PS

      
= 12 − 9

      = 3 cm

Page No 346:

Answer:

Given: AB || DC, AP = PD and BQ = CQ

(i) In ∆QCD and ∆QBE, we have:
DQC = ​BQE   (Vertically opposite angles)

DCQ = ∠EBQ     (Alternate angles, as AE || DC)
 BQ = CQ               (P is the midpoints)
∴ ∆QCD ≅ ∆QBE
​Hence, DQ = QE                     (CPCT)

(ii) Now, in ∆ADE, P and Q are the midpoints of AD and DE, respectively.
∴ PQ || AE
PQ || AB || DC 
⇒​ AB || PR || DC


(iii) PQAB and DC are the three lines cut by transversal AD at P such that AP = PD.
These lines 
PQABDC are also cut by transversal BC at Q such that BQ = QC.
​ Similarly, lines PQAB and DC are also cut by AC at R.
∴ AR = RC                   (By intercept theorem)

Page No 346:

Answer:

AD is a median of  ∆ABC.
∴ BD = DC
We know that the line drawn through the midpoint of one side of a triangle and parallel to another side bisects the third side.
Here, in ∆ABC, D is the mid point of BC and DE || BA (given). Then DE bisects AC.
i.e., AE =  EC
E is the midpoint of AC.
BE is the median of ∆ABC.

Page No 346:

Answer:

In  ABC, we have:
 AC = AE + EC     ...(i)
AE = EC               ...(ii)       [BE is the median of  ABC]
∴ AC = 2EC          ...(iii)

​In ∆BECDF || BE.
∴ EF = CF        (By midpoint theorem, as D is the midpoint of BC)
But EC = EF + CF
EC =​ 2 ⨯ ​CF       ...(iv)
From (iii) and (iv), we get:
AC = 2 ⨯ (2 ⨯​ CF)
∴​ CF =  14AC

Page No 346:

Answer:

 ∆ABC is shown below.  D, E and F are the midpoints of sides AB, BC and CA, respectively.

As, D and  E are the mid points of sides AB, and BC of ∆ ABC.
DE ∣∣ AC   (By midpoint theorem) 
Similarly, DF ​∣∣ BC and EF ​∣∣ AB.
Therefore, ADEF, BDFE and DFCE are all parallelograms.
Now, DE is the diagonal of the parallelogram BDFE.
∴ ​∆BDE ≅ ​∆FED
Simiilarly, DF is the ​diagonal of the parallelogram ADEF.

∴ ∆DAF ≅ ∆FED
And, EF is the ​diagonal of the parallelogram DFCE.
∴ ∆EFC ≅ ∆FED
So, all the four triangles are congruent.

Page No 346:

Answer:

 ∆ ABC is shown below.  D, E and F are the midpoints of sides BC, CA and AB, respectively.
As F and E are the mid points of sides AB and AC of ∆ ABC.
∴ FE ∣∣ BC  (By mid point theorem) 
Similarly, DE ​∣∣ FB and FD ​∣∣ AC.
Therefore, AFDE, BDEF and DCEF are all parallelograms.

In parallelogram AFDE, we have:
A = ∠EDF          (Opposite angles are equal)
In parallelogram BDEF, we have:
B = DEF          (Opposite angles are equal)
In parallelogram DCEF, we have:
 C = ∠​ DFE        (Opposite angles are equal)



Page No 347:

Answer:

Let ABCD be the rectangle and P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively. 
Join AC, a diagonal of the rectangle.
In ∆ ABC, we have:
∴ PQ ∣∣ AC and PQ12AC                    [By midpoint theorem]
Again, in ∆ DAC, the points S and R are the mid points of AD and DC, respectively.
∴ SR ∣∣ AC and SR = 12AC                    [By midpoint theorem]
Now, PQ ∣∣ AC and SR ∣∣ AC
 ⇒ PQ ∣∣ SR

Also, PQ = SR                  [Each equal to 12 AC ]         ...(i)
So, PQRS is a parallelogram.
Now, in ∆SAP and QBP, we have:
AS = BQ   
A = ∠B = 90o
AP = BP
i.e., ∆SAP ≅ ∆QBP​
∴ PS = PQ                ...(ii)
Similarly, ∆SDR ≅ ∆QCR​
SR = RQ                ...(iii)
From (i), (ii) and (iii), we have:
PQ = PQ = SR = RQ
Hence, PQRS is a rhombus.

Page No 347:

Answer:



Let ABCD be a rhombus and P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively. 
Join the diagonals, AC and BD.
In ∆ ABC, we have:
PQ ∣∣ AC and PQ = 12AC                    [By midpoint theorem]
Again, in ∆DACthe points S and R are the midpoints of AD and DC, respectively.
∴ SR ∣∣ AC and SR = 12AC                    [By midpoint theorem]
Now, PQ ∣∣ AC and SR ∣∣ AC  ⇒ PQ ∣∣ SR

Also, PQ = SR                  [Each equal to 12 AC ]            ...(i)
So, PQRS is a parallelogram.
We know that the diagonals of a rhombus bisect each other at right angles.
∴ ∠EOF = 90o
​Now, RQ∣∣ DB
 ⇒RE ∣∣ FO
Also, SR∣∣ AC 
FR ∣∣ OE
∴ OERF is a parallelogram.
So, ∠FRE = ∠EOF = 90o​         (Opposite angles are equal)
Thus, PQRS is a parallelogram with ∠R = 90o.
 ∴​ PQRS is a rectangle.

Page No 347:

Answer:


Let ABCD be a square and P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively. 
Join the diagonals AC and BD. Let BD cut SR at F and AC cut RQ at E. Let O be the intersection point of AC and BD.

In ∆ ABC, we have:
∴ PQ ∣∣ AC and PQ = 12AC                    [By midpoint theorem]
Again, in ∆DACthe points S and R are the midpoints of AD and DC, respectively.
∴ SR ∣∣ AC and SR = 12AC                    [By midpoint theorem]
Now, PQ ∣∣ AC and SR ∣∣ AC  ⇒ PQ ∣∣ SR

Also, PQ = SR                  [ Each equal to 12 AC ]             ...(i)
So, PQRS is a parallelogram.

Now, in ∆SAP and QBP, we have:
AS = BQ   
A = ∠B = 90o
AP = BP
i.e.,  ∆SAP ≅ ∆QBP​
∴ PS = PQ                  ...(ii)
Similarly, ∆SDR ≅ ∆RCQ​
∴ SR = RQ                 ...(iii)
From (i), (ii) and (iii), we have:
PQ = PS = SR = RQ           ...(iv)

We know that the diagonals of a square bisect each other at right angles.
∴ ∠​EOF = 90o
​Now, RQ ∣∣ DB 
RE ∣∣ FO
Also, SR ∣∣ AC
FR ∣∣ OE
∴ OERF is a parallelogram. 
So, ∠FRE = ∠EOF = 90o​         (Opposite angles are equal)
Thus, PQRS is a parallelogram with ∠R = 90o and PQ = PS = SR = RQ.

 ∴​ PQRS is a square.

Page No 347:

Answer:

Let ABCD be the quadrilateral in which P, Q, R, and S are the midpoints of sides AB, BC, CD, and DA, respectively.
Join PQ, QR, RS, SP and BD. BD is a diagonal of ABCD.

In ΔABD, S and P are the midpoints of AD and AB, respectively. 

∴​ SP || BD and SP BD ... (i)        (By midpoint theorem)

Similarly in Δ BCD, we have:

QR || BD and QRBD ... (ii)   (By midpoint theorem)

From equations (i) and (ii), we get:
SP ||  BD || QR

∴ SP || QR and SP = QR             [Each equal to  BD]

In quadrilateral SPQR, one pair of the opposite sides is equal and parallel to each other.
∴  SPQR is a parallelogram.

We know that the diagonals of a parallelogram bisect each other.

∴ PR and QS bisect each other.

Page No 347:

Answer:

Given: In quadrilateral ABCDBD = AC and K, L, M and N are the mid-points of AD, CD, BC and AB, respectively.

To prove: KLMN is a rhombus.

Proof:

In ADC,

Since, K and L are the mid-points of sides AD and CD, respectively.

So, KL || AC and KL = 12AC                   ...(1)

Similarly, in 
ABC,

Since, M and N are the mid-points of sides BC and AB, respectively.

So, NM || AC and NM = 12AC               ...(2)

From (1) and (2), we get
 
KL = NM and KL || NM

But this a pair of opposite sides of the quadrilateral KLMN.


So, KLMN is a parallelogram.

Now, iABD,

Since, K and N are the mid-points of sides AD and AB, respectively.

So, KN || BD and KN = 12BD                ...(3)

But BD = AC            (Given)

12BD = 12AC

KN = NM          [From (2) and (3)]

But these are a pair of adjacent sides of the parallelogram KLMN.

Hence, KLMN is a rhombus.

Page No 347:

Answer:

 
Given: In quadrilateral ABCDAC  BDP, Q, R and S are the mid-points of AB, BC, CD and AD, respectively.

To prove: PQRS is a rectangle.

Proof:

In ΔABCP and Q are mid-points of AB and BC, respectively.

∴ PQ || AC and PQ = 12AC       (Mid-point theorem)          ...(1)

Similarly, in ΔACD,

So, R and S are mid-points of sides CD and AD, respectively.


∴ SR || AC and SR = 12AC          (Mid-point theorem)        ...(2)

From (1) and (2), we get

PQ || SR and PQ = SR

But this is a pair of opposite sides of the quadrilateral PQRS,

So, PQRS is parallelogram.

Now, in ΔBCDQ and R are mid-points of BC and CD, respectively.

∴ QR || BD and QR = 12BD       (Mid-point theorem)          ...(3)

From (2) and (3), we get

SR || AC and QR || BD


But, AC  BD      (Given)

∴ RS  QR

Hence, PQRS is a rectangle.

Page No 347:

Answer:



Given: In quadrilateral ABCD, 
AC BD and AC ⊥ BDP, Q, R and S are the mid-points of AB, BC, CD and AD, respectively.

To prove: PQRS is a square.

Construction: Join AC and BD.

Proof:

In ΔABC,

 P
 and Q are mid-points of AB and BC, respectively.

 
 PQ || AC and PQ = 12AC       (Mid-point theorem)           ...(1)

Similarly, in ΔACD,

 R and S are mid-points of sides CD and AD, respectively.
 
 SR || AC and SR = 12AC        (Mid-point theorem)          ...(2)

From (1) and (2), we get

PQ || SR and PQ = SR

But this a pair of opposite sides of the quadrilateral PQRS.

So, PQRS is parallelogram.


Now, in ΔBCD,

 Q and R are mid-points of sides BC and CD, respectively.
 
 QR || BD and QR = 12BD        (Mid-point theorem)          ...(3)

From (2) and (3), we get

RS || AC and QR || BD


But, AC  BD                     (Given)

∴ RS  QR

But this a pair of adjacent sides of the parallelogram PQRS.

So, PQRS is a rectangle.


Again, AC = BD                (Given)

 12AC = 12BD

 RS = QR                   [From (2) and (3)]

But this a pair of adjacent sides of the rectangle PQRS.

Hence, PQRS is a square.



Page No 349:

Answer:

(b) 73°​

Explanation: 
Let the measure of the fourth angle be xo.
Since the sum of the angles of a quadrilateral is 360o, we have:
80o + 95o + 112ox = 360o  
⇒ 287o
x = 360o
x = 73o
Hence, the measure of the fourth angle is 73o.

Page No 349:

Answer:

(b) 60°​

Explanation:
Let ∠A = 3x​, ∠B = 4x, ∠C = 5x and ∠D = 6x.
Since the sum of the angles of a quadrilateral is 360o, we have:

3x + 4x + 5x + 6x = 360o   
⇒ 
18x =
360o ​
⇒ 
x = 20
o
∴ ∠A = 60o​, ∠B = 80o, ∠C = 100o and ∠D = 120o
Hence, the smallest angle is
60°.

Page No 349:

Answer:

(c) 45°

Explanation: 
∠B = 180o − ∠A 
⇒ ∠B = 180o − 75o = 105o
Now, ∠B =​ ∠ABD + ∠CBD 
⇒​​ 105o​ = ∠ABD + 60o
⇒ ∠ABD​ = 105o − 60o = 45o
⇒ ∠ABD = ​∠BDC​ = 45         (Alternate angles)

Page No 349:

Answer:


We know that diagonals of rhombus bisect each other at 90°.

Then, in ΔBOC,

90° + 50° + ∠OBC = 180°        (Angle sum property of triangle)

 ∠OBC = 180° - 140°

 OBC = 40°

But OBC = ADB     (Alternate interior angles)

Thus, 
ADB = 40°

Hence, the corerct option is (a).

Page No 349:

Answer:

(d) Rectangle.

The diagonals of a rectangle are equal.

Page No 349:

Answer:

(d) rhombus

The diagonals of a rhombus bisect each other at right angles.



Page No 350:

Answer:

(a) 10 cm

Explanation:

Let ABCD be the rhombus.
∴ AB = BC = CD = DA
Here, AC and BD are the diagonals of ABCD, where AC = 16 cm and BD = 12 cm.
Let the diagonals intersect each other at O.
We know that the diagonals of a rhombus are perpendicular bisectors of each other.
∴​ ∆​AOB is a right angle triangle, in which OA = AC /2 = 16/2 = 8 cm and OB = BD/2 = 12/2 = 6 cm.
Now, AB2 = OA2 + OB2              [Pythagoras theorem]
⇒ ​AB2 = (8)2 + (6)2
⇒​ AB2 =​ 64 + 36 = 100
⇒​ AB =​ 10 cm

Hence, the side of the rhombus is 10 cm.

Page No 350:

Answer:

(b) 12 cm

Explanation: 


Let ABCD be the rhombus.
∴ AB = BC = CD = DA = 10 cm
Let AC and BD be the diagonals of the rhombus.  
Let 
AC be x and BD be 16 cm and O be the intersection point of the diagonals.
We know that the diagonals of a rhombus are perpendicular bisectors of each other.
∴​ AOB is a right angle triangle in which OA = AC ÷2 = ÷
÷ 2  and OB = BD ÷÷2 = 16 ÷÷ 2 = 8 cm.

Now,
AB2= OA2 + OB2              [Pythagoras theorem]
102 = x22 + 82x22 = 36 = 62x = 2×6 =12 cm

102 = (x2)2 + 82100  64 = x2436 ×4 = 

Page No 350:

Answer:

 
Given: In rectangle ABCD, ∠OAD = 35
°.

Since, ∠BAD = 90°

 ∠OAB = 90° - 35° = 55°

In ΔOAB,

Since, OA = OB      (Diagonals of a rectangle are equal and bisect each other)

 OAB = OBA = 55°     (Angles opposite to equal sides are equal)

Now, in ΔODA,
 

55° + 55° + ∠DOA = 180°       (Angle sum property of a triangle)

 DOA = 180° - 110°

 DOA = 70°

Thus, the acute angle between the diagonals is 70°.

Hence, the correct option is (b).

Page No 350:

Answer:

(c) Rectangle

Explanation:
A = ∠B
Then A + ∠B = 180o
2A = 180o
⇒ ∠A​ = 90o
⇒ ∠A​ =​ ∠B​ =​∠C​ =​​∠D = 90o
∴​ The parallelogram is a rectangle.

Page No 350:

Answer:

(b) 50o

​​Explanation:

C = 70o and ∠D = 30o
Then A + ∠B = 360o - (70 +30)o = 260o
 12(∠A +B) =12 (260o) = 130o
In ∆​ AOB, we have:
AOB​ = 180o - [12(∠A +B)​]
⇒ ∠AOB​ = ​180 - 130 = 50o

Page No 350:

Answer:

(d) 90° 

Explanation:
Sum of two adjacent angles = 180o
Now, sum of angle bisectors of two adjacent angles = 12×180o = 90o
∴ Intersection angle of bisectors of two adjacent angles =  180o − 90o =  90o

Page No 350:

Answer:

(c) Rectangle

The bisectors of the angles of a parallelogram encloses a rectangle.

Page No 350:

Answer:



Given: In quadrilateral ABCDAS, BQ, Cand DS are angle bisectors of angles A, B, C and D, respectively.

QPS = APB        (Vertically opposite angles)          ...(1)

In APB,

APB + PAB + ABP180°        (Angle sum property of triangle.)

 ∠APB + 12A + 12B = 180°

 APB  = 180° – 12(∠A + B)                  ...(2)

From (1) and (2), we get

QPS = 180° – 12(∠+ ∠B)                          ...(3)

Similarly, QRS = 180° – 12(∠C + D)        ...(4)

From (3) and (4), we get

QPS + QRS = 360° – 12(∠A + ∠B + C + D)

                          = 360° – 12(360°)

                          = 360° – 180°

                          = 180°

So, PQRS is a quadrilateral whose opposite angles are supplementary.

Hence, the correct option is (d).

Page No 350:

Answer:

(d) parallelogram

The figure formed by joining the mid points of the adjacent sides of a quadrilateral is a parallelogram.

Page No 350:

Answer:

(b) Square

The figure formed by joining the mid points of the adjacent sides of a square is a square.

Page No 350:

Answer:

(d) parallelogram.
The figure made by joining the mid points of the adjacent sides of a parallelogram is  a parallelogram.

Page No 350:

Answer:

(a) rhombus

The figure formed by joining the mid points of the adjacent sides of a rectangle is a rhombus.



Page No 351:

Answer:

(c) Rectangle

The figure formed by joining the mid points of the adjacent sides of a rhombus is a rectangle.

Page No 351:

Answer:

Since,

The quadrilateral formed by joining the mid-points of the sides of a parallelogram is parallelogram ,

The quadrilateral formed by joining the mid-points of the sides of a rectangle is rhombus,

​The quadrilateral formed by joining the mid-points of the sides of a quadrilateral with diagonals equal is rhombus, and

​The quadrilateral formed by joining the mid-points of the sides of a quadrilateral with diagonals perpendicular to each other is rectangle.

Hence, the correct option is (d).

Page No 351:

Answer:

Given:

The quadrilateral ABCD is a rhombus.

So, the sides AB, BC, CD and AD are equal.

Now, in PQS, we have

DC=12QS     (Using mid-point theorem)          ...(1)

Similarly, in PSR,

BC=12PR                     ...(2)

As, BC = DC

12QS = 12PR         [From (1) and (2)]

So, QS = PR

Thus, the diagonals of PQRS are equal.

Hence, the correct option is (c).

Page No 351:

Answer:

Since,

The quadrilateral formed by joining the mid-points of the sides of a rhombus is rectangle,

The quadrilateral formed by joining the mid-points of the sides of a quadrilateral with diagonals equal is rhombus,

​The quadrilateral formed by joining the mid-points of the sides of a quadrilateral with diagonals perpendicular is rectangle, and

​The quadrilateral formed by joining the mid-points of the sides of a quadrilateral with diagonals equal and perpendicular is square.

Hence, the correct option is (d).

Page No 351:

Answer:

(c) 72°​

Explanation: 
 Let ABCD be a parallelogram. 
  ∴A = ∠C and ∠B = ∠D      (Opposite angles)
   Let Ax and ∠​B23x(4x5)
  ∴ ∠​A + ∠B = 180o                (Adjacent angles are supplementary)
 x 23x 180o
53x=180°
⇒ x = 108o
∴ B = 23× (108o) = 72o
Hence
AC  = 108o  and B = D = 72o
9x5 = 180ox = 100oA = 100o an

Page No 351:

Answer:

(c)112°​

Explanation:
Let ABCD is a parallelogram. 
  ∴ ∠​A = ∠C and ∠​B = ∠D          (Opposite angles)
 Let A be the smallest angle whose measure is x.
 ∴​∠B  = (2x − 24)o
 Now, ∠​A + ∠B = 180o                (Adjacent angles are supplementary) 
   x + 2x − 24o = 180o
   3x =  204o
  ⇒ x = 68o
∴​∠​B = 2 ⨯ 68o − 24o = 112o
 HenceA = C68o and B = D = 112o

Page No 351:

Answer:

(c) Trapezium

Explanation:

Let the angles be (3x), (7x), (6x) and (4x)​.
Then 3x + 7x + 6x + 4x = 360o
x = 18o
​Thus, the angles are 3 ⨯18o = 54o, ​7 ⨯ 18o = 126o, ​6 ⨯ 18o = 108o and ​4 ⨯18o = 72o.
But 54o + 126o = 180o and 72o + 108o  = 180o
ABCD is a trapezium.

Page No 351:

Answer:

(c) Opposite angles are bisected by the diagonals.

Page No 351:

Answer:

(c) Rectangle

If APB and CQD are two parallel lines, then the bisectors of APQ, ∠BPQ, ∠CQP and ∠PQD enclose a rectangle.



Page No 352:

Answer:

(c)​ 60°

Explanation:
∠BAD = ∠BCD = 75o       [Opposite angles are equal]
In ∆ BCD, ∠ C = 75o            
∴ ​∠CBD = 180o (75o + 45o) = 60o

Page No 352:

Answer:

(c) A < B

Explanation:
Let h be the height of parallelogram.
Then clearly, h < b
∴ ​A = a ⨯​ h < a ⨯ b = B
Hence, A < B

Page No 352:

Answer:

(b) AF = 2 AB

Explanation:
I​n parallelogram ABCD, we have:
AB || DC
DCE = ∠​ EBF            (Alternate interior angles)
In ∆ DCE and ​ ∆ BFE, we have:
DCE = ∠ EBF              (Proved above)

DEC = ∠ BEF              (Vertically opposite angles)
BECE           ( Given)
i.e., ∆ DCE ≅​ ∆ BFE     (By ASA congruence rule)
∴  DC = BF         (CPCT)

But DC= AB, as ABCD is a parallelogram.
DC = AB =  BF                 ...(i)

Now, AF = AB + BF             ...(ii)
   From (i), we get:
∴ AF = AB + AB = 2AB

Page No 352:

Answer:


Given: In ∆ABCMND and E are the mid-points of BP, CP, AB and AC, respectivley.

In ∆ABP,

 D and M are the mid-points of AB,and BP, respectively.      (Given)

 BM12AP and BM || AP          (Mid-point theorem)       ...(i)

Again, in ∆ACP,

 E and N are the mid-points of AC,and CP, respectively.      (Given)

 EN = 12AP and EN || AP          (Mid-point theorem)       ...(ii)

From (i) and (ii), we get

BMEN and BM || EN

But this a pair of opposite sides of the quadrilateral DENM.

So, DENM is a parallelgram.

Hence, the correct option is (b).

Page No 352:

Answer:

(b) 12a+b

Explanation:

Suppose ABCD is a trapezium.
Draw EF parallel to AB.
Join BD to cut EF at M.
Now, in ∆ DAB, E is the midpoint of AD and EM || AB.
∴ M is the mid point of BD and EM = 
12a
Similarly, M is the mid point of BD and MF || DC.
i.e., F is the midpoint of BC and MF = 12b

∴ EF =  EM + MF12a+b
 

Page No 352:

Answer:

(d) 12AB - CD

Explanation:

Join CF and produce it to cut AB at G.
Then ∆CDF  ≅ GBF                [∵ DF = BF, ​DCF = ​BGF and ​CDF = ​GBF]
∴ CD = GB
Thus, in 
∆​CAG, the points E and F are the mid points of AC and CG, respectively. 
∴ EF12AG = 12AB - GB =12AB - CD 

Page No 352:

Answer:

(c) ​90°

Explanation:
∠B = ∠D  
12∠B = 12∠D​
⇒​ ∠ADB = ∠​ABD
∴ ∆ABD is an isosceles triangle and M is the midpoint of BD. We can also say that M is the median of ∆ABD.
∴ AM ⊥ BD and, hence, ​∠AMB =​​ 90°

Page No 352:

Answer:

(c) ​AC2 + BD2 = 4AB2

Explanation:
We know that the diagonals of a rhombus bisect each other at right angles.
Here, OA = 12AC, OB = 12BD and ∠​AOB ​= 90°
Now, AB2= OA2 + OB214(AC)214(BD)2
∴ 4AB2 = (AC2 + BD2)



Page No 353:

Answer:

(c) BC2 + AD2 + 2AB.CD


Explanation:

Draw perpendicular from D and C on AB which meets AB at E and F, respectively.
 ∴​ DEFC is a parallelogram and EF = CD.

In ∆ABC, ∠B is acute.
∴ AC2BC2 + AB2 - 2AB.AE
In ∆ABD, ∠A is acute.​
∴ ​BDAD2 + AB2 - 2AB.AF
∴ ​AC2BD2 = (BC2AD2) + (AB2 + AB2 ) - 2AB(AE + BF)
                     = (BC2 + AD2) + 2AB(AB - AE - BF)                [∵ AB = AE + EF + FB and AB - AE =  BE]
                     = (BC2 + AD2) + 2AB(BE - BF)
                     = (BC2 + AD2) + 2AB.EF 
AC2 + BD2 = ​(BC2 + AD2) + 2AB.CD          

Page No 353:

Answer:

(d) 1:1

Area of a parallelogram = base ⨯ height
If both parallelograms stands on the same base and between the same parallels, then their heights are the same.
So, their areas will also be the same.

Page No 353:

Answer:

(b) ⅓ AC

Explanation:

Let G be the mid point of FC. Join DG.
​In BCF, D is the mid point of BC and G is the mid point of FC.
∴ DG || BF 
⇒ DG || EF

​In ∆ ADG, E is the mid point of AD and EF || DG.
i.e., 
F is the mid point of AG.
Now,
AF = FG = GC       [∵ G is the mid point of FC] 
∴ AF =⅓ AC
4AC

Page No 353:

Answer:

(a) 40°

Explanation:
OAD = ​∠OCB = 30o              (Alternate interior angles)
AOB + ∠BOC = 180o              (Linear pair of angles)
∴ ∠BOC = 180o − 70o = 110o       (∠ AOB = 70o)
In ∆BOC, we have:
OBC = 180o − (110o + 30o) = 40o
∴ ​∠DBC = 40o

Page No 353:

Answer:

(c) I and II

Statements I and II are true. Statement III is false, as the triangle formed by joining the midpoints of the sides of an isosceles triangle is an isosceles triangle.

Page No 353:

Answer:

(b) II and III

Clearly, statements II and III are true. Statement I is false, as diagonal of a rectangle does not bisect ∠A and ∠C (∴ adjacent sides are not equal).

Page No 353:

Answer:

Since, the opposite angles of the quadrilateral PQRS are equal.

 Quadrilateral PQRS is a parallelogram.

SR = PQ  (Opposite sides of a parallelogram are equal.)

 SR = PQ = 2 cm



Page No 354:

Answer:

No, the statement is incorrect because the diagonals of a parallelogram bisect each other.

Page No 354:

Answer:

Since, in quadrilateral PQRS,P + ∠S = 180°.

i.e. the sum of adjacent angles is 180°.

So, PQRS is a parallelogram.

Page No 354:

Answer:

No, the statement is false because if all the four angles of a quadrilateral are less than 90°, then the sum of all four angles will be less than 360°.

Page No 354:

Answer:

Yes, the statement is true because all the angles of a quadrilateral such as rectangle and square are right angles.

Page No 354:

Answer:

No, the statement is false because if all angles are greater than 90°, then the sum of four obtuse angles will be greater than 360°.

Page No 354:

Answer:

Since, the sum of all angles (i.e. 70° + 115° + 60° + 120° = 365°).

So, we cannot form a quadrilateral with these angles.

Page No 354:

Answer:

We know that, sum of all angles in a quadrilateral is 360°.

Let each angle of the quadrilateral be x.

x + x = 360°

⇒ 4x = 360°

⇒ x = 90°

⇒ All angles of the quadrilateral are 90°.

Hence, given quadrilateral is a rectangle.

Page No 354:

Answer:

In ABC,

Since, D and E are respectively the mid-points of sides AB and BC    (Given)

So, DE = 12AC       (Uing mid-point theorem)

But AC = 3.6 cm      (Given)


DE 12(3.6)

or, DE = 1.8 cm

Hence, the length of DE is 1.8 cm.

Page No 354:

Answer:

Since, the diagonals PR and QS of quadrilateral PQRS bisect each other.        (Given)

So, PQRS is a parallelogram.

Now, Q+R=180°(Adjacent angles are supplementary.)

56°+R=180°R=180°-56°R=124°

Page No 354:

Answer:



Given: Parallelograms BDEF and AFDE.

Since, BF = DE     (Opposite sides of parallelogram BDEF)       ...(i)

And, AF = DE     (Opposite sides of parallelogram AFDE)       ...(ii)

From (i) and (ii), we get

AF = FB
 

Page No 354:

Answer:

We know that if the diagonals of a ​quadrilateral bisects each other, then it is a parallelogram.
∴ I gives the answer.
If the diagonals of a quadrilateral are equal, then it is not necessarily a ​parallelogram.
∴ II does not give the answer.

Hence, the correct answer is (a).



Page No 355:

Answer:

Clearly, I alone is not sufficient to answer the given question.
Also, II alone is not sufficient to answer the given question.​
However, both I and II together will give the answer.
∴ Hence, the correct answer is (c).

Page No 355:

Answer:

When the diagonals of a parallelogram are equal, it is either a rectangle or a square.
Also, if the diagonals intersects at a right angle, then it is a square.
∴ Both I and II together will give the answer.
Hence, the correct answer is (c).

Page No 355:

Answer:

We know that a quadrilateral is a parallelogram when either I or II holds true.
​Hence, the correct answer is (b).

Page No 355:

Answer:

(a)  Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

Explanation:
Fourth angle = 360o − (130o + 70o + 60o) = 100o
Clearly, reason (R) and assertion (A) are both true and the reason gives the assertion.
Hence, the correct answer is (a).

Page No 355:

Answer:

(a)  Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

​Explanation:
Clearly, reason (R) and assertion (A) are both true and reason (R) gives assertion (A).
Hence, the correct answer is (a).

Page No 355:

Answer:

(b)  Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.

​Explanation:
Clearly, reason (R) and assertion (A) are both true.

But reason (R) does not give assertion (A).
Hence, the correct answer is (b).



Page No 356:

Answer:

(d) Assertion is false and Reason is true.

​Explanation:
We can easily prove reason (R). So, reason (R) is true.

Clearly,  assertion (A) is false (as every parallelogram is not necessarily a rectangle).
Hence, the correct answer is (d).

Page No 356:

Answer:

​(b)  Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.

​Explanation:
Clearly, assertion (A) is true.
We can easily prove reason (R). So, (R) is also true.

But, reason (R) does not give assertion (A).
Hence, the correct answer is (b)

Page No 356:

Answer:

(a) - (q), (b) - (r), (c) - (s), (d) - (p)

Page No 356:

Answer:

​(a) - (r), (b) - (s), (c) - (p), (d) - (q)

Explanation:

(a) PQ12(AB+ CD) = 12(17) = 8.5 cm
 
(b) OR12(PR) = ​12(13) = 6.5 cm



View NCERT Solutions for all chapters of Class 9