RS Aggarwal Solutions Class 9 Maths Chapter 10 Quadrilaterals (Updated For 2021-22)

RS Aggarwal Solutions Class 9 Maths Chapter 10 Quadrilaterals

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RS Aggarwal Solutions Class 9 Maths Chapter 10 Quadrilaterals

 


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Page No 312:

Question 1:

Three angles of a quadrilateral are 75°, 90° and 75°. Find the measure of the fourth angle.

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°75°+90°+75°+x=360°⇒240°+x=360°

x=360°240°x=120°⇒x=360°-240°⇒x=120°

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

Page No 312:

Question 2:

The angles of a quadrilateral are in the ratio  2: 4 : 5 : 7. Find the angles.

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 = 40B = 80C = 100D = 140

Page No 312:

Question 3:

In the adjoining figure, ABCD is a trapezium in which AB || DC. If ∠A = 55° and ∠B = 70°, find ∠C and ∠D.

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 = 180 − A  
∴  D = 180 − 55 = 125

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

Page No 312:

Question 4:

In the adjoining figure, ABCD is a square and ∆EDC is an equilateral triangle. Prove that (i) AE = BE, (ii) ∠DAE = 15°.

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 ∆BCEwe 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 = 12(180°  150°) = 30°2 =15°12180° – 150° = 30°2 =15°
 

Page No 312:

Question 5:

In the adjoining figure, BM ⊥ AC and DN ⊥ AC. If BM = DN, prove that AC bisects BD.

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:

Question 6:

In the given figure, ABCD is a quadrilateral in which AB = AD and BC = DC. Prove that
(i) AC bisects ∠A and ∠C,
(ii) BE = DE,
(iii) ∠ABC = ∠ADC.

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 ∆ADEwe 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:

Question 7:

In the given figure, ABCD is a square and ∠PQR = 90°. If PB = QC = DR, prove that
(i) QB = RC,
(ii) PQ = QR,
(iii) QPR = 45°.

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 = 12(180°  90°) = 90°2 = 4512180° – 90° = 90°2 = 45°

Page No 312:

Question 8:

If O is a point within a quadrilateral ABCD, show that OA + OB + OC + OD > AC + BD.

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 ∆AOCOA + OC > AC

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

Page No 313:

Question 9:

In the adjoining figure, ABCD is a quadrilateral and AC is one of its diagonals. Prove that:
(i) AB + BC + CD + DA > 2AC
(ii) AB + BC + CD > DA
(iii) AB + BC + CD + DA > AC + BD

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 ∆ABCAB + BC > AC            …(1)

In ∆ACDCD + 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 ∆ABCAB + BC > AC
    In ∆ACDCD + DA > AC
    In ∆ BCDBC CD > BD
   In ∆ ABDDA + AB > BD
   ​Adding these inequalities, we get:
   2(AB + BC + CD + DA) > 2(AC + BD)
   ⇒ (AB + BC + CD + DA) > (AC + BD)
 

Page No 313:

Question 10:

Prove that the sum of all the angles of a quadrilateral is 360°.

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:

Question 1:

In the adjoining figure, ABCD is a parallelogram in which ∠A = 72°. Calculate ∠B, ∠C and ∠D.

ANSWER:

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

Page No 328:

Question 2:

In the adjoining figure, ABCD is a parallelogram in which ∠DAB = 80° and ∠DBC = 60°. Calculate ∠CDB and ∠ADB.

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:

Question 3:

In the adjoining figure, M is the midpoint of side BC of a parallelogram ABCD such that ∠BAM = ∠DAM. Prove that AD = 2CD.

ANSWER:



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

To prove: AD = 2CD
 
Proof:

Since, ADBCAD∥BC and AM is the transversal.

So, DAM=AMB∠DAM=∠AMB      (Alternate interior angles)

But, DAM=BAM∠DAM=∠BAM (Given)

Therefore, AMB=BAM∠AMB=∠BAM

AB=BM⇒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⇒2AB=2CD⇒(AB+AB)=2CD

BM+MC=2CD⇒BM+MC=2CD      (AB BM and MC = BM)

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

Page No 329:

Question 4:

In the adjoining figure, ABCD is a parallelogram in which ∠A = 60°. If the bisectors of ∠A and ∠B meet DC at P, prove that (i) ∠APB = 90°, (ii) AD = DP and PB = PC = BC, (iii) DC = 2AD.

ANSWER:

ABCD is a parallelogram.
∴ ​∠A = ​∠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°60°2=30° and ∠PBA = 120°2=60°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 andtherefore, 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:

Question 5:

In the adjoining figure, ABCD is a parallelogram in which ∠BAO = 35°, ∠DAO = 40° and ∠COD = 105°. Calculate (i) ∠ABO, (ii) ∠ODC, (iii) ∠ACB, (iv) ∠CBD.

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) ∠CBD = ABC  ABD            …(i)

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

Page No 329:

Question 6:

In a || gm ABCD, if ∠A = (2x + 25)° and ∠B = (3x − 5)°, find the value of x and the measure of each angle of the parallelogram.

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 =  32 − 5 = 91o
Hence, x = 32o, ∠A = C = 89o and ∠B = D = 91o 

Page No 329:

Question 7:

If an angle of a parallelogram is four-fifths of its adjacent angle, find the angles of the parallelogram.

ANSWER:

 Let ABCD be a parallelogram. 
∴ ∠​A = ∠and B = ∠D           (Opposite angles)
Let A = xo and ∠B = (4x5)4×5°
Now, ∠​A + ∠B = 180o                 (Adjacent angles are supplementary)
  x + 4x5 = 180o9x5 = 180ox = 100oNow, A = 100o and B = (45)×100° 

= 80o⇒x + 4×5 = 180o⇒9×5 = 180o⇒x = 100oNow, ∠A = 100o and ∠B = 45×100° = 80o
 HenceA = C = 100o; B = ∠D = 80o

Page No 329:

Question 8:

Find the measure of each angle of a parallelogram, if one of its angles is 30° less than twice the smallest angle.

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
 HenceA = C = 70o; ∠B = ∠D = 110o

Page No 329:

Question 9:

ABCD is a parallelogram in which AB = 9.5 cm and its perimeter is 30 cm. Find the length of each side of the parallelogram.

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:

Question 10:

In each of the figures given below, ABCD is a rhombus. Find the value of x and y in each case.

ANSWER:

ABCD is a rhombus and a rhombus is also a parallelogram. A rhombus has four equal sides.
(i)​ In ∆ABC, ∠​BAC = ∠BCA = 12(180  110) = 35o12180 – 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​DAB = 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 ∆​ ABCAB = BC, so ∠​BAC = ∠​ACB]
    Also, ∠​BOC = 90o
  ∴ ∠​OBC = 180o − (90o + 62o) = 28o
  Hence, x = 62oy = 28o

Page No 329:

Question 11:

The lengths of the diagonals of a rhombus are 24 cm and 18 cm respectively. Find the length of each side of the rhombus.

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:

Question 12:

Each side of a rhombus is 10 cm long and one of its diagonals measures 16 cm. Find the length of the other diagonal and hence find the area of the rhombus.

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 = (x2)2 + 82100  64 = x2436 ×4 = x⇒102 = x22 + 82⇒100 – 64 = x24⇒36 ×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 cm212×12×16 = 96 cm2

Page No 330:

Question 13:

In each of the figures given below, ABD is a rectangle. Find the values of x and y in each case.

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°) = 35o12×180° – 110° = 35o
∴​ y = ∠BAC 35o                 [Interior alternate angles]
Also, x = 90o − y                          [ ​∵∠C = 90o =  x + y ]
⇒​ x = 90o − 35o = 55o                     
Hence, x = 55o and y = 35o​​

Page No 330:

Question 14:

In a rhombus ABCD, the altitude from D to the side AB bisects AB. Find the angles of the rhombus.

ANSWER:



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

In ΔAED and ΔBED,In ∆AED and ∆BED,

DE=DEDE=DE                         (Common side)

DEA=DEB=90°∠DEA=∠DEB=90°     (Given)

AE=EBAE=EB                          (Given)

 ΔAEDΔBED∴ ∆AED≅∆BED         (By SAS congruence Criteria)

AD=BD⇒AD=BD                    (CPCT)

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

AD=AB=BD⇒AD=AB=BD

Thus, ΔABD ∆ABD is an equilateral triangle.

Therefore, A=60°∠A=60°

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

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

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

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

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

Page No 330:

Question 15:

In the adjoining figure, ABCD is a square. A line segment CX cuts AB at X and the diagonal BD at O such that ∠COD = 80° and ∠OXA = x°. Find the value of x.

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:

Question 16:

In a rhombus ABCD show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.

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 ΔABC∆ABC,

AB = BC  AB = BC         (Sides of rhombus are equal.)

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

Now,

ADBCAD∥BC          (Opposite sides of rhombus are parallel.)

AC is transversal.

So, 1=4∠1=∠4        (Alternate interior angles)          …(2)

From (1) and (2), we get

1=2∠1=∠2 

Thus, AC bisects A∠A.

Similarly,

Since, ABDCAB∥DC and AC is transversal.

So, 2=3∠2=∠3     (Alternate interior angles)     …(3)

From (1) and (3), we get

4=3∠4=∠3

Thus, AC bisects ∠C.

Hence, AC bisects C and A∠C and ∠A

In ΔDAB∆DAB,

AD = AB  AD = AB                   (Sides of rhombus are equal.)

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

Now,

DCABDC∥AB                        (Opposite sides of rhombus are parallel.)

BD is transversal.

So, CDB=DBA∠CDB=∠DBA        (Alternate interior angles)          …(5)

From (4) and (5), we get

ADB=CDB∠ADB=∠CDB 

Thus, DB bisects D∠D.

Similarly,

Since, ADBCAD∥BC and BD is transversal.

So, CBD=ADB∠CBD=∠ADB     (Alternate interior angles)     …(6)

From (4) and (6), we get

CBD=ABD∠CBD=∠ABD

Thus, BD bisects ∠B.

Hence, BD bisects D and B.∠D and ∠B.

Page No 330:

Question 17:

In a parallelogram ABCD, points M and N have been taken on opposite sides AB and CD respectively such that AM CN. Show that AC and MN bisect each other.

ANSWER:


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

Proof:
Since, ABCD is a parallelogram.

ABDCAMNC⇒AB∥DC⇒AM∥NC

Also, AM = CN           (Given)

Thus, AMCN is a parallelogram.

Since, diagonals of a parallelogram bisect each other.

Hence, AC and MN bisect each other.

This is the complete blog on the RS Aggarwal Solutions Class 9 Maths Chapter 10 Quadrilaterals. To know more about the CBSE Class 9 Maths exam, ask in the comments.

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