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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°

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∘ â€‹
x = 20∘
 A = 40∘; B = 80∘C = 100∘; D = 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 = â€‹DCE =  60∘.
To prove:  AE = BE and 
DAE = 15∘​
Proof: In âˆ†ADE and âˆ†BCE, we have:
AD = BC              [Sides of a square]

DE = EC​             [Sides of an equilateral triangle]
ADE BCE = 90∘+  60∘ = 150∘​ 
∴ âˆ†ADE ≅  âˆ†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 âˆ†BCE are isosceles triangles.
i.e., DAE = DEA = 12180° - 150° = 30°2 =15°
 

Page No 312:

Question 5:

In the adjoining figure, BM AC and DNAC. 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 âˆ†OND and âˆ†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 âˆ†ABC and âˆ†ADCwe have:
AB = AD                                                  (Given)

BC = DC                                                 (Given)
AC is common.
i.e., âˆ†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.
∴ âˆ†ABE ≅  âˆ†ADE                          (SAS congruence rule)
⇒ BE = DE                                                 (By CPCT)
 
(iii)  
∆ABC ≅  âˆ†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 = â€‹BC − RC
∴ BQ = RC

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

(iii) ∆RCQ ≅ âˆ†QBP and QR = PQ (Proven above)
∴ In ∆RPQ, ∠QPR = ∠QRP = 12180° - 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 âˆ†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:

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 âˆ†ABC, AB + BC > AC            ...(1)

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

(ii) In âˆ†ABC, we have :
 â€‹  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 âˆ†ACD, we have:​
  AC > |DA − CD|​        ...(2)
   From (1) and (2), we have:
   AB + BC > |DA − CD|​
  ⇒ AB + BC + CD > DA

(iii) In âˆ†ABC, AB + BC > AC
    In âˆ†ACD, CD + DA > AC
    In âˆ† BCD, BC CD > BD
   In âˆ† ABD, DAAB > BD
   â€‹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 âˆ†ABD, we have:
∠1 + ∠2 + â€‹∠A = 180o      ...(i)
In â€‹âˆ†BCD, we have:
∠3 + ∠4 + ∠C = 180o     ...(ii)
On adding (i) and (ii), we get:

(∠1 + ∠3) + ∠A + ∠C + (∠4 + ∠2) â€‹= 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°â€‹.
We know that opposite angles of a parallelogram are equal.
∴∠A â€‹= ∠C and B â€‹= ∠D â€‹ â€‹
∴ ∠C = 72o
A and ∠B are adajcent angles.
i.e., ∠A â€‹+ ∠B​ = 180o
⇒ ∠B = 180o  ∠A
​⇒ ∠B​ = 180o 72o = 108o
∴​ ∠B​ =​ ∠D108o
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°â€‹ 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 â€‹+ ∠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, 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:

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.
∴ â€‹∠A = â€‹∠C and B =​ ∠D (Opposite angles)
And â€‹∠A + ​∠B = 180o                (Adjacent angles are supplementary)
∴ â€‹∠B = 180o − ∠
180o − 60o = 120o             ( âˆµ∠A = 60o)
∴ â€‹∠A = â€‹∠C = 60o and B =​ ∠D​ = 120o

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

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

   In âˆ† PBC, ∠​ PBC = 60o ∠​ BPC = 180o − (90o +30o) = 60oand ∠​ BCP  = 60o (Opposite angle of ∠A)
   ∴ ∠ PBC = ∠​ BPC = ∠​ BCP
   Hence, âˆ†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:

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 â€‹âˆ£âˆ£â€‹ AD

(i) In âˆ†AOB, ∠BAO = 35°, â€‹∠AOB = ∠COD = 105°  (Vertically opposite angels)
∴ â€‹∠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)   â€‹
​⇒∠ABC = 180o − 75o = 105o   
⇒∠CBD = 105o ABD                         (∠ABDABO)
⇒∠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)   â€‹
∴​ (2x + 25)°â€‹ + (3x − 5)°â€‹ = 180
⇒ ​5x +20 = 180
⇒​ 5x = 160
⇒​ x = 32o
∴​∠A = 2 â¨¯ 32 + 25 = 89o and ∠B = 3 ⨯ 32 − 5 = 91o
Hence, x = 32o, ∠AC89o and ∠BD = 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 = ∠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:

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
 HenceAC = 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 âˆ†ABC, ∠​BAC = ∠BCA = 12180 - 110 = 35o
i.e., x = 35o   
Now, ∠​B + ∠C = 180o                 (Adjacent angles are supplementary)  â€‹
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 âˆ†â€‹AOB, ​∠OAB = 40o, ∠AOB = 90o and ∠ABO = 180o − (40o + 90o) = 50o
   ∴ ​x = 50o
   In âˆ†â€‹AB​D, AB = AD
  
So, ∠ABD = ​∠ADB = ​50o
   Hence, x = 50o;  y = 50o


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

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 âˆ†AOB, we have:
      OA = OB   
Now, ∠​OAB = ∠OBA = 12×180° - 110° = 35o
∴​ y = ∠BAC = 35o                 [Interior alternate angles]
Also, x = 90oy                          [ â€‹âˆµ∠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 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:

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 âˆ†BOX, we have:
∠AXO = ∠OBX + â€‹∠BOX        [Exterior angle of âˆ†BOX]
⇒​ ∠AXO = 45o + 80o = 125o
∴ â€‹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 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:

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 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:

Question 18:

In the adjoining figure, ABCD is a parallelogram. If P and Q are points on AD and BC respectively such that AP=13AD and CQ=13BC, prove that AQCP is a parallelogram.

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.
∴ â€‹AP = CQ                                   [∵ AD = BC]
In âˆ†â€‹DPC and âˆ†â€‹BQA, we have:
AB = CD, ∠B = ∠D and DP = QB                        [∵DP23​ AD and QB = 23BC 
i.e., ∆​DPC ≅ ∆​BQA
∴​ PC  = QA

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

Page No 330:

Question 19:

In the adjoining figure, ABCD is a parallelogram whose diagonals intersect each other at O. A line segment EOF is drawn to meet AB at E and DC at F. Prove that OE = OF.

Answer:

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

Page No 330:

Question 20:

The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 60°. Find the angles of the parallelogram.

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:

Question 21:

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.

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:

Question 22:

In the adjoining figure, ABCD is a parallelogram in which AB is produced to E so that BE = AB. Prove that ED bisects BC.

Answer:

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

Page No 331:

Question 23:

In the adjoining figure, ABCD is a parallelogram and E is the midpoint of side BC. If DE and AB when produced meet at F, prove that AF = 2AB.

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 âˆ†DCE and ∆BFE, we have:
 ∠DCE = ∠EBF              (Proved above)

 ∠DEC = ∠BEF              (Vertically opposite angles)
Also, BE = CE           (Given)
∴ âˆ†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:

Question 24:

Two parallel lines l and m are intersected by a transversal t. Show that the quadrilateral formed by the bisectors of interior angles is a rectangle.

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:

Question 25:

K, L, M and N are points on the sides AB, BC, CD and DA respectively of a square ABCD such that AK = BL = CM = DN. Prove that KLMN is a square.

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:

Question 26:

A ∆ABC is given. If lines are drawn through A, B, C, parallel respectively to the sides BC, CA and AB, forming ∆PQR, as shown in the adjoining figure, show that BC=12QR.

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:

Question 27:

In the adjoining figure, ∆ABC is a triangle and through A, B, C, lines are drawn, parallel respectively to BC, CA and AB, intersecting at P, Q and R. Prove that the perimeter of ∆PQR is double the perimeter of ∆ABC.

Answer:

Perimeter of âˆ†â€‹ABC = AB + BC + CA                         ...(i)
Perimeter of âˆ†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 = â€‹12PQ and AB = â€‹12PR

From (i) and (ii), we have:
Perimeter of âˆ†â€‹ABC  = 12QR + 12PQ + 12PR
                              = 12PR + QR + PQ
 
i.e., Perimeter of âˆ†â€‹ABC  = 12 (Perimeter of âˆ†â€‹PQR)
∴ Perimeter of âˆ†PQR = 2 ⨯ Perimeter of âˆ†ABC 



Page No 345:

Question 1:

P, Q, R and S are respectively the midpoints of the sides AB, BC, CD and DA of a quadrilateral ABCD. Show that
(i) PQ || AC and PQ = 12AC
(ii) PQ || SR
(iii) PQRS is a parallelogram.

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:

Question 2:

A square is inscribed in an isosceles  right triangle so that the square and the triangle have one angle common. Show that the vertex of the square opposite the vertex of the common angle bisects the hypotenuse.

Answer:



Given: In an isosceles right âˆ†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, âˆ†ABC is an isosceles with AC = BC.

⇒ AN + NC = CM + MB

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

Now,

In âˆ†ANP and âˆ†PMB,

AN = MB                [From (i)]

ANP = ∠PMB = 90°

PN = PM             (Sides of square CMPN)

∴ By SAS congruence criteria,

∆ANP  âˆ†BMP


Hence, AP = PB         (By CPCT)



Page No 346:

Question 3:

In the adjoining figure, ABCD is a || gm in which E and F are the midpoints of AB and CD respectively. If GH is a line segment that cuts AD, EF and BC at G, P and H respectively, prove that GP = PH.

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:

Question 4:

M and N are points on opposite sides AD and BC of a parallelogram ABCD such that MN passes through the point of intersection O of its diagonals AC and BD. Show that MN is bisected at O.

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:

Question 5:

In the adjoining figure, PQRS is a trapezium in which PQ || SR and M is the midpoint of PS. A line segment MN || PQ meets QR at N. Show that N is the midpoint of QR.

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 âˆ†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 âˆ†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:

Question 6:

In a parallelogram PQRS, PQ = 12 cm and PS = 9 cm. The bisector of ∠P meets SR in M. PM and QR both when produced meet at T. Find the length of RT.

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 âˆ¥ SR

⇒ ∠TMR = ∠TPQ = x.

In â–³TMR, SRQ is an exterior angle.

⇒ ∠SRQ = ∠TMR + ∠MTR

⇒ 2= + ∠MTR

⇒ ∠MTR = x

⇒ â–³TPQ is an isosceles triangle.

TQ = PQ = 12 cm

Now,

RT = TQ QR

      
= TQ PS

      
= 12 − 9

      = 3 cm

Page No 346:

Question 7:

In the adjoining figure, ABCD is a trapezium in which AB || DC and P, Q are the midpoints of AD and BC respectively. DQ and AB when produced meet at E. Also, AC and PQ intersect at R. Prove that (i) DQ = QE, (ii) PR || AB, (iii) AR = RC.

Answer:

Given: AB || DC, AP = PD and BQ = CQ

(i) In âˆ†QCD and ∆QBE, we have:
DQC = â€‹BQE   (Vertically opposite angles)

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

(ii) Now, in âˆ†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:

Question 8:

In the adjoining figure, AD is a median of ∆ABC and DE || BA. Show that BE is also a median of ∆ABC.

Answer:

AD is a median of  âˆ†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 âˆ†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 âˆ†ABC.

Page No 346:

Question 9:

In the adjoining figure, AD and BE are the medians of ∆ABC  and DF || BE. Show that CF=14AC.

Answer:

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

​In âˆ†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:

Question 10:

Prove that the line segments joining the middle points of the sides of a triangle divide it into four congruent triangles.

Answer:

 âˆ†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 âˆ£âˆ£ AC   (By midpoint theorem) 
Similarly, DF â€‹âˆ£âˆ£ BC and EF â€‹âˆ£âˆ£ AB.
Therefore, ADEF, BDFE and DFCE are all parallelograms.
Now, DE is the diagonal of the parallelogram BDFE.
∴ â€‹âˆ†BDE ≅ ​∆FED
Simiilarly, DF is the â€‹diagonal of the parallelogram ADEF.

∴ âˆ†DAF ≅ ∆FED
And, EF is the â€‹diagonal of the parallelogram DFCE.
∴ âˆ†EFC ≅ âˆ†FED
So, all the four triangles are congruent.

Page No 346:

Question 11:

In the adjoining figure, D, E, F are the midpoints of the sides BC, CA and AB respectively, of ∆ABC. Show that ∠EDF = ∠A, ∠DEF = ∠B and ∠DEF = ∠C.
Figure

Answer:

 âˆ† 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 âˆ† ABC.
∴ FE âˆ£âˆ£ BC  (By mid point theorem) 
Similarly, DE â€‹âˆ£âˆ£ 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:

Question 12:

Show that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a rectangle is a rhombus.

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 âˆ† ABC, we have:
∴ PQ âˆ£âˆ£ AC and PQ12AC                    [By midpoint theorem]
Again, in âˆ† DAC, the points S and R are the mid points of AD and DC, respectively.
∴ SR âˆ£âˆ£ AC and SR = 12AC                    [By midpoint theorem]
Now, PQ âˆ£âˆ£ AC and SR âˆ£âˆ£ AC
 ⇒ PQ âˆ£âˆ£ SR

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

Page No 347:

Question 13:

Show that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a rhombus is a rectangle.

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

Question 14:

Show that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a square is a square.

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

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

Now, in âˆ†SAP and âˆ†QBP, we have:
AS = BQ   
A = ∠B = 90o
AP = BP
i.e.,  ∆SAP ≅ âˆ†QBP​
∴ PS = PQ                  ...(ii)
Similarly, ∆SDR ≅ âˆ†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 âˆ£âˆ£ FO
Also, SR ∣∣ AC
FR âˆ£âˆ£ 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:

Question 15:

Prove that the line segments joining the midpoints of opposite sides of a quadrilateral bisect each other.

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:

Question 16:

The diagonals of a quadrilateral ABCD are equal. Prove that the quadrilateral formed by joining the midpoints of its sides is a rhombus.

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 âˆ†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, i∆ABD,

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:

Question 17:

The diagonals of a quadrilateral ABCD are perpendicular to each other. Prove that the quadrilateral formed by joining the midpoints of its sides is a rectangle.

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:

Question 18:

The midpoints of the sides AB, BC, CD and DA of a quadrilateral ABCD are joined to form a quadrilateral. If AC = BD and AC BD then prove that the quadrilateral formed is a square.

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:

Question 1:

Three angles of a quadrilateral are 80°, 95° and 112°. Its fourth angle is
(a) 78°
(b) 73°
(c) 85°
(d) 100°

Answer:

(b) 73°â€‹

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:

Question 2:

The angles of a quadrilateral are in the ratio 3 : 4 : 5 : 6. The smallest of these angles is
(a) 45°
(b) 60°
(c) 36°
(d) 48°

Answer:

(b) 60°â€‹

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 â€‹
⇒ 
x = 20
o
∴ ∠A = 60o​, ∠B = 80o, ∠C = 100o and ∠D = 120o
​Hence, the smallest angle is
60°.

Page No 349:

Question 3:

In the given figure, ABCD is a parallelogram in which BAD = 75° and ∠CBD = 60°. Then, ∠BDC = ?
(a) 60°
(b) 75°
(c) 45°
(d) 50°

Answer:

(c) 45°

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

Page No 349:

Question 4:

ABCD is a rhombus such that ∠ACB = 50°. Then, ∠ADB = ?
(a) 40°
(b) 25°
(c) 65°
(d) 130°

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:

Question 5:

In which of the following figures are the diagonals equal?
(a) Parallelogram
(b) Rhombus
(c) Trapezium
(d) Rectangle

Answer:

(d) Rectangle.

The diagonals of a rectangle are equal.

Page No 349:

Question 6:

If the diagonals of a quadrilateral bisect each other at right angles, then the figure is a
(a) trapezium
(b) parallelogram
(c) rectangle
(d) rhombus

Answer:

(d) rhombus

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



Page No 350:

Question 7:

The lengths of the diagonals of a rhombus are 16 cm and 12 cm. The length of each side of the rhombus is
(a) 10 cm
(b) 12 cm
(c) 9 cm
(d) 8 cm

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.
∴​ âˆ†â€‹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:

Question 8:

The length of each side of a rhombus is 10 cm and one if its diagonals is of length 16 cm. The length of the other diagonal is
(a) 13 cm
(b) 12 cm
(c) 239 cm
(d) 6 cm

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.
∴​ âˆ†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:

Question 9:

A diagonal of a rectangle is inclined to one side of the rectangle at 35°. The acute angle between the diagonals is
(a) 55°
(b) 70°
(c) 45°
(d) 50°

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:

Question 10:

If ABCD is a parallelogram with two adjacent angles A = ∠B, then the parallelogram is a
(a) rhombus
(b) trapezium
(c) rectangle
(d) none of these

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:

Question 11:

In a quadrilateral ABCD, if AO and BO are the bisectors of A and B respectively, ∠C = 70° and ∠D = 30°. Then, ∠AOB = ?
(a) 40°
(b) 50°
(c) 80°
(d) 100°

Answer:

(b) 50o

​​Explanation:

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

Page No 350:

Question 12:

The bisectors of any two adjacent angles of a parallelogram intersect at
(a) 40°
(b) 45°
(c) 60°
(d) 90°

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:

Question 13:

The bisectors of the angles of a parallelogram enclose a
(a) rhombus
(b) square
(c) rectangle
(d) parallelogram

Answer:

(c) Rectangle

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

Page No 350:

Question 14:

If bisectors of ∠A and ∠B of a quadrilateral ABCD intersect each other at P, of ∠B and ∠C at Q, of ∠C and ∠D at R and of ∠D and ∠A at S then PQRS is a
(a) rectangle
(b) parallelogram
(c) rhombus
(d) quadrilateral whose opposite angles are supplementary

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:

Question 15:

The figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a
(a) rhombus
(b) square
(c) rectangle
(d) parallelogram

Answer:

(d) parallelogram

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

Page No 350:

Question 16:

The figure formed by joining the mid-points of the adjacent sides of a square is a
(a) rhombus
(b) square
(c) rectangle
(d) parallelogram

Answer:

(b) Square

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

Page No 350:

Question 17:

The figure formed by joining the mid-points of the adjacent sides of a parallelogram is a
(a) rhombus
(b) square
(c) rectangle
(d) parallelogram

Answer:

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

Page No 350:

Question 18:

The figure formed by joining the mid-points of the adjacent sides of a rectangle is a
(a) rhombus
(b) square
(c) rectangle
(d) parallelogram

Answer:

(a) rhombus

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



Page No 351:

Question 19:

The figure formed by joining the mid-points of the adjacent sides of a rhombus is a
(a) rhombus
(b) square
(c) rectangle
(d) parallelogram

Answer:

(c) Rectangle

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

Page No 351:

Question 20:

The quadrilateral formed by joining the midpoints of the sides of a quadrilateral ABCD, taken in order, is a rectangle, if
(a) ABCD is a Parallelogram
(b) ABCD is rectangle
(c) diagonals of ABCD are equal
(d) diagonals of ABCD are perpendicular to each other.

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:

Question 21:

The quadrilateral formed by joining the midpoints of the sides of a quadrilateral ABCD, taken in order, is a rhombus, if
(a) ABCD is a Parallelogram
(b) ABCD is rhombus
(c) diagonals of ABCD are equal
(4) diagonals of ABCD are perpendicular to each other.

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:

Question 22:

The figure formed by joining the midpoints of the sides of a quadrilateral ABCD, taken in order, is a square, only if
(a) ABCD is a rhombus
(b) diagonals of ABCD are equal
(c) diagonals of ABCD are perpendicular
(d) diagonals of ABCD are equal and perpendicular

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:

Question 23:

If an angle of a parallelogram is two-third of its adjacent angle, the smallest angle of the parallelogram is
(a) 108°
(b) 54°
(c) 72°
(d) 81°

Answer:

(c) 72°â€‹

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)
 x23x180o
53x=180°
⇒ x = 108o
∴ B = 23× (108o) = 72o
Hence
AC  = 108o  and B = D = 72o
9x5 = 180ox = 100oA = 100o an

Page No 351:

Question 24:

If one angle of a parallelogram is 24° less than twice the smallest angle, then the largest angle of the parallelogram is
(a) 68°
(b) 102°
(c) 112°
(d) 136°

Answer:

(c)112°â€‹

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:

Question 25:

If A, ∠B, ∠C and ∠D of a quadrilateral ABCD taken in order, are in the ratio 3 : 7 : 6 : 4 then ABCD is a
(a) rhombus
(b) kite
(c) trapezium
(d) parallelogram

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, â€‹7 â¨¯ 18o = 126o, ​6 â¨¯ 18o = 108o and ​4 â¨¯18o = 72o.
But 54o + 126o = 180o and 72o + 108o  = 180o
ABCD is a trapezium.

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Question 26:

Which of the following is not true for a parallelogram?
(a) Opposite sides are equal.
(b) Opposite angles are equal.
(c) Opposite angles are bisected by the diagonals.
(d) Diagonals bisect each other.

Answer:

(c) Opposite angles are bisected by the diagonals.

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Question 27:

If APB and CQD are two parallel lines, then the bisectors of APQ, ∠BPQ, ∠CQP and ∠PQD enclose a
(a) square
(b) rhombus
(c) rectangle
(d) kite

Answer:

(c) Rectangle

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



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Question 28:

In the given figure, ABCD is a parallelogram in which BDC = 45° and ∠BAD = 75°. Then, ∠CBD = ?
(a) 45°
(b) 55°
(c) 60°
(d) 75°

Answer:

(c)​ 60°

Explanation:
∠BAD = ∠BCD = 75o       [Opposite angles are equal]
In âˆ† BCD, ∠ C = 75o            
∴ â€‹∠CBD = 180o (75o + 45o) = 60o

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Question 29:

If area of a || gm with sides a and b is A and that of a rectangle with sides a and b is B, then
(a) A > B
(b) A = B
(c) A < B
(d) AB

Answer:

(c) A < B

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

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Question 30:

In the given figure, ABCD is a || gm and E is the mid-point of BC. Also, DE and AB when produced meet at F. Then,
(a) AF=32AB
(b) AF = 2AB
(c) AF = 3AB
(d) AF2 = 2AB2

Answer:

(b) AF = 2 AB

Explanation:
I​n parallelogram ABCD, we have:
AB || DC
DCE = ∠​ EBF            (Alternate interior angles)
In âˆ† DCE and â€‹ ∆ 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

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Question 31:

P is any point on the side BC of a ∆ABC. P is joined to A. If D and E are the midpoints of the sides AB and AC respectively and M and N are the midpoints of BP and CP respectively then quadrilateral DENM is
(a) a trapezium
(b) a parallelogram
(c) a rectangle
(d) a rhombus

Answer:


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

In âˆ†ABP,

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

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

Again, in âˆ†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).

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Question 32:

The parallel sides of a trapezium are a and b respectively. The line joining the mid-points of its non-parallel sides will be
(a) 12(a-b)
(b) 12(a+b)
(c) 2ab(a+b)
(d) ab

Answer:

(b) 12a+b

Explanation:

Suppose ABCD is a trapezium.
Draw EF parallel to AB.
Join BD to cut EF at M.
Now, in âˆ† 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
 

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Question 33:

In a trapezium ABCD, if E and F be the mid-point of the diagonals AC and BD respectively. Then, EF = ?
(a) 12AB
(b) 12CD
(c) 12(AB+CD)
(d) 12(AB-CD)

Answer:

(d) 12AB - CD

Explanation:

Join CF and produce it to cut AB at G.
Then ∆CDF  ≅ âˆ†GBF                [∵ DF = BF, â€‹DCF = â€‹BGF and â€‹CDF = â€‹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 

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Question 34:

In the given figure, ABCD is a parallelogram, M is the mid-point of BD and BD bisects B as well as ∠D. Then, ∠AMB = ?
(a) 45°
(b) 60°
(c) 90°
(d) 30°

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°

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Question 35:

In the given figure, ABCD is a rhombus. Then,
(a) AC2 + BD2 = AB2
(b) AC2 + BD2 = 2AB2
(c) AC2 + BD2 = 4AB2
(d) 2(AC2 + BD2) = 3AB2

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)



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Question 36:

In a trapezium ABCD, if AB || CD, then (AC2 + BD2) = ?
(a) BC2 + AD2 + 2BCAD
(b) AB2 + CD2+ 2ABCD
(c) AB2 + CD2 + 2ADBC
(d) BC2 + AD2 + 2ABCD

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 âˆ†ABC, ∠B is acute.
∴ AC2BC2 + AB2 - 2AB.AE
In âˆ†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:

Question 37:

Two parallelograms stand on equal bases and between the same parallels. The ratio of their areas is
(a) 1 : 2
(b) 2 : 1
(c) 1 : 3
(d) 1 : 1

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.

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Question 38:

In the given figure, AD is a median of ∆ABC and E is the mid-point of AD. If BE is joined and produced to meet AC in F, then AF = ?
(a) 12AC
(b) 13AC
(3) 23AC
(4) 34AC

Answer:

(b) â…“ AC

Explanation:

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

​In âˆ† 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 =â…“ AC​
4AC

Page No 353:

Question 39:

The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O such that DAC = 30° and ∠AOB = 70°. Then, ∠DBC = ?
(a) 40°
(b) 35°
(c) 45°
(d) 50°

Answer:

(a) 40°

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

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Question 40:

Three statements are given below:
I. In a || gm, the angle bisectors of two adjacent angles enclose a right angle.
II. The angle bisectors of a || gm form a rectangle.
III. The triangle formed by joining the mid-points of the sides of an isosceles triangle is not necessarily an isosceles triangle.
Which is true?
(a) I only
(b) II only
(c) I and II
(d) II and III

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.

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Question 41:

Three statements are given below:
I. In a rectangle ABCD, the diagonal AC bisects A as well as ∠C.
II. In a square ABCD, the diagonal AC bisects ∠A as well as ∠C.
III. In a rhombus ABCD, the diagonal AC bisects ∠A as well as ∠C.
(a) I only
(b) II and III
(c) I and III
(d) I and II

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).

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Question 42:

In a quadrilateral PQRS, opposite angles are equal. If SR = 2 cm and PR = 5 cm then determine PQ.

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



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Question 43:

Diagonals of a parallelogram are perpendicular to each other. Is this statement true? Give reasons for your answer.

Answer:

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

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Question 44:

What special name can be given to a quadrilateral PQRS if ∠P + ∠S = 180°?

Answer:

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

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

So, PQRS is a parallelogram.

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Question 45:

All the angles of a quadrilateral can be acute. Is this statement true? Give reasons for your answer.

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:

Question 46:

All the angles of a quadrilateral can be right angles. Is this statement true? Give reasons for your answer.

Answer:

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

Page No 354:

Question 47:

All the angles of a quadrilateral can be obtuse. Is this statement true? Give reasons for your answer.

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:

Question 48:

Can we form a quadrilateral whose angles are 70°, 115°, 60° and 120°? Give reasons for your answer.

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:

Question 49:

What special name can be given to a quadrilateral whose all angles are equal?

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:

Question 50:

If D and E are respectively the midpoints of the sides AB and BC of ∆ABC in which AB = 7.2 cm, BC = 9.8 cm and AC = 3.6 cm then determine the length of DE.

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:

Question 51:

In a quadrilateral PQRS, the diagonals PR and QS bisect each other. If ∠Q = 56°, determine ∠R.

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:

Question 52:

In the adjoining figure, BDEF and AFDE are parallelograms. Is AF = FB? Why or why not

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:

Question 53:

Is quadrilateral ABCD a || gm?
I. Diagonals AC and BD bisect each other.
II. Diagonals AC and BD are equal.
(a) if the question can be answered by one of the given statements alone and not by the other;
(b) if the question can be answered by either statement alone;
(c) if the question can be answered by both the statements together but not by any one of the two;
(d) if the question cannot be answered by using both the statements together.

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 â€‹parallelogram.
∴ II does not give the answer.

Hence, the correct answer is (a).



Page No 355:

Question 54:

Is quadrilateral ABCD a  rhombus?
I. Quad. ABCD is a || gm.
II. Diagonals AC and BD are perpendicular to each other.
(a) if the question can be answered by one of the given statements alone and not by the other;
(b) if the question can be answered by either statement alone;
(c) if the question can be answered by both the statements together but not by any one of the two;
(d) if the question cannot be answered by using both the statements together.

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:

Question 55:

Is || gm ABCD a square?
I. Diagonals of || gm ABCD are equal.
II. Diagonals of || gm ABCD intersect at right angles.
(a) if the question can be answered by one of the given statements alone and not by the other;
(b) if the question can be answered by either statement alone;
(c) if the question can be answered by both the statements together but not by any one of the two;
(d) if the question cannot be answered by using both the statements together.

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:

Question 56:

Is quad. ABCD a parallelogram?
I. Its opposite sides are equal.
II. Its opposite angles are equal.
(a) if the question can be answered by one of the given statements alone and not by the other;
(b) if the question can be answered by either statement alone;
(c) if the question can be answered by both the statements together but not by any one of the two;
(d) if the question cannot be answered by using both the statements together.

Answer:

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

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Question 57:

Assertion: If three angles of a quadrilateral are 130°, 70° and 60°, then the fourth angles is 100°.
Reason: The sum of all the angle of a quadrilateral is 360°.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

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:

Question 58:

Assertion: ABCD is a quadrilateral in which P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Then, PQRS is a parallelogram.
Reason: The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

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:

Question 59:

Assertion: In a rhombus ABCD, the diagonal AC bisects A as well as ∠C.
Reason: The diagonals of a rhombus bisect each other at right angles.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

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:

Question 60:

Assertion: Every parallelogram is a rectangle.
Reason: The angle bisectors of a parallelogram form a rectangle.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

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:

Question 61:

Assertion: The diagonals of a || gm bisect each other.
Reason: If the diagonals of a || gm are equal and intersect at right angles, then the parallelogram is a square.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

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:

Question 62:

Match the following columns:

Column I Column II
(a) Angle bisectors of a parallelogram form a (p) parallelogram
(b) The quadrilateral formed by joining the mid-points of the pairs of adjacent sides of a square is a (q) rectangle
(c) The quadrilateral formed by joining the mid-points of the pairs of adjacent sides of a rectangle is a (r) square
(d) The figure formed by joining the mid-points of the pairs of adjacent sides of a quadrilateral is a (s) rhombus
(a) .....,
(b) .....,
(c) .....,
(d) .....,

Answer:

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

Page No 356:

Question 63:

Match the following columns:

Column I Column II
(a) In the given figure, ABCD is a trapezium in which AB = 10 cm and CD = 7 cm. If P and Q are the mid-points of AD and BC respectively, then PQ =
(p) equal
(b) In the given figure, PQRS is a || gm whose diagonals intersect at O. If PR = 13 cm, then OR =
(q) at right angles
(c) The diagonals of a square are (r) 8.5 cm
(d) The diagonals of a rhombus bisect each other (s) 6.5 cm
(a) ......,
(b) ......,
(c) ......,
(d) ......,

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



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