116. In the figure, BD and CD are angle bisectors of ∠ ABC and ∠ ACE, respectively. Then ∠ BDC is equal to :
     
     
     

1. 1. ∠BAC

   
   
   

   2. 2∠BAC

   
   
   

   3. 1	∠BAC
      2

   
   
   

   4. 1	∠BAC
      3

   
   
   

   5. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   From above given figure , we have
   In ΔABC, ∠ACE = ∠ABC + ∠BAC
   Similarly in ΔBCD, ∠BDC = ∠DCE − ∠DBC [Ext. angle prop. of a Δ]
   

   But ∠DCE =	1	∠ACE and
   	2

   
   

   ⇒	1	∠DBC =	1	∠ABC
   	2		2

   
   

   Correct Option: C

   From above given figure , we have
   In ΔABC, ∠ACE = ∠ABC + ∠BAC
   Similarly in ΔBCD, ∠BDC = ∠DCE − ∠DBC [Ext. angle prop. of a Δ]
   

   But ∠DCE =	1	∠ACE and
   	2

   
   

   ⇒	1	∠DBC =	1	∠ABC
   	2		2

   
   Now ∠BDC = ∠DCE - ∠DBC
   

   =	1	(∠ACE - ∠ABC)
   	2

   
   

   =	1	(∠ACE + ∠ABC - ∠ACE)
   	2

   ∴ ∠BDC =	1	∠BAC
   	2

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117. In a Δ ABC, the sides AB and AC are produced to P and Q respectively. The bisectors of ∠ OBC and ∠ QCB intersect at a point O. Then ∠ BOC is equal to:

1. 1. 90 ° +	1	∠A
      	2

   
   
   

   2. 90 ° -	1	∠A
      	2

   
   
   

   3. 120° +	1	∠A
      	2

   
   
   

   4. 120° -	1	∠A
      	2

   
   
   

   5. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   From given figure , We have ∠B + ∠CBP = 180° (Linear pair)
   

   ⇒	1	∠B + ∠CBP = 90°
   	2

   ⇒	1	∠B + ∠1 = 90°
   	2

   ⇒ ∠1 = 90° -	1	∠B ............ ( 1 )
   	2

   Similarly ∠2 = 90° -	1	∠C
   	2

   
   

   
   

   Correct Option: B

   From given figure , We have ∠B + ∠CBP = 180° (Linear pair)
   

   ⇒	1	∠B + ∠CBP = 90°
   	2

   ⇒	1	∠B + ∠1 = 90°
   	2

   ⇒ ∠1 = 90° -	1	∠B ............ ( 1 )
   	2

   Similarly ∠2 = 90° -	1	∠C
   	2

   
   In ΔOBC, we have ∠1 + ∠2 + ∠BOC = 180° (Angle sum prop.)
   

   ⇒

   90° -

   1

   ∠B

   +

   90° -

   1

   ∠C

   + ∠BOC = 180°

   2

   2

   ⇒ ∠BOC =	1	(∠B + ∠C)
   	2

   ∠BOC =	1	(∠A + ∠B + ∠C) -	1	∠A
   	2		2

   ∠BOC =	1	×180° -	1	∠A { ∴ ∠A + ∠B + ∠C = 180° }
   	2		2

   ∠BOC = 90° -	1	∠A
   	2

   
   

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118. In a Δ ABC, the bisectors of ∠ B and ∠ C intersect each other at a point O. Then ∠ BOC is equal to :

1. 1. 90 ° -	1	∠A
      	2

   
   
   

   2. 120° +	1	∠A
      	2

   
   
   

   3. 90 ° +	1	∠A
      	2

   
   
   

   4. 120° -	1	∠A
      	2

   
   
   

   5. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   In ΔABC, we know that ∠A + ∠B + ∠C = 180°
   

   1	∠ A +	1	∠B +	1	∠C = 90°
   2		2		2

   1	∠A + ∠1 + ∠2 = 90°
   2

   ∠1 + ∠2 = 90° -	1	∠A
   	2

   
   Now , In ΔBOC, ∠1 + ∠2 + ∠BOC = 180° ................ ( 1 )
   

   	90° -	1	∠A		+ ∠BOC = 180° { using (i) }
   		2

   ⇒ ∠BOC = 90° +	1	∠A
   	2

   
   

   Correct Option: C

   In ΔABC, we know that ∠A + ∠B + ∠C = 180°
   

   1	∠ A +	1	∠B +	1	∠C = 90°
   2		2		2

   1	∠A + ∠1 + ∠2 = 90°
   2

   ∠1 + ∠2 = 90° -	1	∠A
   	2

   
   Now , In ΔBOC, ∠1 + ∠2 + ∠BOC = 180° ................ ( 1 )
   

   	90° -	1	∠A		+ ∠BOC = 180° { using (i) }
   		2

   ⇒ ∠BOC = 90° +	1	∠A
   	2

   
   

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119. In the fig. XY || AC and XY divides triangular region ABC into two part equal in area.

     Then	AX	is equal to :
     	AB

     
     
     

1. 1. 1

      √2

   
   
   

   2. √2 + 2

      √2

   
   
   

   3. 1

      2

   
   
   

   4. √2 - 2

      √2

   
   
   

   5. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   From above given figure , we can see that
   ar △XBY = ar trap. XYCA (Given)
   ar(ΔABC) = 2 ar(ΔXBY)
   

   ar(∆XBY)	=	1
   ar(∆ABC)		2

   
   But ΔXBY ∼ ΔABC (∴ XY || AC)
   

   ∴	ar(∆XBY)	=	XB2	( Areas of similar triangle )
   	ar(∆ABC)		AB2

   Correct Option: D

   From above given figure , we can see that
   ar △XBY = ar trap. XYCA(Given)
   ar(ΔABC) = 2 ar(ΔXBY)
   

   ar(∆XBY)	=	1
   ar(∆ABC)		2

   
   But ΔXBY ∼ ΔABC (∴ XY || AC)
   

   ∴	ar(∆XBY)	=	XB2	( Areas of similar triangle )
   	ar(∆ABC)		AB2

   
   

   ∴	1	=	XB2
   	2		AB2

   
   

   ∴	XB	=	1
   	AB		√2

   AB - AX	=	1	{ ∴ AB - AX = XB }
   AB		√2

   
   

   ∴ 1 -	AX	=	1
   	AB		√2

   
   

   ⇒	AX	= 1 -	1
   	AB		√2

   
   

   ⇒	AX	=	√2 - 1
   	AB		√2

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120. Two poles of ht. a and b meters are p meters apart (b > a). The height of the point of intersection of the lines joining the top of each pole to the foot of the opposite pole is :

1. 1. a+b

      ab

   
   
   

   2. p

      a+b

   
   
   

   3. ab

      a+b

   
   
   

   4. a+b

      p

   
   
   

   5. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   As per given figure , we know that
   

   1	=	1	+	1
   h		a		b

   
   

   Correct Option: C

   As per given figure , we know that
   

   1	=	1	+	1
   h		a		b

   
   

   1	=	a+b
   h		ab

   
   

   ∴ h =	ab
   	a+b

   
   

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