631. ABC is a right angled triangle, B being the right angle. Mid-points of BC and AC are respectively B' and A'. The ratio of the area of the quadrilateral AA' B'B to the area of the triangle ABC is
     

1. 1. 1 : 2
      

   
   
   

   2. 2 : 3
      

   
   
   

   3. 3 : 4
      

   
   
   

   4. None of the above

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   A'B'C ~ ∆ABC
   ∠C = ∆C,
   

   CA'	=	1	and	CB'	=	1
   CA		2		CB		2

   
   and
   

   ∴	∆ABC	=	(2x)²	=	4
   	∆A'B'C		x²		1

   
   [ ∵ BC = 2B'C and AC = 2A'C]
   

   ⇒	∆A'B'C	=	1
   	∆ABC		4

   
   

   ⇒ 1 -	∆A'B'C	= 1 -	1
   	∆ABC		4

   
   

   ⇒	∎AA'B'C	=	3	= 3 : 4
   	∆ABC		4

   Correct Option: C

   
   A'B'C ~ ∆ABC
   ∠C = ∆C,
   

   CA'	=	1	and	CB'	=	1
   CA		2		CB		2

   
   and
   

   ∴	∆ABC	=	(2x)²	=	4
   	∆A'B'C		x²		1

   
   [ ∵ BC = 2B'C and AC = 2A'C]
   

   ⇒	∆A'B'C	=	1
   	∆ABC		4

   
   

   ⇒ 1 -	∆A'B'C	= 1 -	1
   	∆ABC		4

   
   

   ⇒	∎AA'B'C	=	3	= 3 : 4
   	∆ABC		4

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632. Two triangles ABC and PQR are congruent. If the area of ∆ABC is 60 sq. cm, then area of ∆PQR will be
     

1. 1. 60 sq.cm
      

   
   
   

   2. 30 sq.cm
      

   
   
   

   3. 15 sq.cm
      

   
   
   

   4. 120 sq.cm

   
   
   

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1. View Hint View Answer Discuss in Forum

   Both the triangles are congruent.
   ∴ ∆ABC = 60 sq.cm.
   ∆PQR = 60 sq.cm.

   Correct Option: A

   Both the triangles are congruent.
   ∴ ∆ABC = 60 sq.cm.
   ∆PQR = 60 sq.cm.

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633. In ∆PQR, the line drawn from the vertex P intersects QR at a point S. If QR = 4.5 cm and SR = 1.5 cm then the ratios of the area of triangle PQS and triangle PSR is
     

1. 1. 4 : 1
      

   
   
   

   2. 3 : 1
      

   
   
   

   3. 3 : 2
      

   
   
   

   4. 2 : 1

   
   
   

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1. View Hint View Answer Discuss in Forum

   Using Rule 1,
   
   QR = 4.5 cm
   SR = 1.5 cm
   ∴ QS = 4.5 – 1.5 = 3 cm
   

   ∆PQS

   ∆PSR

   
   

   =	1	× h × QS
   	2
   	1	× h × SR
   	2

   
   

   =	3	= 2 : 1
   	1.5

   Correct Option: D

   Using Rule 1,
   
   QR = 4.5 cm
   SR = 1.5 cm
   ∴ QS = 4.5 – 1.5 = 3 cm
   

   ∆PQS

   ∆PSR

   
   

   =	1	× h × QS
   	2
   	1	× h × SR
   	2

   
   

   =	3	= 2 : 1
   	1.5

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634. ABCD is parallelogram. P and Q are the mid-points of sides BC and CD respectively. If the area of ∆ABC is 12 cm⊂2, then the area of ∆APQ is
     

1. 1. 12 cm²
      

   
   
   

   2. 8 cm²
      

   
   
   

   3. 9 cm²
      

   
   
   

   4. 10 cm²

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   

   ∆ APQ =	3	(∎ABCD)
   	8

   
   

   =	3	∆ABC
   	4

   
   

   =	3	× 12 = 9 sq.cm.
   	4

   Correct Option: C

   
   

   ∆ APQ =	3	(∎ABCD)
   	8

   
   

   =	3	∆ABC
   	4

   
   

   =	3	× 12 = 9 sq.cm.
   	4

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635. ABC is a right angled triangle. B being the right angle. Mid-points of BC and AC are respectiveely B' and A'. Area of ∆A'B'C' is
     

1. 1. 1	× area of ∆ABC
      2

   
   
   

   2. 2	× area of ∆ABC
      3

   
   
   

   3. 1	× area of ∆ABC
      4

   
   
   

   4. 1	× area of ∆ABC
      8

   
   
   

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1. View Hint View Answer Discuss in Forum

   Using Rule 1,
   
   In ∆ABC and ∆A'B'C A'B' || AB
   ∠B' = ∠B, ∠A' = ∠A
   ∴ ∆ABC ~ ∆A'B'C
   

   ∆ A'B' =	1	AB.
   	2

   
   

   ∴ Area of ∆A'B'C =	1	× B'C × A'B'
   	2

   
   

   =	1	×	1	BC ×	1	AB
   	2		2		2

   
   

   =	1	×	1	(BC × AB)
   	4		2

   
   

   =	1	× Area of ∆ABC
   	4

   Correct Option: C

   Using Rule 1,
   
   In ∆ABC and ∆A'B'C A'B' || AB
   ∠B' = ∠B, ∠A' = ∠A
   ∴ ∆ABC ~ ∆A'B'C
   

   ∆ A'B' =	1	AB.
   	2

   
   

   ∴ Area of ∆A'B'C =	1	× B'C × A'B'
   	2

   
   

   =	1	×	1	BC ×	1	AB
   	2		2		2

   
   

   =	1	×	1	(BC × AB)
   	4		2

   
   

   =	1	× Area of ∆ABC
   	4

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