Q 1 :    

If ABC ~ EDF and ABC is not similar to DEF, then which of the following is not true ?

  • BC. EF = AC. FD

     

  • AB. EF = AC. DE

     

  • BC. DE = AB. EF

     

  • BC. DE = AB. FD

     

(3)

Since, ABC ~ EDF

Therefore, the ratio of their corresponding sides is proportional.

BCEF=ABDEBC.DEAB.EF

 

 



Q 2 :    

In the ABC, DE || BC and AD = 3x − 2, AE = 5x − 4, BD = 7x − 5, CE = 5x − 3, then find the value of x

  • 1

     

  • 7/10

     

  • both (a) & (b)

     

  • none of these

     

(3)

Given that, AD = 3x − 2, AE = 5x − 4, BD = 7x − 5, CE = 5x − 3

By Basic Proportionality theorem, we have ADBD=AEEC

3x-27x-5=5x-45x-3(3x-2)(5x-3)=(5x-4)(7x-5)

15x2-19x+6=35x2-53x+20

20x2-34x+14=010x2-17x+7=0

(x-1)(10x-7)=0x=1,x=710



Q 3 :    

In the given figure, DE  BC, AE = a units, EC = b units, DE = x units and BC = y units. Which of the following is true?

[IMAGE]

  • x=a+bay

     

  • y=axa+b

     

  • x=aya+b

     

  • xy=ab

     

(3)

[IMAGE]

As DEBC

 AEAC=DEBC  [From BPT]

aa+b=xy

x(a+b)=ayx=aya+b

 



Q 4 :    

ABCD is a trapezium with ADBC and AD = 4cm. If the diagonals AC and BD intersect each other at O such that AO/OC = DO/OB =1/2, then BC =

  • 6 cm

     

  • 7 cm

     

  • 8 cm

     

  • 9 cm

     

(3)

[IMAGE]

ADBC

and AOOC=DOOB=12

  ADBC=AOOC4BC=12BC=8 cm

 



Q 5 :    

ΔABC~ΔPQR. If AM and PN are altitudes of ΔABC and ΔPQR respectively and AB2:PQ2 = 4 : 9, then AM : PN =

  • 3 : 2

     

  • 16 : 81

     

  • 4 : 9

     

  • 2 : 3

     

(4)

[Image]

We have, ABC~PQR

         ABPQ=BCQR=CARP=AMPN  (corresponding sides of similar triangles)

But AB2PQ2=49ABPQ=23

i.e., ABPQ=AMPN=23

Hence, AM:PN=2:3



Q 6 :    

The perimeters of two similar triangles ABC and PQR are 56 cm and 48 cm respectively. PQ/AB is equal to

  • 7/8

     

  • 6/7

     

  • 7/6

     

  • 8/7

     

(2)

The ratio of the corresponding sides of similar triangles is same as the ratio of their perimeter

  ABC~PQR or PQR~ABC

PQAB=QRBC=PRAC=4856PQAB=67



Q 7 :    

In the given figure, if M and N are points on the sides OP and OS respectively of OPS, such that MNPS, then the length of OP is :

[IMAGE]

  • 6.8 cm

     

  • 17 cm

     

  • 15.3 cm

     

  • 9.6 cm

     

(3)     15.3 cm

 



Q 8 :    

In the given figure ABC is shown. DE is parallel to BC. If AD = 5 cm, DB = 2.5 cm and BC = 12 cm, then DE is equal to

[IMAGE]

  • 10 cm

     

  • 6 cm

     

  • 8 cm

     

  • 7.5 cm

     

(3)

AD=5 cm

DB=2.5 cm

BC=12 cm

DEBC

ABC~ADE

ADAB=DEBC57.5=DE12DE=607.5=8 cm