SSC CGL 6 06 2019 Shift 2 Paper

Question : 72 of 100

Marks: +1, -0

ABCD is a trapezium in which AB ∥ DC and itsdiagonals intersect at P. If AP= (3x - 1)cm , PC= (5x -3 )cm, BP= (2x + 1) cm and PD= (6x -5 )cm, then the length of DB is:

14 cm
12cm
10 cm
16 cm

Validate

Solution:

\ {AP}=3 \ {x}-1

\ {PC}=5 \ {x}-3

\ {BP}=2 \ {x}+1

\ {PD}=6 \ {x}-5

\ {A P}/{P C}=\ {B P}/{P D}

(3 \ {x}-1)(6 \ {x}-5)=(2 \ {x}+1)(5 \ {x}-3)

18 \ {x}^{2}-21 \ {x}+5=10 \ {x}^{2}-\ {x}-3

8 \ {x}^{2}-20 \ {x}+8=0

2 \ {x}^{2}-5 \ {x}+2=0

2 \ {x}^{2}-4 \ {x}-\ {x}+2=0

2 \ {x}(\ {x}-2)-1(\ {x}-2)=0

\ {x}=2,1 / 2

\ {PD}=12-5=7

\ {BP}=5

\ {DB}=12 \ {cm}

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