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Question
ABCD is a square with each side 12 cm. P is a point on BC such that area of ΔABP: area of trapezium APCD = 1: 5. Find the length of CP.
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Solution
The figure is shown below:
`"Area of ΔABP"/"Area of trapezium APCD" = 1/5`
⇒ `[ 1/2 xx 12 xx ( 12 - "CP" )]/[ 1/2 xx ( 12 +"CP") xx 12] = 1/5`
⇒ 60 - 5CP = 12 + CP
⇒ 6CP = 48
⇒ CP = 8 cm.
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