KVPY SA Exam 19-Nov-2017 Question Paper

Question : 61 of 80

Marks: +1, -0

Part - II

Mathematics

Let ABCD be a trapezium with parallel sides AB and CD such that the circle S with AB as its diameter touches CD. Further, the circle S passes through the midpoints of the diagonals AC and BD of the trapezium. The smallest angle of the trapezium is

$π/3$
$π/4$
$π/5$
$π/6$

Validate

Solution:

Join AN
∵ ANB = 90°
In

∆\ANB

\cos θ={BN}/{2r}

BN = 2r \cos θ

BD = 2BN = 4r \cos θ

∆\BQD

\sin θ ={BQ}/{BD}=r/{4r\cosθ}

\sin 2θ =1/2

θ = 15°
Now similarly α = 15° = θ & AC = 4r cos α
∴ Trapezium will be isosceles
∵ ∠ ADB = 30°

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