KVPY SX SB Exam 2015 Question Paper

Question : 8 of 120

Marks: +1, -0

Suppose ABCDEF is a hexagon such that AB = BC = CD = 1 and DE = EF = FA = 2. If the vertices A, B, C, D, E, F are concylic, the radius of the circle passing through them is -

$√{ 5/2}$
$√{ 7/3}$
$√{11/5 }$
$√2$

Validate

Solution:

From the figure:

\sinθ= 1/{2r } \ & \sinθ =1/ r

θ+ α=60°

Now,

cos(θ + α) =1/2

⇒

\cosθ. \cosα – \sinθ. \sinα = 1/2

⇒

√{1-1/{4r^2}}√{1-1/r^2}-1/{2r}.{1/r}=1/2

⇒

√{4r^2- 1 }√{r^2 - 1} - 1 = r^2

⇒

(4r^2-1) (r^2 - 1) =(r^2 + 1)^2

⇒

4r^4 - 5r^2 + 1= r^4 + 2r^2 +1

⇒

3r^4 = 7r^2

⇒

r^2=7/3

⇒

r =√{7/3}

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