KVPY 2020-21 SB SX Exam Previous Papers

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Question : 85 of 120

Marks: +1, -0

In a triangle

{ABC}

, the angle bisector

{BD}

of

\∠{B}

intersects AC in D. Suppose

{BC}=2, {CD}=1

and

{BD}={3}/{\√{2}}

. The perimeter of the triangle

{ABC}

is

${17}/{2}$
${15}/{2}$
${17}/{4}$
${15}/{4}$

Solution:

{2 a c}/{a+c} \cos {B}/{2}={3}/{\√{2}}

\⇒{4 c}/{2+c}[{4+{9}/{2}-1}/{2 \×2 \×{3}/{\√{2}}}]={3}/{\√{2}}

\⇒c=3

Now,

{c}/{a}={A D}/{D C} \⇒A D={3}/{2}

\⇒ b={5}/{2}

\⇒

Perimeter

={15}/{2}

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