KVPY 2020-21 SB SX Exam Previous Papers

Question : 88 of 120

Marks: +1, -0

Six consecutive sides of an equiangular octagon are

6,9,8,7,10,5

in that order. The integer nearest to the sum of the remaining two sides is

Solution:

Let ABCDEFGH be the equiangular octagon as shown PQ = SR

\⇒{y}/{\√{2}}+6+{9}/{\√{2}}={5}/{\√{2}}+10+{7}/{\√{2}}

\⇒y=3+4 \√{2}

Also

: P S=Q R

\⇒{y}/{\√{2}}+x+{5}/{\√{2}}={9}/{\√{2}}+8+{7}/{\√{2}}

\⇒x=4+4 \√{2}

\∴x+y=7+8 \√{2}=18.313

\∴

Nearest integer

=18 .

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