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  4. 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

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Q. 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

KVPYKVPY 2020

Solution:

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Let ABCDEFGH be the equiangular octagon as shown $PQ = SR$
$\Rightarrow \frac{y}{\sqrt{2}}+6+\frac{9}{\sqrt{2}}=\frac{5}{\sqrt{2}}+10+\frac{7}{\sqrt{2}}$
$\Rightarrow y=3+4 \sqrt{2}$
Also : PS $= QR$
$\Rightarrow \frac{y}{\sqrt{2}}+x+\frac{5}{\sqrt{2}}=\frac{9}{\sqrt{2}}+8+\frac{7}{\sqrt{2}}$
$\Rightarrow x=4+4 \sqrt{2}$
$\therefore x+y=7+8 \sqrt{2}=18.313$
$\therefore $ Nearest integer $=18$