Question

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

• A. 17
• B. 18
• C. 19
• D. 20

Answer 1

Answer: (B)

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 $=18$