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(B) Let the remaining two sides be $a$ and $b$. An equiangular octagon has interior angles of $135^\circ$. By extending the sides to form a rectangle

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(B) Let the remaining two sides be $a$ and $b$. An equiangular octagon has interior angles of $135^\circ$. By extending the sides to form a rectangle $ABCD$,the sides of the octagon are related to the sides of the rectangle.From the geometry of the figure:Horizontal sides of the rectangle: $\frac{9}{\sqrt{2}} + 6 + \frac{b}{\sqrt{2}} = \frac{7}{\sqrt{2}} + 10 + \frac{5}{\sqrt{2}}$$\frac{b}{\sqrt{2}} = 4 + \frac{3}{\sqrt{2}} \implies b = 4\sqrt{2} + 3$Vertical sides of the rectangle: $\frac{9}{\sqrt{2}} + 8 + \frac{7}{\sqrt{2}} = \frac{b}{\sqrt{2}} + a + \frac{5}{\sqrt{2}}$Substituting $b = 4\sqrt{2} + 3$:$\frac{9}{\sqrt{2}} + 8 + \frac{7}{\sqrt{2}} = \frac{4\sqrt{2} + 3}{\sqrt{2}} + a + \frac{5}{\sqrt{2}}$$8\sqrt{2} + 8 = 4 + \frac{3}{\sqrt{2}} + a + \frac{5}{\sqrt{2}}$$8\sqrt{2} + 4 = a + \frac{8}{\sqrt{2}} = a + 4\sqrt{2}$$a = 4\sqrt{2} + 4$Sum $a + b = (4\sqrt{2} + 4) + (4\sqrt{2} + 3) = 8\sqrt{2} + 7$Using $\sqrt{2} \approx 1.414$,$a + b \approx 8(1.414) + 7 = 11.312 + 7 = 18.312$.The nearest integer is $18$.

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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