Question

How do we deduce to Area of regular hexagon $=3\frac{\sqrt{3a}}{2}sq$ units Area of regular octagon $=2(1+\sqrt{2})a$

Answer 1

Hexagon:-

A Hexagan can be divided in 6 equilateral triangle having side = a units
Area of equilateral triangle $=\frac{\sqrt{3}{a}^2}{4}$
$\therefore$ Area of Hexagan
$=6^3\times \frac{\sqrt{3}.9^2}{4_2}=\frac{3\sqrt{3}{9}^2}{2}sq.units$

Octagon :

Consider a square KLMN is which octagon ABCDEFGH is inscribed CD is hypothenses of $\Delta CLD=a$ side of octagon.
$CL=LD=x$
$C{C}^{2}HC{D}^2=C{D}^2$
$d{x}^2=a^{2}orx=\frac{a}{\sqrt{2}}$
$\therefore$ Side of square = 2x + a
$=\frac{\sqrt{2}a}{\sqrt{2}}+a=a(1+\sqrt{2})$
Area of square $=a^2(1+\sqrt{2}{)}^2$
Area of octagon = Area of square $-4\times Areaof\Delta CLD$
$=a^2(1+\sqrt{2}{)}^2-4\times \frac{1}{2}\times x^2=a^2(1+\sqrt{2}{)}^2-2\frac{a^2}{2}=a^2\left\{\right(1+\sqrt{2}{)}^2\}=\sqrt{2}(d+\sqrt{2})a^2$