What is the area of a regular octagon, whose permineter is equal to 72 cm?

Guest May 28, 2014

#1**+5 **

b = 72/8 cm = 9cm

tan(22.5°)=(b/2)/h so h = (b/2)/tan(22.5°) cm

Area of 1/8 of Octagon = (b/2)*h = (b/2)^{2}/tan(22.5°)

Area of Octagon = 8*(b/2)^{2}/tan(22.5°)= 2b^{2}/tan(22.5°)

$${\mathtt{Area}} = {\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{9}}}^{{\mathtt{2}}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{22.5}}^\circ\right)}}} \Rightarrow {\mathtt{Area}} = {\mathtt{391.102\: \!597\: \!104\: \!531\: \!143\: \!5}}$$

Area ≈ 391.1 cm^{2}

Alan
May 28, 2014

#1**+5 **

Best Answer

b = 72/8 cm = 9cm

tan(22.5°)=(b/2)/h so h = (b/2)/tan(22.5°) cm

Area of 1/8 of Octagon = (b/2)*h = (b/2)^{2}/tan(22.5°)

Area of Octagon = 8*(b/2)^{2}/tan(22.5°)= 2b^{2}/tan(22.5°)

$${\mathtt{Area}} = {\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{9}}}^{{\mathtt{2}}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{22.5}}^\circ\right)}}} \Rightarrow {\mathtt{Area}} = {\mathtt{391.102\: \!597\: \!104\: \!531\: \!143\: \!5}}$$

Area ≈ 391.1 cm^{2}

Alan
May 28, 2014