+0

# area of a circle circumscribed

0
451
1

find the area of a circle cirbumscribed about a regular octagon with a perimeter of 80 inches.

Guest Jun 2, 2015

#1
+26402
+15

Length of one side of the octagon must be 80/8 = 10 inches.

Angle of one side subtended at the centre = 360°/8 = 45°

Length of side of isosceles triangle formed by lines from centre to one side of the octagon = 5/sin(45/2°) inches.  This is also the radius of the circumscribed circle, so:

Area of circle = pi*(5/sin(22.5°))2  in2

$${\mathtt{Area}} = {\mathtt{\pi}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{5}}}{\underset{\,\,\,\,^{{360^\circ}}}{{sin}}{\left({\mathtt{22.5}}^\circ\right)}}}\right)}^{{\mathtt{2}}} \Rightarrow {\mathtt{Area}} = {\mathtt{536.303\: \!412\: \!267\: \!149\: \!253\: \!5}}$$

Area ≈ 536.3 in2

.

Alan  Jun 3, 2015
Sort:

#1
+26402
+15

Length of one side of the octagon must be 80/8 = 10 inches.

Angle of one side subtended at the centre = 360°/8 = 45°

Length of side of isosceles triangle formed by lines from centre to one side of the octagon = 5/sin(45/2°) inches.  This is also the radius of the circumscribed circle, so:

Area of circle = pi*(5/sin(22.5°))2  in2

$${\mathtt{Area}} = {\mathtt{\pi}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{5}}}{\underset{\,\,\,\,^{{360^\circ}}}{{sin}}{\left({\mathtt{22.5}}^\circ\right)}}}\right)}^{{\mathtt{2}}} \Rightarrow {\mathtt{Area}} = {\mathtt{536.303\: \!412\: \!267\: \!149\: \!253\: \!5}}$$

Area ≈ 536.3 in2

.

Alan  Jun 3, 2015

### 13 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details