geometry help again?

Question

a. find the apothem of the regular octagon with each side being 10

b. find the area of the same octagon

show work please and thank you!

Answer 1

The sum of the interior angles of a polygon has (n-2)*180 degrees where n is the number of sides, so the octagon angles add up to (8-2)*180 = 1080. Since it is a regular octagon, all the interior angles are the same and they measure 1080/8 or 135 degrees. Since the apothem joins the center at the midpoint of a side, you have a right triangle with one leg measuring 10/2 or 5 and an interior angle of 135/2 or 67.5 degrees. Now you have tan(67.5)=a/5 where a is the apothem you seek. Solving for a gives you 12.0711

The area of the triangle is 1/2bh or 1/2 *12.0711*5 which is 30.1777. Multiply by 16 to get the total area to be 482.843

Answer 2

Since this is a regular octagon, the central angle between adjacent vertices is 360°/8 or 45°. Dividing one of the triangles made by connecting the center with the vertices in two yields an angle between the apothem and the radius of 22.5°. The length of the segment connecting the far end of the apothem to the radius is 1/2 the length of a side, or 5. The length of the apothem itself is then cot 22.5° * 5. Now, remembering that tan θ/2 = sin θ/(1+cos θ), we have that cot 22.5° = 1/tan 22.5° = (1+cos 45°)/sin 45° = (1+√2/2)/(√2/2) = ((2+√2)/2)/(√2/2) = √2 + 1. So we have that the apothem is then 5√2+5.

The area of each triangle formed by connecting the center wit h the vertices is 1/2 * base * height = 1/2 * side * apothem = 1/2 * 10 * (5√2+5) = 25√2+25. Since there are eight such triangles, the total area of the octagon is then 8 times this, that is 200√2+200