A regular octagon with sides of length 7 & an apothem of length 8.45 has an area of _____ square units?
2007-06-05 09:26:22 · 5 answers · asked by JoAnna 2 in Science & Mathematics ➔ Mathematics
S=8*7*8.45/2= 236.6 square units
2007-06-05 09:30:33 · answer #1 · answered by santmann2002 7 · 0⤊ 0⤋
A = 1/2 ap where a is the apothem and p is the perimeter
a=8.45 and p=56 (7*8)
A = 1/2 * 8.45 * 56
A = 236.6
2007-06-05 09:30:55 · answer #2 · answered by hrhbg 3 · 0⤊ 0⤋
The area of a triangle formed by a side and the lines joining its vertices to the centre is:
7 * 8.45 / 2
There are eight such triangles.
The area of the octagon is therefore:
8 * 7 * 8.45 / 2
= 236.6.
2007-06-05 09:33:38 · answer #3 · answered by Anonymous · 0⤊ 0⤋
You can form eight triangles. each with area = ½ * base * height:
½ * 7 * 8.45 = 29.575
Now, multiply this by eight, and you have your answer:
29.575 * 8 = 236.6 square units.
2007-06-05 09:31:54 · answer #4 · answered by Dave 6 · 0⤊ 0⤋
hi
area = per*apo/2 = n*side*apo/2
area = 8*7*8.45/2 = 236.6
bye
2007-06-05 09:30:51 · answer #5 · answered by railrule 7 · 0⤊ 0⤋