A regular octagon has side length of 10.9 in. The perimeter of the octagon is 87.2 in & the area is 392.4 in sq. A second octagon has side lengths equal to 16.35 in. Find the area of the 2nd octagon.
A)882.9
B)717.06
C)642.66
D)461.60

Question

A regular octagon has side length of 10.9 in. The perimeter of the octagon is 87.2 in & the area is 392.4 in sq. A second octagon has side lengths equal to 16.35 in. Find the area of the 2nd octagon.
A)882.9
B)717.06
C)642.66
D)461.60

campbell_st · Answer

the ratio of the side lengths is needed

10.9: 16.35 is the same as 2:3

to compare area square the ratios 

2^2:3^2   or 4:9

the area of the larger octagons is 9/4 the smaller area

so  Larger Octagon Area = 9/4 * 392.4
                                          = 887.4 in^2

anonymous · Answer

That wasnt one of the choices they gave me.

campbell_st · Answer

oops my typo... 9/4 * 392.4 = 882.9

anonymous · Answer

Area of a regular octagon is given by $$ 2(1+\sqrt{2})a^2$$ Thus, a  regular octagon has side length of 10.9 in will have an area of  = $$ 2(1+\sqrt{2})(10.9)^2)= 573.665 \neq 392.4 $$

campbell_st · Answer

its a question about know the scaling factor for length. and area... not so much about area of a polygon

anonymous · Answer

The information seems incorrect.

campbell_st · Answer

yep... lots of those type of things in textbooks.... "not to scale"...lol

anonymous · Answer

wait so is the answer 882.9