Question

If the perimeter of the regular octagon is 48in what’s is the area ? ( round it to the nearest tenths)

Answer 1

Answer:

The area of the octagon is $173.8\ in^{2}$

Step-by-step explanation:

we know that

The area of a regular octagon is equal to the area of eight isosceles triangle

The base of each isosceles triangle is equal to the length side of the regular octagon

The vertex angle of each isosceles triangle is equal to

$360\°/8=45\°$

The area of each isosceles triangle is equal to

$A=\frac{1}{2}bh$

where

b is the length side of the regular octagon

h is the height of each isosceles triangle

Find the length side of the regular octagon b

The perimeter of a octagon is equal to

$P=8b$

$P=48\ in$

so

$48=8b$

$b=6\ in$

Find the height of each isosceles triangle h

$tan(45\°/2)=(b/2)/h$

$h=(b/2)/tan(45\°/2)$

substitute the values

$h=(6/2)/tan(22.5\°)=7.24\ in$

Find the area of the octagon

$A=8[\frac{1}{2}bh]$

$A=8[\frac{1}{2}(6)(7.24)]=173.8\ in^{2}$