The regular pentagon shown has an apothem with a length of 2 centimeters. To the nearest tenth, what is the area of the regular pentagon in square cintimeters

11.8cm²

Step-by-step explanation:

The area of the regular pentagon is expressed as;

A = pa/2

p is the perimeter of the pentagon

a is the apothem

Since p = 5s

s is the side length, hence;

A = 5sa/2

Get the side length s;

central angle = 360/5

central angle = 72degrees

angle in the right angle triangle =72/2 = 36degrees

Using SOH CAH TOA

sin 36 = x/2

x = 2sin36

x = 1.1755

s = 2x

s = 2.35cm

Get the area

A = 5(2.35)(2)/2

A = 5(2.35)

A = 11.8cm²

Hence the area of the regular pentagon is 11.8cm²

i think the length is 12 and the width is 17

isocolese triangle

we know that the line connect a and b is the hyptonuse

so first find the hypotonuse length using the distance formula, call the length c

since it is an isocolese triangle, the lengths of the legs are the same

using pythagorean theorem,

in this case, a=b since isocolese triangle so we get

or

we can find c so we can find the distances

we know that if we draw 2 circles, each has center at a and b, and find where they intersect, we can the 2 possible locations for c

if a has center at (xa,ya) and b has center at (xb,yb), then we can solve where

and

intersect to find where they intersect

i'm not going to put an example because i have to go to class. i hope this explanation is enough