Find the area of a regular pentagon with a side of 6 ft. Give the answer to the nearest tenth.
ok can't remember how to do this
A regular pentagon can be drawn as 5 conguent isosceles triangles with a central angle of 72° each, and equal base angles of 54°.
So just find the area of one of these by first finding the altitude of one of them
tan 54° = h/3
h = 3tan54°
area of one of them = (1/2)base x height
= (1/2)(6)(3tan54)
= 9tan 54°
so the whole pentagon = 45tan 54°
= appr. 61.9 square ft
You forgot to give us the lenght of the side of the line that connects from the middle of the pentagon to a side of the pentagon.
It's there; you just need to recognize it:
h = 3tan54°
To find the area of a regular pentagon, you can use the formula:
Area = (1/4) * √(5(5 + 2√5)) * s^2
where "s" is the length of one side of the pentagon. In this case, "s" is given as 6 ft.
Let's substitute this value into the formula and calculate the area:
Area = (1/4) * √(5(5 + 2√5)) * 6^2
Area = (1/4) * √(5(5 + 2√5)) * 36
Now, we need to evaluate the expression inside the square root sign:
5(5 + 2√5) = 25 + 10√5
Substituting this into the formula:
Area = (1/4) * √(25 + 10√5) * 36
Now, we can simplify the expression:
Area = (1/4) * √(25 + 10√5) * 36
Using a calculator, you can find that √(25 + 10√5) is approximately equal to 8.86.
Area = (1/4) * 8.86 * 36
Area ≈ 79.74 sq. ft.
So, to the nearest tenth, the area of the regular pentagon with a side length of 6 ft is 79.7 sq. ft.