Area of a Regular Pentagon

**purplec16** · February 28th 2010, 08:52 AM

The perimeter of the building has the shape of a regular pentagon with each side of length 921 ft. Find the area enclosed by the perimeter of the building.

**Archie Meade** · February 28th 2010, 10:11 AM

Originally Posted by purplec16

The perimeter of the building has the shape of a regular pentagon with each side of length 921 ft. Find the area enclosed by the perimeter of the building.

Hi purplec16,

if you split this pentagon into 5 identical triangles, you can solve for the area of one triangle,
using triangle area=0.5(base)(height)

$x=\frac{360}{5}^o$

$Tan\left(\frac{180-x}{2}\right)=\frac{h}{0.5(921)}$

$h=0.5(921)Tan\left(\frac{180-x}{2}\right)$

Area of the pentagon is

$5(0.5)(921)h$

**purplec16** · February 28th 2010, 11:04 AM

I dont really understand what you did

**Archie Meade** · February 28th 2010, 11:19 AM

hi purplec16,

were you able to open the adobe file ?
and have you covered trigonometry to the extent
that you can divide an isosceles triangle in two parts
and calculate it's height from an angle and it's base length?

In steps, these are the calculations...

1. The regular pentagon may be inscribed in a circle.
2. Divide 360 degrees by 5 to find x.
3. The 5 triangles are isosceles because 2 sides are the circle radius.
4. Both corner angles of any of the isosceles triangles are equal.
5. All 3 angles within the triangle sum to 180 degrees, so x+2A=180, so 2A=180-x, A=0.5(180-x).
6. To calculate h, use TanA and half the length of a side, by dividing the isosceles triangle into a pair of right-angled triangles.
7. When you have h, use the triangle area formula to find triangle area.
8. The pentagon contains 5 identical triangles, so multiply the area of one of them by 5.

**purplec16** · February 28th 2010, 11:35 AM

The answer in the book is $1,459,349 ft^2$ I dont see how I can get that with what u show me can you explain please.

**masters** · February 28th 2010, 11:57 AM

Originally Posted by purplec16

The answer in the book is $1,459,349 ft^2$ I dont see how I can get that with what u show me can you explain please.

Hi purplec16,

Let me try to break it down for you.

[1] The area of a regular polygon is given by $A=\frac{1}{2}aP$ , where a = apothem, and P = perimeter.

Now, we get the Perimeter by multiplying 5 times 921 = 4605.

[2] Now find the apothem (a). Recall the apothem is the perpendicular drawn from the center of the regular polygon to a side.

Draw two radii from the center of the pentagon to two vertices forming an isosceles triangle.
The vertex angle will measure $\frac{360}{5}=72$ .

From the center of the pentagon, draw the perpendicular bisector of one of its sides. This also bisects the vertex angle into two 36 degree angles. The base of the right triangle is $\frac{921}{2}=460.5$

Next, use a little trig to find the apothem and base of the right triangle we just formed.

$\tan 36=\frac{460.5}{a}$

$a=\frac{460.5}{\tan 36}$

$P=4605$

Area of the pentagon = $\frac{1}{2}aP=.5\left(\frac{460.5}{\tan 36}\right) \cdot 4605$

Area of the pentagon = $1,459,379.471 ft^2$

The discrepency in my answer and yours has to do with you rounding values early.

**purplec16** · February 28th 2010, 12:02 PM

Thanks to the both of you, I get it now the formula helped

**Archie Meade** · February 28th 2010, 12:27 PM

Hi purplec16,

sorry i was away for a while,
the method i offered yields the same as masters solution.

Thanks to masters for his help!

Thread: Area of a Regular Pentagon

LinkBack

Thread Tools

Display

Area of a Regular Pentagon

Similar Math Help Forum Discussions

euclidean construction of the regular pentagon

Need Help Regular Pentagon

Regular Pentagon

What is the area of a regular pentagon if its apothem has a length of 4 feet and each

5 points of a regular pentagon

Search Tags