We have learned that the area of a regular polygon with n sides is given by the formula below.
A=1/2aP = 1/2a* ns
In our case, n=5 and s=5. Thus, the is P=5* 5 = 25. Now, we just need to calculate the .
The measure of ∠ CPD, which is a , is 360^(∘)5=72^(∘). Since △ PCD is , the PX coincides with the of ∠ CPD.
m∠ XPC = m∠ CPD/2 = 72^(∘)/2 = 36^(∘)
By the
m∠ XPC_(36^(∘)) + m∠ XCP = 90^(∘)
⇓
m∠ XCP = 54^(∘)
Besides this, PX also coincides with the of DC. Thus, XC = 52=2.5.
Since △ XPC is a , we can apply the . By using the tangent ratio of the 54^(∘) angle, we can find the apothem.
tan 54^(∘) = a/2.5 ⇒
a = 2.5tan 54^(∘)
Let's substitute the expression above into the formula written at the beginning to find the area of ABCDE.
A=1/2aP
A=1/2( 2.5tan 54^(∘))( 25)
A = 43.01
Dividing the Pentagon Into Smaller Polygons
We can also find the area of the pentagon by adding the areas of △ ABE and the area of the trapezoid BCDE.
A_(pentagon) = A_(△ ABE) + A_(BCDE)
We will find each area separately.
Area of △ ABE
Let's consider BE as the base and we will draw the altitude from A. Notice that △ ABE is isosceles, which implies that the altitude from A is the of BE.
The tells us that the sum of the interior angles of pentagon ABCDE is 3* 180^(∘) = 540^(∘). This means that m∠ BAE = 108^(∘). Since AQ coincides with the angle bisector of ∠ BAE, we have that m∠ QAE = 54^(∘).
As before, since △ QAE is a right triangle, we can use the trigonometric ratios to find the height and the base.
| Dimension
|
Trigonometric Ratio
|
Expression
|
| Base
|
sin 54^(∘) = EQ/5
|
EQ = 5sin 54^(∘)
|
| Height
|
cos 54^(∘) = AQ/5
|
AQ = 5cos 54^(∘)
|
Notice that the base of △ ABE is twice EQ. We now are ready to find the area of △ ABE.
A_(△ ABE) = 1/2bh
A_(△ ABE) = 1/2* 2EQ* AQ
A_(△ ABE) = EQ* AQ
A_(△ ABE) = 5sin 54^(∘)* 5cos 54^(∘)
A_(△ ABE) = 5* 0.81 * 5* 0.59
A_(△ ABE) = 11.9
Area of Trapezoid BCDE
From the previous part we already know that EB = 2EQ, which is the same as 10sin 54^(∘). Also, we know that DC=5, and so we know the length of the two bases of trapezoid BCDE.
From the previous part we also know that m∠ D = 108^(∘). Next, let's find the height of the trapezoid. To do that we will draw the height from E.
As we can see, ∠ CDE and ∠ EDR form a and are angles, which implies that m∠ EDR = 72^(∘). One more time, we can apply a trigonometric ratio to △ RDE.
sin 72^(∘) = h/5 ⇒
h = 5sin 72^(∘)
Now, we have all that is needed to find the area of the trapezoid.
A_(BCDE) = (b_1+b_2)h/2
A_(BCDE) = ( EB+ CD)h/2
A_(BCDE) = ( 10sin 54^(∘)+ 5)h/2
A_(BCDE) = (10sin 54^(∘) + 5) 5sin 72^(∘)/2
A_(BCDE) = (10* 0.81 + 5)5* 0.95/2
A_(BCDE) = (8.1 + 5)4.75/2
A_(BCDE) = (13.1)4.75/2
A_(BCDE) = 62.225/2
A_(BCDE) = 31.11
Adding the Areas
We are now ready to find the area of the pentagon by adding the area of △ ABE and the area of trapezoid BCDE.
A_(pentagon) = A_(△ ABE)_(12) + A_(BCDE)_(31)
⇓
A_(pentagon) = 43
As we can see, the area of the pentagon is about 43 square units.
Conclusion
We have seen that both methods yield the same value, however, the first one requires fewer computations. This leads us to conclude that it is better to use the formula to find the area of a regular polygon. Nevertheless, your answer may vary.
