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The convex pentagon $ABCDE$ has $\angle A = \angle B = 120^\circ$, $EA = AB = BC = 2$ and $CD = DE = 4$. What is the area of $ABCDE$?

michaelcai  Aug 24, 2017
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4+0 Answers

 #1
avatar+76108 
+4

 

Here's one approach.....but possibly not the fastest !!

 

Draw EB

 

Now since EA  = AB, then triangle EAB is isoceles....and its area  = (1/2)(2)(2)sin(120)  =

sqrt (3) units^2

 

Using the Law of Sines...we can find EB

 

EB / sin EAB  = EA / sin EBA

 

EB / sin (120)  = EA /sin(30)

 

EB / [sqrt(3)/2]  = 2 / (1/2)

 

EB /sqrt (3)  = 2

 

EB = 2sqrt(3)  = sqrt (12)

 

 

Now..draw EC... and  triangle  ECB  is a right triangle with angle EBC  being right

 

So....the area of EBC  = (1/2)(sqrt(12)(2)  =  sqrt(12)  = 2sqrt(3) units^2

 

And since ECB is right....EC   = sqrt ( EB^2 + EC^2)  =   sqrt  ( 12  + 4)  = sqrt (16)  = 4

 

So....triangle DEC is eqruilateral because DE = EC = CD  = 4

 

And its area   = (1/2)4^2sin(60)  = 4sqrt (3)unit's^2

 

So....the total area  of ABCDE =  [sqrt (3)  + 2sqrt(3)  + 4sqrt(3) ]  = 7sqrt(3) units^2

 

 

cool cool cool

CPhill  Aug 24, 2017
edited by CPhill  Aug 24, 2017
 #2
avatar+4478 
+5

Here's another way...but it looks a bit longer than CPhill's!

 

First let's extend line  AB .

Then....

Draw a line perpendicular to  AB  that goes through point  C  .

Draw a line perpendicular to AB  that goes through point  E .

Draw a line parallel to  AB  (and perpendicular to the previous two lines) that goes through point  D .

 

Now let's label these new points of intersection W, X, Y, and Z , like the picture,

and put the information from the problem.

 

∠EAZ  =  ∠CBY  =  180° - 120°  =  60°

 

Since triangles AEZ and BCY are 30-60-90 triangles where the side across from the 90° angle is 2....

the side across from the 30° angle  =  2/2  =  1

the side across from the 60° angle  =  √3

 

And...

WD + DX  =  2 + 1 + 1             And  WD = DX , so we can substitute  WD  in for  DX .

WD + WD  =  2 + 1 + 1

2WD  =  4

WD  =  2

 

Now we can use the Pythagorean theorem to find  CX .

22 + CX2  =  42

4 + CX2  =  16             Subtract  4  from both sides of the equation.

CX2  =  12                   Take the positive square root of both sides.

CX  =  2√3

 

 

And.....

area of ABCDE   =   area of WXYZ  -  2(area of △EDW)  -  2(area of △AEZ)

                           =   (width)(length)  -  2(1/2)(base)(height)  -  2(1/2)(base)(height)

                           =   (2 + 2)(2√3 + √3)  -  2(1/2)(2)(2√3)  -  2(1/2)(1)(√3)

                           =   12√3  -  4√3  -  √3

                           =   7√3  square units

hectictar  Aug 24, 2017
edited by hectictar  Aug 25, 2017
 #3
avatar+76108 
+1

 

Thanks, hectitctar....!!!

 

Perhaps X2 has another method, too....!!!

 

 

cool cool cool

CPhill  Aug 24, 2017
 #4
avatar+1099 
+3

I have figured out an alternate method of solving this problem, too! Shout out to Cphill and hectictar for attempting this! I have created a diagram using the programs of paint and GeoGebra. The diagram consolidates all the given information, and I have added a few segments to reference as I solve. Note that \(\overline{BG}\perp\overline{EC}\hspace{1mm}\text{and}\hspace{1mm}\overline{AF}\perp\overline{EC}\):

 

 

If we could figure out the length of \(\overline{EC}\), then we would only need to apply a few formulas to figure out the total area. I have not marked it in the diagram, but we also know by given info that \(m\angle EAB=m\angle CBA=120^{\circ}\).

 

For now, let's only worry about figure \(ABCE\). Since both \(\overline{AF}\) and \(\overline{BG}\) are altitudes (a segment that extends perpendicularly from a vertex to a segment opposite of that vertex) we can conclude that \(\overline{AB}\parallel\overline{EC}\). Since we have a quadrilateral that has exactly one pair of opposite sides parallel, then the quadrilateral is a trapezoid, by definition. We can actually go a step further than this, actually! Since the trapezoid's nonparallel sides are of equal length, this is an isosceles trapezoid. Classifying this quadrilateral this specifically actually aides us in solving this problem.

 

Isosceles trapezoids have a few properties associates with them. One of them is that adjacent angles along the nonparallel sides are supplementary. In other words, \(m\angle BAE+m\angle AEF=180^{\circ}\) and \(m\angle ABC+m\angle BCG=180^{\circ}\). Let's solve for those unknown angles. 

 

\(m\angle BAE+m\angle AEF=180^{\circ}\)\(m\angle ABC+m\angle BCG=180^{\circ}\)Plug in the known values and solve.
\(120+m\angle AEF=180^{\circ}\)\(120+m\angle BCG=180^{\circ}\) 
\(m\angle AEF=60^{\circ}\)\(m\angle BCG=60^{\circ}\) 

 

Great! By using the triangle sum theorem, we can figure out the \(m\angle EAF\) and \(m\angle CBG\). Let's do that:

 

\(m\angle{AFE}+m\angle{FEA}+m\angle EAF=180^{\circ}\)\(m\angle BGC+m\angle GCB+ m\angle CBG=180^{\circ}\)This is the traingle sum theorem in use. Now, solve for the unknown.
\(90^{\circ}+60^{\circ}+m\angle EAF=180^{\circ}\)\(90^\circ+60^{\circ}+ m\angle CBG=180^{\circ}\) 
\(150^{\circ}+m\angle EAF=180^{\circ}\)\(150^{\circ}+m\angle CBG=180^{\circ}\) 
\(m\angle EAF=30^{\circ}\)\(m\angle CBG=30^{\circ}\) 
   

 

Do you notice something interesting about these angle measures? If not, I will attempt to make it clearer by extracting one of the triangles from the diagram, as done below:

 

 

Yes, it is a 30-60-90 triangle! We can use this to our advantage. In a 30-60-90 triangle, the ratio of the lengths of the sides is \(1:\sqrt{3}:2\). This means that we can determine the length of both \(\overline{EF}\) and \(\overline{AF}\). Let's solve for both lengths now.

 

\(2EF=AE\)This equation is derived from the constant ratio of a 30-60-90 triangle. Of course, AE=2, so substitute that in.
\(2EF=2\)Divide by 2 on both sides to determine the length of EF.
\(EF=1\) 

 

Now, let's solve for AF.

 

\(AF=EF\sqrt{3}\)EF=1, so plug that in.
\(AF=1*\sqrt{3}\)Simplify.
\(AF=\sqrt{3}\) 

 

The same phenomenon occurs for \(\triangle BCG\)\(BG=\sqrt{3}\) and \(CG=1\). This means that \(EC=2+1+1=4\), the length of the opposite base.

 

Now, we can calculate the area of this trapezoid by using the following formula \(A_{trap.}=\frac{1}{2}(b_1+b_2)h\)

 

A = area of a trapezoid

b1 = the length of one base

b2 = the length of the other base

h = perpendicular height

 

We already have all this information, so we can solve for the area of the trapezoid.

 

\(A_{trap.}=\frac{1}{2}(b_1+b_2)h\)As aforementioned, this is the formula. Now, plug in all the values.
\(A_{trap.}=\frac{1}{2}(2+4)\sqrt{3}\)Simplify in the parentheses first.
\(A_{trap.}=\frac{1}{2}*6\sqrt{3}\)1/2 and 6 are like terms.
\(A_{trap.}=(3\sqrt{3})units^2\)This is the area of the trapezoid in siplest radical form. Of course, dont forget units!
  

 

We are not done yet because we have not considered \(\triangle DEC\) at all yet; remember, this is a composite figure! I have a zoomed-in diagram that extracts the triangle. 

 

 

Equilateral triangles also have a formula to them, too. This simplifies the process significantly. The formula is the following 

 

\(A_{\triangle}=\frac{\sqrt{3}}{4}s^2\)

 

A = area of an equilateral triangle

s = side length

 

Of course, let's apply it now!

\(A_{\triangle}=\frac{\sqrt{3}}{4}s^2\)Plug in 4 for s.
\(A_{\triangle}=\frac{\sqrt{3}}{4}*4^2\)Simplify the exponent first.
\(A_{\triangle}=\frac{\sqrt{3}}{4}*\frac{16}{1}\)Notice that 4 and 16 has a GCF of 4.
\(A_{\triangle}=\frac{\sqrt{3}}{1}*\frac{4}{1}=(4\sqrt{3})units^2\) 
  

 

The last thing to do is to add both areas of the individual shapes in this composite figure together, and that is your final answer!

 

\(A=3\sqrt{3}+4\sqrt{3}=(7\sqrt{3})units^2\)

 

This is your final answer!

TheXSquaredFactor  Aug 25, 2017

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