+0  
 
0
121
6
avatar

In hexagon ABCDEF, AB = DE = 2, BC = EF = 3, and CD = AF = 8, and all the interior angles are equal. Find the area of hexagon ABCDEF.

 

The answers are not 22sqrt3, 16sqrt3, 20sqrt3, or 27sqrt2.

 

Thank you!

 Jan 13, 2020
 #1
avatar+109061 
0

If all the interior angles are equal, they must all be  = 60°

 

So.....we have six equilateral triangles....and the area  of  any two triangles with the same side =  (side^2) √ (3)/ 2

 

So...the area  is

 

 (2^2 + 3^2 + 8^2) *√3 /2  =  77√3 / 2   

 

cool cool cool

 Jan 13, 2020
 #2
avatar
+1

I'm sorry but that says incorrect also. Maybe check your work? I don't know and I'm very confused.

Guest Jan 13, 2020
 #3
avatar+108624 
+1

I have not looked much but the angles will all be 120 degrees.

 

Use cosine rule to find distance AE

triangle AFE and triangle DBC are congruent

so AE = BD

Area of AFE can be determined.

 

ABDE is a parallelogram. I am pretty sure it is a rectangle but i have not looked at it enough to be sure.

If it is a rectangle then the area is easily obtained.

 

That is a rough outline of some of the steps you will need to use.

 Jan 14, 2020
 #4
avatar+515 
+2

Area of hexagon  ABCDEF ≈ 39.84 u²     or     Area = sqrt(3) * 23    indecision

 

 Jan 14, 2020
edited by Dragan  Jan 14, 2020
edited by Dragan  Jan 14, 2020
edited by Dragan  Jan 14, 2020
 #5
avatar
0

The answer is 15*sqrt(2).

 Jan 14, 2020
 #6
avatar+24344 
+3

In hexagon ABCDEF, AB = DE = 2, BC = EF = 3, and CD = AF = 8, and all the interior angles are equal.

Find the area of hexagon ABCDEF.

 

\(\begin{array}{|rcll|} \hline \mathbf{A} &=& \mathbf{\dbinom{0}{0}} \\\\ \mathbf{B} &=& 2\dbinom{\cos(120^\circ)}{\sin(120^\circ)}= \mathbf{\dbinom{-2\cos(60^\circ)}{2\sin(60^\circ)}} \\\\ \mathbf{C} &=& 3\dbinom{\cos(120^\circ-60^\circ)}{\sin(120^\circ-60^\circ)} + \mathbf{\dbinom{-2\cos(60^\circ)}{2\sin(60^\circ)}}\\ &=& \dbinom{3\cos(60^\circ)}{3\sin(60^\circ)}+\dbinom{-2\cos(60^\circ)}{2\sin(60^\circ)} = \mathbf{\dbinom{\cos(60^\circ)}{5\sin(60^\circ)}} \\\\ \mathbf{D} &=& 8\dbinom{\cos(120^\circ-120^\circ)}{\sin(120^\circ-120^\circ)} + \mathbf{\dbinom{\cos(60^\circ)}{5\sin(60^\circ)}} \\ &=& \dbinom{8\cos(0^\circ)}{8\sin(0^\circ)} + \dbinom{\cos(60^\circ)}{5\sin(60^\circ)} = \mathbf{ \dbinom{8+\cos(60^\circ)}{5\sin(60^\circ) } } \\\\ \mathbf{E} &=& 2\dbinom{\cos(120^\circ-180^\circ)}{\sin(120^\circ-180^\circ)} + \mathbf{ \dbinom{8+\cos(60^\circ)}{5\sin(60^\circ) } } \\ &=& \dbinom{2\cos(60^\circ)}{-2\sin(60^\circ)} + \dbinom{8+\cos(60^\circ)}{5\sin(60^\circ)} = \mathbf{ \dbinom{8+3\cos(60^\circ)}{3\sin(60^\circ) } } \\\\ \mathbf{F} &=& 3\dbinom{\cos(120^\circ-240^\circ)}{\sin(120^\circ-240^\circ)} + \mathbf{ \dbinom{8+3\cos(60^\circ)}{3\sin(60^\circ) } } \\ &=& \dbinom{-3\cos(60^\circ)}{-3\sin(60^\circ)} + \dbinom{8+3\cos(60^\circ)}{3\sin(60^\circ) } = \mathbf{\dbinom{8}{0}} \\ \hline \end{array}\)

 

The area of hexagon ABCDEF

\(\begin{array}{|crr|} \hline \text{Point} & x & y \\ \hline A & 0 & 0 \\ B & -2\cos(60^\circ) & 2\sin(60^\circ) \\ C & \cos(60^\circ) & 5\sin(60^\circ) \\ D & 8+\cos(60^\circ) & 5\sin(60^\circ) \\ E & 8+3\cos(60^\circ) & 3\sin(60^\circ) \\ F & 8 & 0 \\ A & 0 & 0 \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline 2F &=& -2\cos(60^\circ)5\sin(60^\circ)-\cos(60^\circ)2\sin(60^\circ) \\ &+& \cos(60^\circ)5\sin(60^\circ)-(8+\cos(60^\circ))5\sin(60^\circ) \\ &+& (8+\cos(60^\circ))3\sin(60^\circ)-(8+3\cos(60^\circ))5\sin(60^\circ) \\ &+& (8+3\cos(60^\circ))*0-8*3\sin(60^\circ) \\\\ 2F &=& -10\sin(60^\circ)\cos(60^\circ)-2\sin(60^\circ)\cos(60^\circ) \\ &+& 5\sin(60^\circ)\cos(60^\circ)-40\sin(60^\circ)-5\sin(60^\circ))\cos(60^\circ) \\ &+& 24\sin(60^\circ)+3\sin(60^\circ)\cos(60^\circ)-40\sin(60^\circ)-15\sin(60^\circ)\cos(60^\circ) \\ &-& 24\sin(60^\circ) \\\\ 2F &=& -10\sin(60^\circ)\cos(60^\circ)-2\sin(60^\circ)\cos(60^\circ) -40\sin(60^\circ) \\ &+& 3\sin(60^\circ)\cos(60^\circ)-40\sin(60^\circ)-15\sin(60^\circ)\cos(60^\circ) \\\\ 2F &=& -24\sin(60^\circ)\cos(60^\circ) - 80\sin(60^\circ) \quad | \quad \cos(60^\circ) = \dfrac{1}{2} \\ 2F &=& -24\sin(60^\circ)\dfrac{1}{2} - 80\sin(60^\circ) \\ 2F &=& -12\sin(60^\circ)- 80\sin(60^\circ) \\ 2F &=& -92\sin(60^\circ) \quad | \quad : 2 \\ F &=& -46\sin(60^\circ) \quad | \quad \sin(60^\circ) = \dfrac{\sqrt{3}}{2} \\ F &=& -46\dfrac{\sqrt{3}}{2} \\ F &=& |-23 \sqrt{3}| \\ \mathbf{F} &=& \mathbf{23 \sqrt{3}} \\ \hline \end{array}\)

 

laugh

 Jan 14, 2020

14 Online Users