+0  
 
+8
854
6
avatar+1068 

The area of the equilateral triangle ABE = sin600(cm2). Line segment AI = 5cm. BF = EF =CI =DI ; find the area of the polygon DEGHI.

 

 May 29, 2015

Best Answer 

 #1
avatar+21860 
+18

The area of the equilateral triangle ABE = sin600(cm2). Line segment AI = 5cm. BF = EF ; find the area DEGHI.

 

AI = 5 cm

 

$$\mathbf{
\sin{(60\ensurement{^{\circ}})} = \frac{1}{2} \cdot \sqrt{3}
}\\
\mathbf{
\overline{AB} = \overline{BE} = \sqrt{2} \qquad
\overline{BF} = \overline{EF} =\frac{ \sqrt{2}} {2}
}\\
\mathbf{
\overline{AF} = \sqrt{2} \cdot \sin{(60\ensurement{^{\circ}})} = \frac{\sqrt{2} }{2} \cdot \sqrt{3}
}\\
\mathbf{
\overline{FI}= \overline{AI} - \overline{AF} =
5- \frac{\sqrt{2} }{2} \cdot \sqrt{3}
}\\$$

 

If the perpendicular of H is $$\small{\text{$H_0$}}$$ on line ED:

 

$$\mathbf{
\overline{BE} : \overline{HH_0} = 1 : \dfrac{2}{3}\qquad
\overline{ED} : \overline{EH_0} = 1 : \dfrac{2}{3}
}\\
\mathbf{
\overline{EH_0} = \dfrac{2}{3} \cdot ( \overline{AI} - \overline{AF} )\qquad \overline{HH_0} = \dfrac{2}{3} \cdot \overline{BE}
}\\$$

 

The area DEGHI = A:

 

$$\small{\text{
$
\mathbf{
2\cdot A =\dfrac{2}{3}\cdot (\overline{AI}-\overline{AF}) \cdot ( \dfrac{2}{3}\cdot \overline{BE} )+
\dfrac{1}{3}\cdot (\overline{AI}-\overline{AF}) \cdot ( \dfrac{2}{3}\cdot \overline{BE} - \overline{BF} )+\textcolor[rgb]{1,0,0}{2}\cdot
\dfrac{1}{3}\cdot (\overline{AI}-\overline{AF}) \cdot \overline{BF}
}
$}}\\\\
\small{\text{
$
\mathbf{
2\cdot A =(\overline{AI}-\overline{AF}) \cdot \left[
\dfrac{2}{3}\cdot\dfrac{2}{3}\cdot \sqrt{2}+
\dfrac{1}{3}\cdot ( \dfrac{1}{6}\cdot \sqrt{2} )+
\textcolor[rgb]{1,0,0}{2}\cdot
\dfrac{1}{3}\cdot \dfrac{\sqrt{2}}{2}
\right]
}
$}}\\\\
\small{\text{
$
\mathbf{
2\cdot A =(\overline{AI}-\overline{AF}) \cdot \dfrac{\sqrt{2}}{3} \cdot \left(
\dfrac{4}{3} + \dfrac{1}{6} + \dfrac{\textcolor[rgb]{1,0,0}{2}}{2}
\right)
}
$}}\\\\
\small{\text{
$
\mathbf{
2\cdot A =(\overline{AI}-\overline{AF}) \cdot \dfrac{\sqrt{2}}{3} \cdot \textcolor[rgb]{1,0,0}{\dfrac{5}{2}}
}
$}}\\\\
\small{\text{
$
\mathbf{
A =( \overline{AI}-\overline{AF}) \cdot \sqrt{2}\cdot\textcolor[rgb]{1,0,0}{\dfrac{5}{12}}
}
$}}\\\\
\small{\text{
$
\mathbf{
A =(5- \frac{\sqrt{2} }{2} \cdot \sqrt{3}) \cdot \sqrt{2}\cdot\textcolor[rgb]{1,0,0}{\dfrac{5}{12}}
}
$}}\\\\
\small{\text{
$
\mathbf{
A =( 5\cdot\sqrt{2} - \sqrt{3} ) \cdot \textcolor[rgb]{1,0,0}{\dfrac{5}{12}}
}
$}}\\\\
\small{\text{
$
\mathbf{
A =2.22459041846~\rm{cm^2}
}
$}}$$

 

P.S.

$$\small{\text{
$
\vec{EH}
\mathbf{
=\binom{ \overline{EH_0} }{ \overline{HH_0} }
=\binom{ 0 }{ \overline{BE} }
+ \lambda \binom{ \overline{AI}-\overline{AF} }
{ -\frac{ \overline{BE} } {2} }
= \mu \binom{ \overline{AI}-\overline{AF} }
{ \overline{BE} }\qquad
\mathbf{ \overline{EH_0} =\lambda ( \overline{AI}-\overline{AF} ) }\qquad
\mathbf{ \overline{HH_0} = \overline{BE} - \lambda \frac{ \overline{BE} } {2} }
}
$}}\\\\
\small{\text{
$
\begin{array}{lrcl}
&&\\
\lambda~?\\
&&\\
1.\\
& 0+\lambda( \overline{AI}-\overline{AF} ) &=& \mu ( \overline{AI}-\overline{AF} ) \\
& \lambda &=& \mu
&&\\
2.\\
&\overline{BE} - \lambda \frac{ \overline{BE} } {2} &=& \mu \overline{BE} \\\\
&\overline{BE} - \lambda \frac{ \overline{BE} } {2} &=& \lambda \overline{BE} \qquad | \qquad \mu = \lambda\\\\
&\overline{BE} - \lambda \frac{ \overline{BE} } {2} &=& \lambda \overline{BE} \qquad | \qquad : \overline{BE} \\\\
& 1 - \frac{ \lambda } {2} &=& \lambda \\\\
& \lambda + \frac{ \lambda } {2} &=& 1 \\\\
& \frac{ 3 } {2} \lambda &=& 1 \\\\
& \lambda &=& \frac{ 2 } {3} \\\\
\end{array}
$}}\\\\
\small{\text{
$
\mathbf{
\overline{EH_0} =\lambda ( \overline{AI}-\overline{AF} )
= \frac{2}{3} ( \overline{AI}-\overline{AF} )
}
$}}\\\\
\small{\text{
$
\mathbf{
\overline{HH_0} = \overline{BE} - \lambda \frac{ \overline{BE} } {2}
=\overline{BE} - \frac{2}{3} \frac{ \overline{BE} } {2}
= \frac{2}{3} \overline{BE}
}
$}}\\\\$$

 

 May 29, 2015
 #1
avatar+21860 
+18
Best Answer

The area of the equilateral triangle ABE = sin600(cm2). Line segment AI = 5cm. BF = EF ; find the area DEGHI.

 

AI = 5 cm

 

$$\mathbf{
\sin{(60\ensurement{^{\circ}})} = \frac{1}{2} \cdot \sqrt{3}
}\\
\mathbf{
\overline{AB} = \overline{BE} = \sqrt{2} \qquad
\overline{BF} = \overline{EF} =\frac{ \sqrt{2}} {2}
}\\
\mathbf{
\overline{AF} = \sqrt{2} \cdot \sin{(60\ensurement{^{\circ}})} = \frac{\sqrt{2} }{2} \cdot \sqrt{3}
}\\
\mathbf{
\overline{FI}= \overline{AI} - \overline{AF} =
5- \frac{\sqrt{2} }{2} \cdot \sqrt{3}
}\\$$

 

If the perpendicular of H is $$\small{\text{$H_0$}}$$ on line ED:

 

$$\mathbf{
\overline{BE} : \overline{HH_0} = 1 : \dfrac{2}{3}\qquad
\overline{ED} : \overline{EH_0} = 1 : \dfrac{2}{3}
}\\
\mathbf{
\overline{EH_0} = \dfrac{2}{3} \cdot ( \overline{AI} - \overline{AF} )\qquad \overline{HH_0} = \dfrac{2}{3} \cdot \overline{BE}
}\\$$

 

The area DEGHI = A:

 

$$\small{\text{
$
\mathbf{
2\cdot A =\dfrac{2}{3}\cdot (\overline{AI}-\overline{AF}) \cdot ( \dfrac{2}{3}\cdot \overline{BE} )+
\dfrac{1}{3}\cdot (\overline{AI}-\overline{AF}) \cdot ( \dfrac{2}{3}\cdot \overline{BE} - \overline{BF} )+\textcolor[rgb]{1,0,0}{2}\cdot
\dfrac{1}{3}\cdot (\overline{AI}-\overline{AF}) \cdot \overline{BF}
}
$}}\\\\
\small{\text{
$
\mathbf{
2\cdot A =(\overline{AI}-\overline{AF}) \cdot \left[
\dfrac{2}{3}\cdot\dfrac{2}{3}\cdot \sqrt{2}+
\dfrac{1}{3}\cdot ( \dfrac{1}{6}\cdot \sqrt{2} )+
\textcolor[rgb]{1,0,0}{2}\cdot
\dfrac{1}{3}\cdot \dfrac{\sqrt{2}}{2}
\right]
}
$}}\\\\
\small{\text{
$
\mathbf{
2\cdot A =(\overline{AI}-\overline{AF}) \cdot \dfrac{\sqrt{2}}{3} \cdot \left(
\dfrac{4}{3} + \dfrac{1}{6} + \dfrac{\textcolor[rgb]{1,0,0}{2}}{2}
\right)
}
$}}\\\\
\small{\text{
$
\mathbf{
2\cdot A =(\overline{AI}-\overline{AF}) \cdot \dfrac{\sqrt{2}}{3} \cdot \textcolor[rgb]{1,0,0}{\dfrac{5}{2}}
}
$}}\\\\
\small{\text{
$
\mathbf{
A =( \overline{AI}-\overline{AF}) \cdot \sqrt{2}\cdot\textcolor[rgb]{1,0,0}{\dfrac{5}{12}}
}
$}}\\\\
\small{\text{
$
\mathbf{
A =(5- \frac{\sqrt{2} }{2} \cdot \sqrt{3}) \cdot \sqrt{2}\cdot\textcolor[rgb]{1,0,0}{\dfrac{5}{12}}
}
$}}\\\\
\small{\text{
$
\mathbf{
A =( 5\cdot\sqrt{2} - \sqrt{3} ) \cdot \textcolor[rgb]{1,0,0}{\dfrac{5}{12}}
}
$}}\\\\
\small{\text{
$
\mathbf{
A =2.22459041846~\rm{cm^2}
}
$}}$$

 

P.S.

$$\small{\text{
$
\vec{EH}
\mathbf{
=\binom{ \overline{EH_0} }{ \overline{HH_0} }
=\binom{ 0 }{ \overline{BE} }
+ \lambda \binom{ \overline{AI}-\overline{AF} }
{ -\frac{ \overline{BE} } {2} }
= \mu \binom{ \overline{AI}-\overline{AF} }
{ \overline{BE} }\qquad
\mathbf{ \overline{EH_0} =\lambda ( \overline{AI}-\overline{AF} ) }\qquad
\mathbf{ \overline{HH_0} = \overline{BE} - \lambda \frac{ \overline{BE} } {2} }
}
$}}\\\\
\small{\text{
$
\begin{array}{lrcl}
&&\\
\lambda~?\\
&&\\
1.\\
& 0+\lambda( \overline{AI}-\overline{AF} ) &=& \mu ( \overline{AI}-\overline{AF} ) \\
& \lambda &=& \mu
&&\\
2.\\
&\overline{BE} - \lambda \frac{ \overline{BE} } {2} &=& \mu \overline{BE} \\\\
&\overline{BE} - \lambda \frac{ \overline{BE} } {2} &=& \lambda \overline{BE} \qquad | \qquad \mu = \lambda\\\\
&\overline{BE} - \lambda \frac{ \overline{BE} } {2} &=& \lambda \overline{BE} \qquad | \qquad : \overline{BE} \\\\
& 1 - \frac{ \lambda } {2} &=& \lambda \\\\
& \lambda + \frac{ \lambda } {2} &=& 1 \\\\
& \frac{ 3 } {2} \lambda &=& 1 \\\\
& \lambda &=& \frac{ 2 } {3} \\\\
\end{array}
$}}\\\\
\small{\text{
$
\mathbf{
\overline{EH_0} =\lambda ( \overline{AI}-\overline{AF} )
= \frac{2}{3} ( \overline{AI}-\overline{AF} )
}
$}}\\\\
\small{\text{
$
\mathbf{
\overline{HH_0} = \overline{BE} - \lambda \frac{ \overline{BE} } {2}
=\overline{BE} - \frac{2}{3} \frac{ \overline{BE} } {2}
= \frac{2}{3} \overline{BE}
}
$}}\\\\$$

 

heureka May 29, 2015
 #2
avatar+1068 
0

I'm too tired; I'll work on this one tomorrow; I'm going to Image result for sleep

 

@heureka:/  This time your numbers are correct! I came up with the same answer --with a little help from you. Thanks for showing me how to find a side of the equilateral triangle when only the area is known!

I guess..., I can go back to (see above image) 

 May 30, 2015
 #3
avatar+21860 
+5
.
 May 30, 2015
 #4
avatar+21860 
+13

how to find a side of the equilateral triangle when only the area A is known!

AE = AB = BE ( equilateral triangle )

P.S.

$$\mathbf{
2A = \overline{AB}\cdot \overline{BE}\cdot \sin{(60\ensurement{^{\circ}})}
}\\\\
\small{\text{
\begin{array}{rcl}
2A &=& \overline{AB}\cdot \overline{BE}\cdot \sin{(60\ensurement{^{\circ}})} \qquad | \qquad \overline{AB} = \overline{BE}\\\\
2A &=& \overline{BE}\cdot \overline{BE}\cdot \sin{(60\ensurement{^{\circ}})}\\\\
2A &=& \overline{BE}^2\cdot \sin{(60\ensurement{^{\circ}})}
\qquad | \qquad A = \sin{ (60\ensurement{^{\circ}}) } \\\\
2 \cdot \sin{ (60\ensurement{^{\circ}}) } &=& \overline{BE}^2\cdot \sin{(60\ensurement{^{\circ}})} \\\\
2 &=& \overline{BE}^2 \\\\
\overline{BE}^2 &=& 2 \qquad | \qquad \sqrt{} \\\\
\overline{BE} &=& \sqrt{2}
\end{array}
$
$}}$$

.
 May 30, 2015
 #5
avatar+575 
+8

.
 May 30, 2015
 #6
avatar+575 
+13

.
 May 30, 2015

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