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# The area of the equilateral triangle ABE = sin60o(cm2). Line segment AI = 5cm. BF = EF ; find the area DEGHI.

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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

#1
+26367
+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:

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

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
+26367
+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:

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

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
+1694
0

I'm too tired; I'll work on this one tomorrow; I'm going to

@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
+26367
+5
heureka May 30, 2015
#4
+26367
+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
+583
+8

fiora May 30, 2015
#6
+583
+13

fiora May 30, 2015