+0

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

+8
442
6
+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.

civonamzuk  May 29, 2015

#1
+18843
+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
Sort:

#1
+18843
+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
+1068
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)

civonamzuk  May 30, 2015
#3
+18843
+5
heureka  May 30, 2015
#4
+18843
+13

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

AE = AB = BE ( equilateral triangle )

P.S.

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

heureka  May 30, 2015
#5
+519
+8

fiora  May 30, 2015
#6
+519
+13

fiora  May 30, 2015

### 19 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details