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Square ABCD has area 12.25.

 

E is the intersection of BD and the semicircle of diameter AD.

 

CF is a segment drawn from C and tangnet to the semicircle at F.

 

Find the area of triangle DEF.

 

 Feb 20, 2020
 #1
avatar+24430 
+3

Square ABCD has area 12.25.

E is the intersection of BD and the semicircle of diameter AD.

CF is a segment drawn from C and tangnet to the semicircle at F.

Find the area of triangle DEF.

 

\(\begin{array}{|lll|} \hline\mathbf{Point }~E: \text{ Intersection circle and line }BD \\ \hline \begin{array}{rcll} \mathbf{y_E} &=& \mathbf{-x_E+2r} \\ \mathbf{y_E^2} &=& \mathbf{2rx_E-x_E^2} \\ (-x_E+2r)^2 &=& 2rx_E-x_E^2 \\ x_E^2 - 4rx_E+4r^2 &=& 2rx_E-x_E^2 \\ 2x_E^2-6rx_E+4r^2 &=& 0 \quad | \quad :2 \\ \mathbf{x_E^2-3rx_E+2r^2} &=& \mathbf{0} \\\\ x_E^2 &=& \dfrac{3r\pm \sqrt{9r^2-4*(2r^2)} }{2} \\ x_E^2 &=& \dfrac{3r\pm \sqrt{r^2} }{2} \\ x_E^2 &=& \dfrac{3r\pm r} {2} \\\\ x_E &=& \dfrac{3r-r}{2} \\ x_E &=& \dfrac{2r}{2} \\ \mathbf{x_E} &=& \mathbf{r}\ \checkmark \\\\ x_E &=& \dfrac{3r+r}{2} \\ x_E &=& \dfrac{4r}{2} \\ \mathbf{x_E} &=& \mathbf{2r} \quad | \quad \text{Pont }E ~\neq~ \text{Pont }D \\\\ y_E &=& -x_E+2r \quad | \quad x_E= r\\ y_E &=& -r+2r \\ \mathbf{y_E} &=& \mathbf{r}\ \checkmark \\\\ \mathbf{E} &=& \mathbf{(r,r)} \\ \end{array}\\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{MH}: \\ \hline r^2 &=& MH*CM \quad | \quad CM=r\sqrt{5} \\ r^2 &=& MH*r\sqrt{5} \\ r &=& MH\sqrt{5} \\ \mathbf{MH} &=& \mathbf{\dfrac{r}{\sqrt{5}} } \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{FH}: \\ \hline FH^2 &=& MH*(CM-MH) \\ && \boxed{MH=\dfrac{r}{\sqrt{5}},\ CM=r\sqrt{5}} \\ FH^2 &=& \dfrac{r}{\sqrt{5}}*\left(r\sqrt{5}-\dfrac{r}{\sqrt{5}}\right) \\\\ FH^2 &=& r^2-\dfrac{r^2}{5} \\\\ FH^2 &=& \dfrac{4r^2}{5} \\\\ \mathbf{FH} &=& \mathbf{\dfrac{2r}{\sqrt{5}}} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{x_F}: \\ \hline FD^2 &=& (AD-x_F)*AD \quad | \quad FD=2FH,\ AD=2r \\ (2FH)^2 &=& (2r-x_F)*2r \quad | \quad FH =\dfrac{2r}{\sqrt{5}} \\ \left(\dfrac{4r}{\sqrt{5}}\right)^2 &=& (2r-x_F)*2r \\ \dfrac{16r^2}{5} &=& 4r^2-2rx_F \\ 2rx_F &=& 4r^2 - \dfrac{16r^2}{5} \quad | \quad :2r \\ x_F &=& 2r - \dfrac{8r}{5} \\ \mathbf{x_F} &=& \mathbf{\dfrac{2r}{5}} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \mathbf{y_F}: \\ \hline y_F^2 &=& x_F(AD-x_F) \quad | \quad x_F=\dfrac{2r}{5},\ AD=2r \\\\ y_F^2 &=& \dfrac{2r}{5}(2r-\dfrac{2r}{5}) \\\\ y_F^2 &=& \dfrac{4r^2}{5} - \dfrac{4r^2}{25} \\\\ y_F^2 &=& \dfrac{20r^2}{5} - \dfrac{4r^2}{25} \\\\ y_F^2 &=& \dfrac{16r^2}{25} \\\\ \mathbf{y_F} &=& \mathbf{\dfrac{4r}{5}} \\ \hline \end{array}\)

 

\(\mathbf{F} = \mathbf{(\dfrac{2r}{5},\dfrac{4r}{5})} \)

 

\(\begin{array}{|lll|} \hline \mathbf{\text{The area of triangle }~DEF}: \\ \begin{array}{lccc} \hline \text{Point} & x & y & \text{cross product} \\ \hline D: & 2r & 0 \\ E: & r & r & 2r*r-r*0 \\ F: & \dfrac{2r}{5} & \dfrac{4r}{5} & r *\dfrac{4r}{5} - \dfrac{2r}{5} * r \\ D: & 2r & 0 & \dfrac{2r}{5} * 0 - 2r * \dfrac{4r}{5} \\ \hline & & & \text{sum }~=2r^2 + \dfrac{4r^2}{5} - \dfrac{2r^2}{5} - \dfrac{8r^2}{5} \\ & & & =2r^2\left(1 + \dfrac{2}{5} - \dfrac{1}{5} - \dfrac{4r}{5}\right) \\ & & & =2r^2\left(1 + \dfrac{2}{5} - 1\right) \\ & & & =2r^2\left( \dfrac{2}{5}\right) \\ & & & =\dfrac{4r^2}{5} \\ \hline \end{array}\\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \text{The area of triangle }~DEF &=& \dfrac{\text{sum}}{2} \\\\ &=& \dfrac{\dfrac{4r^2}{5}}{2} \\\\ &=& \dfrac{2r^2}{5} \\\\ && \text{Square }~ ABCD &= AB*AD \quad | \quad AB=AD=2r \\ && \text{Square }~ ABCD &= (2r)^2 \\ && \text{Square }~ ABCD &= 4r^2 \quad | \quad \text{Square }~ ABCD= 12.25 \\ && 12.25 &= 4r^2 \quad | \quad :4 \\ && 3.0625 &= r^2 \\ && \mathbf{r^2} &= \mathbf{3.0625} \\\\ &=& \dfrac{2*3.0625}{5} \\\\ \mathbf{\text{The area of triangle }~DEF} &=& \mathbf{1.225} \\ \hline \end{array}\)

 

Source: https://en.wikipedia.org/wiki/Geometric_mean_theorem

https://www.youtube.com/watch?v=AByG1nTvA_Q

 

laugh

 Feb 20, 2020
 #2
avatar+24430 
+3

Square ABCD has area 12.25.
E is the intersection of BD and the semicircle of diameter AD.
CF is a segment drawn from C and tangnet to the semicircle at F.
Find the area of triangle DEF.

 

 

 

 

\(\begin{array}{|lll|} \hline\mathbf{Point }~E: \text{ Intersection circle and line }BD \\ \hline \begin{array}{rcll} \mathbf{y_E} &=& \mathbf{-x_E+2r} \\ \mathbf{y_E^2} &=& \mathbf{2rx_E-x_E^2} \\ (-x_E+2r)^2 &=& 2rx_E-x_E^2 \\ x_E^2 - 4rx_E+4r^2 &=& 2rx_E-x_E^2 \\ 2x_E^2-6rx_E+4r^2 &=& 0 \quad | \quad :2 \\ \mathbf{x_E^2-3rx_E+2r^2} &=& \mathbf{0} \\\\ x_E^2 &=& \dfrac{3r\pm \sqrt{9r^2-4*(2r^2)} }{2} \\ x_E^2 &=& \dfrac{3r\pm \sqrt{r^2} }{2} \\ x_E^2 &=& \dfrac{3r\pm r} {2} \\\\ x_E &=& \dfrac{3r-r}{2} \\ x_E &=& \dfrac{2r}{2} \\ \mathbf{x_E} &=& \mathbf{r}\ \checkmark \\\\ x_E &=& \dfrac{3r+r}{2} \\ x_E &=& \dfrac{4r}{2} \\ \mathbf{x_E} &=& \mathbf{2r} \quad | \quad \text{Pont }E ~\neq~ \text{Pont }D \\\\ y_E &=& -x_E+2r \quad | \quad x_E= r\\ y_E &=& -r+2r \\ \mathbf{y_E} &=& \mathbf{r}\ \checkmark \\\\ \mathbf{E} &=& \mathbf{(r,r)} \\ \end{array}\\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{MH}: \\ \hline r^2 &=& MH*CM \quad | \quad CM=r\sqrt{5} \\ r^2 &=& MH*r\sqrt{5} \\ r &=& MH\sqrt{5} \\ \mathbf{MH} &=& \mathbf{\dfrac{r}{\sqrt{5}} } \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \mathbf{FH}: \\ \hline FH^2 &=& MH*(CM-MH) \\ && \boxed{MH=\dfrac{r}{\sqrt{5}},\ CM=r\sqrt{5}} \\ FH^2 &=& \dfrac{r}{\sqrt{5}}*\left(r\sqrt{5}-\dfrac{r}{\sqrt{5}}\right) \\\\ FH^2 &=& r^2-\dfrac{r^2}{5} \\\\ FH^2 &=& \dfrac{4r^2}{5} \\\\ \mathbf{FH} &=& \mathbf{\dfrac{2r}{\sqrt{5}}} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{x_F}: \\ \hline FD^2 &=& (AD-x_F)*AD \quad | \quad FD=2FH,\ AD=2r \\ (2FH)^2 &=& (2r-x_F)*2r \quad | \quad FH =\dfrac{2r}{\sqrt{5}} \\ \left(\dfrac{4r}{\sqrt{5}}\right)^2 &=& (2r-x_F)*2r \\ \dfrac{16r^2}{5} &=& 4r^2-2rx_F \\ 2rx_F &=& 4r^2 - \dfrac{16r^2}{5} \quad | \quad :2r \\ x_F &=& 2r - \dfrac{8r}{5} \\ \mathbf{x_F} &=& \mathbf{\dfrac{2r}{5}} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{y_F}: \\ \hline y_F^2 &=& x_F(AD-x_F) \quad | \quad x_F=\dfrac{2r}{5},\ AD=2r \\\\ y_F^2 &=& \dfrac{2r}{5}(2r-\dfrac{2r}{5}) \\\\ y_F^2 &=& \dfrac{4r^2}{5} - \dfrac{4r^2}{25} \\\\ y_F^2 &=& \dfrac{20r^2}{5} - \dfrac{4r^2}{25} \\\\ y_F^2 &=& \dfrac{16r^2}{25} \\\\ \mathbf{y_F} &=& \mathbf{\dfrac{4r}{5}} \\ \hline \end{array}\)

 

\(\mathbf{F} = \mathbf{(\dfrac{2r}{5},\dfrac{4r}{5})} \)

 

\(\begin{array}{|lll|} \hline \mathbf{\text{The area of triangle }~DEF}: \\ \begin{array}{lccc} \hline \text{Point} & x & y & \text{cross product} \\ \hline D: & 2r & 0 \\ E: & r & r & 2r*r-r*0 \\ F: & \dfrac{2r}{5} & \dfrac{4r}{5} & r *\dfrac{4r}{5} - \dfrac{2r}{5} * r \\ D: & 2r & 0 & \dfrac{2r}{5} * 0 - 2r * \dfrac{4r}{5} \\ \hline & & & \text{sum }~=2r^2 + \dfrac{4r^2}{5} - \dfrac{2r^2}{5} - \dfrac{8r^2}{5} \\ & & & =2r^2\left(1 + \dfrac{2}{5} - \dfrac{1}{5} - \dfrac{4r}{5}\right) \\ & & & =2r^2\left(1 + \dfrac{2}{5} - 1\right) \\ & & & =2r^2\left( \dfrac{2}{5}\right) \\ & & & =\dfrac{4r^2}{5} \\ \hline \end{array}\\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \text{The area of triangle }~DEF &=& \dfrac{\text{sum}}{2} \\\\ &=& \dfrac{\dfrac{4r^2}{5}}{2} \\\\ &=& \dfrac{2r^2}{5} \\\\ && \text{Square }~ ABCD &= AB*AD \quad | \quad AB=AD=2r \\ && \text{Square }~ ABCD &= (2r)^2 \\ && \text{Square }~ ABCD &= 4r^2 \quad | \quad \text{Square }~ ABCD= 12.25 \\ && 12.25 &= 4r^2 \quad | \quad :4 \\ && 3.0625 &= r^2 \\ && \mathbf{r^2} &= \mathbf{3.0625} \\\\ &=& \dfrac{2*3.0625}{5} \\\\ \mathbf{\text{The area of triangle }~DEF} &=& \mathbf{1.225} \\ \hline \end{array}\)

 

source:
https://en.wikipedia.org/wiki/Geometric_mean_theorem

https://www.youtube.com/watch?v=AByG1nTvA_Q

 

laugh

 Feb 20, 2020
 #3
avatar+531 
+2

AD = 3.5

ND = 1.75

BD = 3.13

DG = 2.97

FG = 0.99

DE = 2.475

EG = 0.495

Area of  Δ DFG = 1.47 u²

Area of  Δ EFG = 0.245 u² 

Area of  Δ DEF = 1.225 u²     wink

 

 Feb 20, 2020
edited by Dragan  Feb 20, 2020
edited by Dragan  Feb 21, 2020

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