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Given that triangle ABCAis an irregular triangle, AD is the median which bisects BC, and the green, blue and red regions are squares. If the total area of green region and blue region is 9, find the area of the red region.

 

 Feb 18, 2020

Best Answer 

 #1
avatar+24430 
+2

Given that triangle ABCAis an irregular triangle, AD is the median which bisects BC, and the green, blue and red regions are squares.

If the total area of green region and blue region is 9, find the area of the red region.

 

\(b^2m+c^2n=a(d^2+mn)\)

Source: https://en.wikipedia.org/wiki/Cevian#Ratio_properties

 

\(\begin{array}{|rcll|} \hline b^2m+c^2n &=& a(d^2+mn) \quad | \quad m=n, a=2n \\ b^2n+c^2n &=& 2n(d^2+n^2) \\ (b^2+c^2)n &=& 2n(d^2+n^2) \\ {\color{red}b^2}+{\color{red}c^2} &=& 2({\color{blue}d^2}+{\color{green}n^2}) \\ {\color{red}b^2}+{\color{red}c^2} &=& 2*9 \\ \mathbf{{\color{red}b^2}+{\color{red}c^2}} &=& \mathbf{18} \\ \hline \end{array}\)

 

The area of the red region is 18

 

laugh

 Feb 18, 2020
 #1
avatar+24430 
+2
Best Answer

Given that triangle ABCAis an irregular triangle, AD is the median which bisects BC, and the green, blue and red regions are squares.

If the total area of green region and blue region is 9, find the area of the red region.

 

\(b^2m+c^2n=a(d^2+mn)\)

Source: https://en.wikipedia.org/wiki/Cevian#Ratio_properties

 

\(\begin{array}{|rcll|} \hline b^2m+c^2n &=& a(d^2+mn) \quad | \quad m=n, a=2n \\ b^2n+c^2n &=& 2n(d^2+n^2) \\ (b^2+c^2)n &=& 2n(d^2+n^2) \\ {\color{red}b^2}+{\color{red}c^2} &=& 2({\color{blue}d^2}+{\color{green}n^2}) \\ {\color{red}b^2}+{\color{red}c^2} &=& 2*9 \\ \mathbf{{\color{red}b^2}+{\color{red}c^2}} &=& \mathbf{18} \\ \hline \end{array}\)

 

The area of the red region is 18

 

laugh

heureka Feb 18, 2020
 #2
avatar+109740 
+1

Very nice, heureka  !!!!

 

 

 

cool cool cool

CPhill  Feb 18, 2020
 #3
avatar+24430 
+2

Thank you, CPhill !

 

laugh

heureka  Feb 19, 2020

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