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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

#1
+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 Feb 18, 2020

#1
+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 heureka Feb 18, 2020
#2
+1

Very nice, heureka  !!!!   CPhill  Feb 18, 2020
#3
+2

Thank you, CPhill ! heureka  Feb 19, 2020