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.
+26367 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
+26367 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
