Question

In $△ABC,BC$$=8$, CA$=6$ and AB$=10$. A line diving the triangle ABC into two regions of equal area is perpendicular to AB at point $X$. Find the value of $\frac{BX}{\sqrt{2}}$.

Answer 1

$tanB=\frac{AC}{BC}=\frac{n}{m}$

$\frac{6}{8}=\frac{n}{m}$

$\frac{3}{4}=\frac{n}{m}$

$Areaof△BXY=\frac{1}{2}\times n\times m=48$

$m\times n=\frac{48}{2}=24$

$m\times \frac{3}{4}\times m=24$

$m^2=32=4\sqrt{2}$

$BX=4\sqrt{2}$

$\frac{BX}{\sqrt{2}}=4$