Cuttting a Triangle in Half

What distance from the base can we cut a triangle so that the triangle is cut in half?

Draw a line PQ parallel to the base as shown and let the height of this line above the base be  

The triangles ABC and PBQ are similar so  
\[\frac{h}{b}=\frac{10}{8} \rightarrow h=\frac{5b}{4}\]

The area of triangle ABC is  
\[\frac{8} \times 10=40\]
  and the area of triangle PBQ is half of this, so  
\[\frac{1}{2}b \times h=20 \rightarrow bh=40\]

We now have simultaneous equations,  
\[h=\frac{5b}{4}, \: hb=40\]
Substitute the second of these into the first to give  
\[\frac{5b}{4} \times b =40 \rightarrow b^2 =\frac{40}{5/4}=32 \rightarrow b =\sqrt{32}=4 \sqrt{2}\]
\[h=\frac{5b}{4}= \frac{5 \times 4 \sqrt{2}}{4}=5 \sqrt{2}\]
\[x=10-5 \sqrt{2}=5(2- \sqrt{2})\]

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