## 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
$x$
.

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
$\frac{5b}{4} \times b =40 \rightarrow b^2 =\frac{40}{5/4}=32 \rightarrow b =\sqrt{32}=4 \sqrt{2}$
.
Then
$h=\frac{5b}{4}= \frac{5 \times 4 \sqrt{2}}{4}=5 \sqrt{2}$
.
Then
$x=10-5 \sqrt{2}=5(2- \sqrt{2})$
.

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