Given 3 heights : $h_1=5\mathrm{cm}$ ; $h_2=7\mathrm{cm}$ ; $h_3=8\mathrm{cm}$ ... It is required to draw that triangle using only compass and ruler !
N.B.: It is not allowed to calculate the area then the sides: the measure of the sides won't be exact.
Thanks all
Given the heights (and an arbitrary "unit" length) we can construct lengths $\frac1{h_1}$, $\frac1{h_2}$, $\frac1{h_3}$. Construct a triangle $APQ$ with $PQ=\frac1{h_1}$, $AQ=\frac1{h_2}$, $AP=\frac1{h_3}$. This triangle has heights $h_A\sim h_1$, $h_P\sim h_2$, $h_Q\sim h_3$. All we need is to scale appropriately. Let $g$ be the line through $A$ perpendicular to $PQ$. Find $R$ on $g$ with $AR=h_1$. The line through $R$ perpendicular to $g$ (or equivalently: parallel to $PQ$) intersects $AP$ in $B$ and $AQ$ in $C$. Then in triangle $ABC$ we have $h_A=h_1$, $h_B=h_2$, $h_C=h_3$.