Question Number 72362 by Maclaurin Stickker last updated on 27/Oct/19
$${Calculate}\:{the}\:{sides}\:{of}\:{a}\:{triangle} \\ $$$${knowing}\:{the}\:{heights}\:{h}_{\mathrm{a}\:} =\frac{\mathrm{1}}{\mathrm{9}} \\ $$$${h}_{\mathrm{b}} =\frac{\mathrm{1}}{\mathrm{7}}\:{and}\:{h}_{\mathrm{c}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Answered by mr W last updated on 27/Oct/19
$${let}\:\Delta={area}\:{of}\:{triangle} \\ $$$${ah}_{{a}} ={bh}_{{b}} ={ch}_{{c}} =\mathrm{2}\Delta={k}\:\left({say}\right) \\ $$$$\Rightarrow{a}=\frac{{k}}{{h}_{{a}} }=\mathrm{9}{k} \\ $$$$\Rightarrow{b}=\frac{{k}}{{h}_{{b}} }=\mathrm{7}{k} \\ $$$$\Rightarrow{c}=\frac{{k}}{{h}_{{c}} }=\mathrm{4}{k} \\ $$$${s}=\frac{{a}+{b}+{c}}{\mathrm{2}}=\mathrm{10}{k} \\ $$$$\Delta=\sqrt{{s}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)}=\sqrt{\mathrm{10}×\mathrm{1}×\mathrm{3}×\mathrm{6}}{k}^{\mathrm{2}} =\mathrm{6}\sqrt{\mathrm{5}}{k}^{\mathrm{2}} \\ $$$$\mathrm{2}\Delta=\mathrm{12}\sqrt{\mathrm{5}}{k}^{\mathrm{2}} ={k} \\ $$$$\Rightarrow{k}=\frac{\mathrm{1}}{\mathrm{12}\sqrt{\mathrm{5}}}=\frac{\sqrt{\mathrm{5}}}{\mathrm{60}} \\ $$$$\Rightarrow{a}=\frac{\mathrm{9}\sqrt{\mathrm{5}}}{\mathrm{60}}=\frac{\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{20}} \\ $$$$\Rightarrow{b}=\frac{\mathrm{7}\sqrt{\mathrm{5}}}{\mathrm{60}} \\ $$$$\Rightarrow{c}=\frac{\mathrm{4}\sqrt{\mathrm{5}}}{\mathrm{60}}=\frac{\sqrt{\mathrm{5}}}{\mathrm{15}} \\ $$
Commented by Maclaurin Stickker last updated on 27/Oct/19
$${Why}\:{a}=\mathrm{9}{k}? \\ $$
Commented by mr W last updated on 27/Oct/19
$${a}×{h}_{{a}} =\mathrm{2}\Delta={k}\:{with}\:\Delta={area} \\ $$$${a}=\frac{\mathrm{2}\Delta}{{h}_{{a}} }=\frac{{k}}{{h}_{{a}} }=\mathrm{9}{k} \\ $$
Commented by Maclaurin Stickker last updated on 27/Oct/19
$${Oh},\:{thank}\:{you}. \\ $$