Question Number 76272 by Maclaurin Stickker last updated on 25/Dec/19
$${Find}\:{the}\:{area}\:{S}\:{of}\:{a}\:{triangle}\:{ABC} \\ $$$${as}\:{a}\:{function}\:{of}\:{the}\:{heights} \\ $$$${h}_{{a}} ,\:{h}_{{b}} \:{and}\:{h}_{{c}} . \\ $$
Commented by mr W last updated on 25/Dec/19
$${i}\:{got}\:{this}\:{result}: \\ $$$${S}=\frac{\mathrm{1}}{\:\sqrt{\left(\frac{\mathrm{1}}{{h}_{{a}} }+\frac{\mathrm{1}}{{h}_{{b}} }+\frac{\mathrm{1}}{{h}_{{c}} }\right)\left(−\frac{\mathrm{1}}{{h}_{{a}} }+\frac{\mathrm{1}}{{h}_{{b}} }+\frac{\mathrm{1}}{{h}_{{c}} }\right)\left(\frac{\mathrm{1}}{{h}_{{a}} }−\frac{\mathrm{1}}{{h}_{{b}} }+\frac{\mathrm{1}}{{h}_{{c}} }\right)\left(\frac{\mathrm{1}}{{h}_{{a}} }+\frac{\mathrm{1}}{{h}_{{b}} }−\frac{\mathrm{1}}{{h}_{{c}} }\right)}} \\ $$
Commented by Maclaurin Stickker last updated on 25/Dec/19
$${Thank}\:{you}.\:{I}\:{got}\:{the}\:{same}\:{result} \\ $$
Answered by Maclaurin Stickker last updated on 25/Dec/19
Commented by Maclaurin Stickker last updated on 25/Dec/19
$${let}\:{AB}={c},\:{BC}={a},\:{AC}={b},\:{BE}={h}_{\mathrm{b}} , \\ $$$${AD}={h}_{{a}\:} \:{and}\:{CF}={h}_{{c}} . \\ $$$${By}\:{Heron}'\:{formula},\:{we}\:{have}: \\ $$$${a}=\frac{\mathrm{2}{S}}{{h}_{{a}} },\:{b}=\frac{\mathrm{2}{S}}{{h}_{{b}} }\:{and}\:{c}=\frac{\mathrm{2}{S}}{{h}_{{c}} } \\ $$$${a}+{b}+{c}=\mathrm{2}{p} \\ $$$${p}=\left(\frac{{S}}{{h}_{{a}} }+\frac{{S}}{{h}_{{b}} }+\frac{{S}}{{h}_{{c}} }\right) \\ $$$${For}\:{area}: \\ $$$${S}=\sqrt{\left(\frac{{S}}{{h}_{{a}} }+\frac{{S}}{{h}_{{b}} }+\frac{{S}}{{h}_{{c}} }\right)\left(\frac{{S}}{{h}_{{b}} \:}+\frac{{S}}{{h}_{{c}} }−\frac{{S}}{{h}_{{a}} }\right)\left(\frac{{S}}{{h}_{{a}} }−\frac{{S}}{{h}_{{b}} }+\frac{{S}}{{h}_{{c}} }\right)\left(\frac{{S}}{{h}_{{a}} }+\frac{{S}}{{h}_{{b}} }−\frac{{S}}{{h}_{{c}} }\right)} \\ $$$$\frac{\mathrm{1}}{{S}^{\mathrm{2}} }=\left(\frac{\mathrm{1}}{{h}_{{a}} }+\frac{\mathrm{1}}{{h}_{{b}} }+\frac{\mathrm{1}}{{h}_{{c}} }\right)\left(\frac{\mathrm{1}}{{h}_{{b}} \:}+\frac{\mathrm{1}}{{h}_{{c}} }−\frac{\mathrm{1}}{{h}_{{a}} }\right)\left(\frac{\mathrm{1}}{{h}_{{a}} }−\frac{\mathrm{1}}{{h}_{{b}} }+\frac{\mathrm{1}}{{h}_{{c}} }\right)\left(\frac{\mathrm{1}}{{h}_{{a}} }+\frac{\mathrm{1}}{{h}_{{b}} }−\frac{\mathrm{1}}{{h}_{{c}} }\right) \\ $$$${S}=\frac{\mathrm{1}}{\:\sqrt{\left(\frac{\mathrm{1}}{{h}_{{a}} }+\frac{\mathrm{1}}{{h}_{{b}} }+\frac{\mathrm{1}}{{h}_{{c}} }\right)\left(\frac{\mathrm{1}}{{h}_{{b}} }+\frac{\mathrm{1}}{{h}_{{c}} }−\frac{\mathrm{1}}{{h}_{{a}} }\right)\left(\frac{\mathrm{1}}{{h}_{{a}} }−\frac{\mathrm{1}}{{h}_{{b}} }+\frac{\mathrm{1}}{{h}_{{c}} }\right)\left(\frac{\mathrm{1}}{{h}_{{a}} }+\frac{\mathrm{1}}{{h}_{{b}} }−\frac{\mathrm{1}}{{h}_{{c}} }\right)}} \\ $$$${S}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}\left(\frac{\mathrm{1}}{{h}_{{a}} ^{\mathrm{2}} {h}_{{b}} ^{\mathrm{2}} }+\frac{\mathrm{1}}{{h}_{{a}} ^{\mathrm{2}} {h}_{{c}} ^{\mathrm{2}} }+\frac{\mathrm{1}}{{h}_{{b}} ^{\mathrm{2}} {h}_{{c}} ^{\mathrm{2}} }\right)−\left(\frac{\mathrm{1}}{{h}_{{a}} ^{\mathrm{4}} }+\frac{\mathrm{1}}{{h}_{{b}} ^{\mathrm{4}} }+\frac{\mathrm{1}}{{h}_{{c}} ^{\mathrm{4}} }\right)}} \\ $$$$ \\ $$
Commented by benjo last updated on 26/Dec/19
$$\mathrm{sir}\:,\:\mathrm{what}\:\mathrm{the}\:\mathrm{application}\:\mathrm{for} \\ $$$$\mathrm{drawing}\:\mathrm{this}\:\mathrm{triangle}\:\mathrm{in}\:\mathrm{a}\:\mathrm{cellphone}? \\ $$
Commented by Maclaurin Stickker last updated on 26/Dec/19
$${MediBang} \\ $$