Question Number 223553 by fantastic last updated on 29/Jul/25
$${the}\:{length}\:{of}\:\mathrm{3}\:{meadians}\:{of}\:{a}\:{triangle}\:{is}\:{given} \\ $$$${how}\:{to}\:{calculate}\:{the}\:{area}? \\ $$
Commented by fantastic last updated on 29/Jul/25
$${please}\:{help} \\ $$
Answered by mr W last updated on 29/Jul/25
Commented by mr W last updated on 29/Jul/25
$${say}\:{the}\:{lengthes}\:{of}\:{the}\:{medians}\:{of}\: \\ $$$${the}\:{triangle}\:{ABC}\:{are}\:{m}_{{a}} ,\:{m}_{{b}} ,\:{m}_{{c}} . \\ $$$${we}\:{can}\:{see}\:{and}\:{prove}\:{that}\:{the}\:{green} \\ $$$${hatched}\:{triangle}\:{has}\:{the}\:{sides}\: \\ $$$$\frac{{m}_{{a}} }{\mathrm{3}},\:\frac{{m}_{{b}} }{\mathrm{3}},\:\frac{{m}_{{c}} }{\mathrm{3}}\:{and}\:{its}\:{area}\:{is}\:\frac{\mathrm{1}}{\mathrm{12}}\:{of}\:{the} \\ $$$${area}\:{of}\:\Delta{ABC}. \\ $$$$\Delta_{{green}} =\frac{\mathrm{1}}{\mathrm{9}}\:{of}\:{the}\:{area}\:{of}\:{triangle}\: \\ $$$${with}\:{sides}\:{m}_{{a}} ,{m}_{{b}} ,{m}_{{c}} . \\ $$$$\Delta_{{green}} =\frac{\mathrm{1}}{\mathrm{12}}\:{of}\:{area}\:{of}\:\Delta{ABC} \\ $$$$\Rightarrow\Delta{ABC}=\frac{\mathrm{12}}{\mathrm{9}}×{triangle}\:{with}\:{sides}\:{m}_{{a}} ,{m}_{{b}} ,{m}_{{c}} . \\ $$$$\Rightarrow\Delta{ABC}=\frac{\mathrm{4}\sqrt{{s}_{{m}} \left({s}_{{m}} −{m}_{{a}} \right)\left({s}_{{m}} −{m}_{{b}} \right)\left({s}_{{m}} −{m}_{{c}} \right)}}{\mathrm{3}} \\ $$$${with}\:{s}_{{m}} =\frac{{m}_{{a}} +{m}_{{b}} +{m}_{{c}} }{\mathrm{2}} \\ $$
Commented by fantastic last updated on 29/Jul/25
$${thank}\:{you}\:{very}\:{much}\:{sir} \\ $$