Question Number 102638 by bemath last updated on 10/Jul/20
$${to}\:{Mr}\:{W} \\ $$$${i}\:{forgot}\:{the}\:{formula}\:{to}\:{find} \\ $$$${the}\:{area}\:{of}\:{a}\:{triangle}\:{by} \\ $$$${using}\:{the}\:{three}\:{height}\:{line} \\ $$
Commented by Ar Brandon last updated on 10/Jul/20
$${Permit}\:{me}\:{to}\:{intervain} \\ $$$$\:\mathrm{Maybe}\:\mathrm{you}\:\mathrm{should}\:\mathrm{just}\:\mathrm{search}\:\boldsymbol{\mathrm{triangle}}. \\ $$$$\mathrm{It}\:\mathrm{is}\:\mathrm{possible}\:\mathrm{with}\:\mathrm{the}\:\mathrm{recent}\:\mathrm{update}. \\ $$
Commented by bemath last updated on 10/Jul/20
$${how}\:{sir}? \\ $$
Commented by mr W last updated on 10/Jul/20
$${do}\:{you}\:{mean}\:{Q}\mathrm{97236}? \\ $$
Commented by bemath last updated on 10/Jul/20
$${yes}\:{sir}.\:{thank}\:{you} \\ $$$$ \\ $$
Commented by Ar Brandon last updated on 10/Jul/20
I meant in "search question"��
Commented by bemath last updated on 10/Jul/20
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Answered by bramlex last updated on 10/Jul/20
$$\mathrm{area}\:=\:\frac{\mathrm{1}}{\sqrt{\left(\frac{\mathrm{1}}{\mathrm{h}_{\mathrm{1}} }+\frac{\mathrm{1}}{\mathrm{h}_{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{h}_{\mathrm{3}} }\right)\left(−\frac{\mathrm{1}}{\mathrm{h}_{\mathrm{1}} }+\frac{\mathrm{1}}{\mathrm{h}_{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{h}_{\mathrm{3}} }\right)\left(\frac{\mathrm{1}}{\mathrm{h}_{\mathrm{1}} }−\frac{\mathrm{1}}{\mathrm{h}_{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{h}_{\mathrm{3}} }\right)\left(\frac{\mathrm{1}}{\mathrm{h}_{\mathrm{1}} }+\frac{\mathrm{1}}{\mathrm{h}_{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{h}_{\mathrm{3}} }\right)}} \\ $$