Question Number 216279 by BaliramKumar last updated on 02/Feb/25 | ||
![]() | ||
Commented by mairatonny last updated on 02/Feb/25 | ||
![]() | ||
Commented by mairatonny last updated on 02/Feb/25 | ||
someone please do these maths please | ||
Commented by AntonCWX last updated on 03/Feb/25 | ||
![]() | ||
$${You}\:{shouldn}'{t}\:{send}\:{your}\:{questions}\: \\ $$$${here}\:{in}\:{the}\:{other}\:{people}'{s}\:{questions}. \\ $$$${Send}\:{yourself}\:{again}. \\ $$ | ||
Answered by mr W last updated on 02/Feb/25 | ||
![]() | ||
$${A}={area} \\ $$$${A}=\frac{\mathrm{10}{a}}{\mathrm{2}}=\frac{\mathrm{12}{b}}{\mathrm{2}}=\frac{\mathrm{15}{c}}{\mathrm{2}} \\ $$$$\Rightarrow{a}=\frac{{A}}{\mathrm{5}},\:{b}=\frac{{A}}{\mathrm{6}},\:{c}=\frac{{A}}{\mathrm{7}.\mathrm{5}} \\ $$$${s}=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{A}}{\mathrm{5}}+\frac{{A}}{\mathrm{6}}+\frac{{A}}{\mathrm{7}.\mathrm{5}}\right)=\frac{{A}}{\mathrm{4}} \\ $$$${A}=\sqrt{\frac{{A}}{\mathrm{4}}\left(\frac{{A}}{\mathrm{4}}−\frac{{A}}{\mathrm{5}}\right)\left(\frac{{A}}{\mathrm{4}}−\frac{{A}}{\mathrm{6}}\right)\left(\frac{{A}}{\mathrm{4}}−\frac{{A}}{\mathrm{7}.\mathrm{5}}\right)} \\ $$$$\mathrm{1}=\frac{{A}\sqrt{\mathrm{7}}}{\mathrm{240}} \\ $$$$\left.\Rightarrow{A}=\frac{\mathrm{240}}{\:\sqrt{\mathrm{7}}}\:\:\:\Rightarrow{answer}\:{b}\right) \\ $$ | ||