Question Number 192560 by Ari last updated on 20/May/23
Commented by dubylee last updated on 22/May/23
$${what}\:{is}\:{the}\:{question}\:{here}? \\ $$
Commented by Skabetix last updated on 21/May/23
$$“{find}\:{the}\:{area}\:{of}\:{the}\:{triangle}'' \\ $$
Commented by Ari last updated on 22/May/23
$$\mathrm{given}\:\mathrm{the}\:\mathrm{median}\:\mathrm{of}\:\mathrm{then} \\ $$$$\mathrm{tria}{n}\mathrm{gle}\:\mathrm{find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{th}{is} \\ $$$$\mathrm{tr}{i}\mathrm{angle} \\ $$
Answered by som(math1967) last updated on 21/May/23
Commented by som(math1967) last updated on 21/May/23
$${produce}\:{BP}\:{to}\:{Q}\:{such}\:{GP}={PQ} \\ $$$${join}\:{A},{Q}\:{and}\:{Q},{C} \\ $$$$\:{GP}={PQ}\:\:\:{AP}={PC} \\ $$$${AQCG}\:{parallelogram} \\ $$$$\therefore\:\:{ar}\bigtriangleup{AGQ}={ar}\bigtriangleup{AGC} \\ $$$${AG}=\frac{\mathrm{2}}{\mathrm{3}}{AN}=\frac{\mathrm{2}}{\mathrm{3}}×\mathrm{12}=\mathrm{8}\:{unit} \\ $$$${GC}=\frac{\mathrm{2}}{\mathrm{3}}{MC}={AQ}=\mathrm{10}\:{unit} \\ $$$${GP}=\frac{\mathrm{1}}{\mathrm{3}}×{BP}=\frac{\mathrm{1}}{\mathrm{3}}×\mathrm{9}=\mathrm{3}{unit} \\ $$$${GQ}=\mathrm{2}×\mathrm{3}=\mathrm{6}{unit} \\ $$$$\because\:\mathrm{10}^{\mathrm{2}} =\mathrm{6}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} \\ $$$$\therefore{area}\:{of}\bigtriangleup{AGQ}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{6}×\mathrm{8}=\mathrm{24}{squ} \\ $$$$\therefore{ar}\bigtriangleup{AGC}=\mathrm{24}{squ} \\ $$$$\therefore{ar}\:\bigtriangleup{ABC}=\mathrm{24}×\mathrm{3}=\mathrm{72}{squ} \\ $$