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Question Number 192560 by Ari last updated on 20/May/23

Commented by dubylee last updated on 22/May/23

$$whatisthequestionhere?$$

Commented by Skabetix last updated on 21/May/23

$$‘‘findtheareaofthe{triangle}^{″}$$

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

$$produceBPtoQsuchGP=PQ$$$$joinA,QandQ,C$$$$GP=PQAP=PC$$$$AQCGparallelogram$$$$\therefore ar△AGQ=ar△AGC$$$$AG=\frac{2}{3}AN=\frac{2}{3}\times 12=8unit$$$$GC=\frac{2}{3}MC=AQ=10unit$$$$GP=\frac{1}{3}\times BP=\frac{1}{3}\times 9=3unit$$$$GQ=2\times 3=6unit$$$$\because {10}^2=6^2+8^2$$$$\therefore areaof△AGQ=\frac{1}{2}\times 6\times 8=24squ$$$$\therefore ar△AGC=24squ$$$$\therefore ar△ABC=24\times 3=72squ$$

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