Question Number 148880 by DELETED last updated on 01/Aug/21
Answered by DELETED last updated on 01/Aug/21
$$\mid\mathrm{AC}\mid=\sqrt{\mathrm{10}^{\mathrm{2}} +\mathrm{10}^{\mathrm{2}} }=\mathrm{10}\sqrt{\mathrm{2}}\:\mathrm{cm} \\ $$$$\mid\mathrm{CG}\mid=\mathrm{10}\:\mathrm{cm} \\ $$$$\rightarrow\mathrm{Luas}\:\Delta\mathrm{ACG}: \\ $$$$\:\:\:\:=\:\:\frac{\mathrm{10}×\mathrm{10}\sqrt{\mathrm{2}}\:}{\mathrm{2}}=\mathrm{50}\sqrt{\mathrm{2}}\:\mathrm{cm}^{\mathrm{2}} \\ $$