Question Number 192828 by Ari last updated on 28/May/23
Commented by Ari last updated on 28/May/23
Can we find the side x in the problem given in the figure?
Answered by MM42 last updated on 28/May/23
$${cosE}=\frac{\mathrm{25}+\mathrm{64}−\mathrm{49}}{\mathrm{2}×\mathrm{5}×\mathrm{8}}=\frac{\mathrm{40}}{\mathrm{80}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} =\mathrm{64}+\mathrm{16}−\mathrm{2}×\mathrm{8}×\mathrm{4}×\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{48} \\ $$$$\Rightarrow{x}=\mathrm{4}\sqrt{\mathrm{3}}\:\checkmark \\ $$
Answered by HeferH last updated on 29/May/23
Commented by HeferH last updated on 29/May/23
$${Say}\:{AC}=\mathrm{13}{k};\:\:\:\mathrm{5}{k}=\:\frac{\mathrm{35}}{\mathrm{13}};\:\mathrm{8}{k}=\frac{\mathrm{56}}{\mathrm{13}} \\ $$$${Bisector}\:{of}\:{AEC}=\sqrt{\mathrm{40}−\frac{\mathrm{35}\centerdot\mathrm{56}}{\mathrm{169}}}\:\:=\:\frac{\sqrt{\mathrm{169}\centerdot\mathrm{40}−\mathrm{49}\centerdot\mathrm{40}}}{\mathrm{13}} \\ $$$$\:=\:\frac{\mathrm{40}\sqrt{\mathrm{3}}}{\mathrm{13}} \\ $$$$\:{AD}=\:\frac{\mathrm{40}\sqrt{\mathrm{3}}}{\mathrm{13}}\centerdot\frac{\mathrm{13}}{\mathrm{5}}\:=\:\mathrm{8}\sqrt{\mathrm{3}} \\ $$$$\:{congruent}\:{triangles}\:\Rightarrow\:{x}\:=\:\frac{\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{4}\sqrt{\mathrm{3}} \\ $$