Question Number 185271 by Mingma last updated on 19/Jan/23
Answered by HeferH last updated on 19/Jan/23
Commented by HeferH last updated on 19/Jan/23
$$\frac{\mathrm{4}\:+{x}\sqrt{\mathrm{3}}}{\mathrm{2}}\:+\:\mathrm{9}\:=\:\mathrm{4}\:+\:\mathrm{10}\:−\:{x}\sqrt{\mathrm{3}} \\ $$$$\:\mathrm{3}{x}\sqrt{\mathrm{3}}\:=\:\mathrm{6} \\ $$$$\:{x}\:=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:\Rightarrow\:{AB}\:=\:{AC}\:=\:\mathrm{14}\:−\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\centerdot\sqrt{\mathrm{3}}\:=\:\mathrm{12}\:{u}\: \\ $$