Question Number 192704 by Mingma last updated on 25/May/23
Answered by HeferH last updated on 25/May/23
Commented by HeferH last updated on 25/May/23
$$\mathrm{12}+\mathrm{12}\sqrt{\mathrm{3}}−{x}\:=\mathrm{24} \\ $$$$\:{x}=\mathrm{12}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right) \\ $$$$\:{shaded}\:=\:\frac{\mathrm{12}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)\centerdot\mathrm{6}}{\mathrm{2}}\:=\:\mathrm{36}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)\:{u}^{\mathrm{2}} \\ $$
Commented by Mingma last updated on 25/May/23
Perfect ��
Answered by MM42 last updated on 25/May/23
$${BC}=\mathrm{24}{sin}\mathrm{15}\:\:\&\:\:<{BCD}=\mathrm{105}^{\mathrm{0}} \:\: \\ $$$$\bigtriangleup{BCD}\:\::\:\frac{\mathrm{12}}{{sin}\mathrm{105}}\:=\frac{\mathrm{24}{sin}\mathrm{15}}{{sinD}}\: \\ $$$$\Rightarrow{sinD}=\mathrm{2}{sin}\mathrm{105}{sin}\mathrm{15}=\mathrm{2}{cos}\mathrm{15}{sin}\mathrm{15}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow<{D}=\mathrm{30}^{\mathrm{0}} \:\Rightarrow<{CBD}=\mathrm{45}^{\mathrm{0}} \\ $$$$\Rightarrow{S}_{{BCD}} =\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{12}×\mathrm{24}×{sin}\mathrm{15}×{sin}\mathrm{45}=\mathrm{36}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)\:\checkmark \\ $$
Commented by MM42 last updated on 25/May/23
Commented by Mingma last updated on 25/May/23
Perfect ��