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Question Number 193099 by Mingma last updated on 04/Jun/23

Answered by som(math1967) last updated on 04/Jun/23

Commented by som(math1967) last updated on 04/Jun/23

b^2 =((16)/( (√3))) , c^2 =((36)/( (√3))) cos60=((b^2 +c^2 −a^2 )/(2bc)) ⇒(1/2)=((((16)/( (√3)))+((36)/( (√3)))−a^2 )/(2×((4×6)/( (√3))))) ⇒((24)/( (√3)))=((52)/( (√3))) −a^2 ⇒a^2 =((52−24)/( (√3))) ∴((√3)/4)×a^2 =((28)/( (√3)))×((√3)/4)=7squ =red area

$$\:{b}^{\mathrm{2}} =\frac{\mathrm{16}}{\:\sqrt{\mathrm{3}}}\:\:,\:\:{c}^{\mathrm{2}} =\frac{\mathrm{36}}{\:\sqrt{\mathrm{3}}} \\ $$$$\:{cos}\mathrm{60}=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}=\frac{\frac{\mathrm{16}}{\:\sqrt{\mathrm{3}}}+\frac{\mathrm{36}}{\:\sqrt{\mathrm{3}}}−{a}^{\mathrm{2}} }{\mathrm{2}×\frac{\mathrm{4}×\mathrm{6}}{\:\sqrt{\mathrm{3}}}} \\ $$$$\Rightarrow\frac{\mathrm{24}}{\:\sqrt{\mathrm{3}}}=\frac{\mathrm{52}}{\:\sqrt{\mathrm{3}}}\:−{a}^{\mathrm{2}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} =\frac{\mathrm{52}−\mathrm{24}}{\:\sqrt{\mathrm{3}}} \\ $$$$\therefore\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}×{a}^{\mathrm{2}} =\frac{\mathrm{28}}{\:\sqrt{\mathrm{3}}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}=\mathrm{7}{squ}\:={red}\:{area} \\ $$

Commented by Mingma last updated on 04/Jun/23

Perfect ��

Commented by Mingma last updated on 04/Jun/23

What's your solution for the green area?

Commented by math1234 last updated on 04/Jun/23

Commented by math1234 last updated on 04/Jun/23

Commented by math1234 last updated on 04/Jun/23

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