Question-224003

Question Number 224003 by mr W last updated on 13/Aug/25

Commented by mr W last updated on 13/Aug/25

$${the}\:{areas}\:{of}\:{two}\:{equilaterals}\:{are} \\ $$$${known}.\:{find}\:{the}\:{area}\:{of}\:{the}\:{third} \\ $$$${triangle}. \\ $$

Answered by A5T last updated on 13/Aug/25

((s_1 ^2 (√3))/4)=4 and ((s_2 ^2 (√3))/4)=9 ⇒s_1 =(√((16(√3))/3)) and s_2 =(√((36(√3))/3)) ⇒Area=((s_1 s_2 ×sin60°)/2)=(√((16×36)/3))×((√3)/4)=6

$$\frac{\mathrm{s}_{\mathrm{1}} ^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{4}}=\mathrm{4}\:\mathrm{and}\:\frac{\mathrm{s}_{\mathrm{2}} ^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{4}}=\mathrm{9} \\ $$$$\Rightarrow\mathrm{s}_{\mathrm{1}} =\sqrt{\frac{\mathrm{16}\sqrt{\mathrm{3}}}{\mathrm{3}}}\:\:\mathrm{and}\:\mathrm{s}_{\mathrm{2}} =\sqrt{\frac{\mathrm{36}\sqrt{\mathrm{3}}}{\mathrm{3}}} \\ $$$$\Rightarrow\mathrm{Area}=\frac{\mathrm{s}_{\mathrm{1}} \mathrm{s}_{\mathrm{2}} ×\mathrm{sin60}°}{\mathrm{2}}=\sqrt{\frac{\mathrm{16}×\mathrm{36}}{\mathrm{3}}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}=\mathrm{6} \\ $$

Commented by mr W last updated on 14/Aug/25

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Answered by mr W last updated on 14/Aug/25

Commented by mr W last updated on 14/Aug/25

(A_2 /A_1 )=(b/a) (A_2 /A_3 )=(a/b) ⇒(A_2 ^2 /(A_1 A_3 ))=1 ⇒A_2 =(√(A_1 A_3 ))=(√(4×9))=6

$$\frac{{A}_{\mathrm{2}} }{{A}_{\mathrm{1}} }=\frac{{b}}{{a}} \\ $$$$\frac{{A}_{\mathrm{2}} }{{A}_{\mathrm{3}} }=\frac{{a}}{{b}} \\ $$$$\Rightarrow\frac{{A}_{\mathrm{2}} ^{\mathrm{2}} }{{A}_{\mathrm{1}} {A}_{\mathrm{3}} }=\mathrm{1}\:\Rightarrow{A}_{\mathrm{2}} =\sqrt{{A}_{\mathrm{1}} {A}_{\mathrm{3}} }=\sqrt{\mathrm{4}×\mathrm{9}}=\mathrm{6} \\ $$

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