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Question Number 185900 by Mingma last updated on 29/Jan/23

Answered by mr W last updated on 29/Jan/23

say AB=BC=a ∠ABQ=15° DQ=a(1−tan 15°) DR=((DQ)/(tan 15°)) orange=((DQ×DR)/2)=((a^2 (1−tan 15°)^2 )/(2 tan 15°)) =a^2 ×(((1−2+(√3))^2 )/(2(2−(√3))))=a^2 ×((2(2−(√3)))/(2(2−(√3))))=a^2 =yellow ✓

$${say}\:{AB}={BC}={a} \\ $$$$\angle{ABQ}=\mathrm{15}° \\ $$$${DQ}={a}\left(\mathrm{1}−\mathrm{tan}\:\mathrm{15}°\right) \\ $$$${DR}=\frac{{DQ}}{\mathrm{tan}\:\mathrm{15}°} \\ $$$${orange}=\frac{{DQ}×{DR}}{\mathrm{2}}=\frac{{a}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{tan}\:\mathrm{15}°\right)^{\mathrm{2}} }{\mathrm{2}\:\mathrm{tan}\:\mathrm{15}°} \\ $$$$={a}^{\mathrm{2}} ×\frac{\left(\mathrm{1}−\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)}={a}^{\mathrm{2}} ×\frac{\mathrm{2}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)}{\mathrm{2}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)}={a}^{\mathrm{2}} \\ $$$$={yellow}\:\checkmark \\ $$

Answered by HeferH last updated on 29/Jan/23

let AB = a ∠ BQA = 75° , ∠ABQ = 15°⇒ AQ = a(2−(√3))⇒ QD = a − a(2−(√3))= a((√3)−1)⇒ DR = ((a((√3)−1))/(2−(√3))) Orange area = ((QD∙DR)/2) = [((a^2 ((√3)−1)^2 )/(2−(√3)))](1/2) = ((a^2 (4−2(√3)))/(4−2(√3))) = a^2 = Yellow area

$${let}\:{AB}\:=\:{a} \\ $$$$\angle\:{BQA}\:=\:\mathrm{75}°\:,\:\angle{ABQ}\:=\:\mathrm{15}°\Rightarrow \\ $$$$\:{AQ}\:=\:{a}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\Rightarrow \\ $$$$\:{QD}\:=\:{a}\:−\:{a}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)=\:{a}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)\Rightarrow \\ $$$$\:{DR}\:=\:\frac{{a}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)}{\mathrm{2}−\sqrt{\mathrm{3}}} \\ $$$$\:{Orange}\:{area}\:=\:\frac{{QD}\centerdot{DR}}{\mathrm{2}}\:=\:\left[\frac{{a}^{\mathrm{2}} \left(\sqrt{\mathrm{3}}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}−\sqrt{\mathrm{3}}}\right]\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:=\:\:\frac{{a}^{\mathrm{2}} \left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}\right)}{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}}\:=\:\:{a}^{\mathrm{2}} \:=\:{Yellow}\:{area}\: \\ $$

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