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Question Number 127547 by Mathgreat last updated on 30/Dec/20

Answered by mr W last updated on 31/Dec/20

Commented by mr W last updated on 31/Dec/20

ΔAD′B≡ΔCDB AB=BC=(√(1^2 +2^2 ))=(√5) A_(ΔABC) =(((√5)×(√5))/2)=(5/2) A_(ΔADC) =A_(ΔABC) −(A_(ΔCDB) +A_(ΔABD) ) =A_(ΔABC) −A_(AD′BD) =A_(ΔABC) −(A_(ΔAD′D) +A_(ΔDD′B) ) =(5/2)−((((√2)×(√2))/2)+((1×1)/2)) =(5/2)−(3/2) =1

$$\Delta{AD}'{B}\equiv\Delta{CDB} \\ $$$${AB}={BC}=\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }=\sqrt{\mathrm{5}} \\ $$$${A}_{\Delta{ABC}} =\frac{\sqrt{\mathrm{5}}×\sqrt{\mathrm{5}}}{\mathrm{2}}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$${A}_{\Delta{ADC}} ={A}_{\Delta{ABC}} −\left({A}_{\Delta{CDB}} +{A}_{\Delta{ABD}} \right) \\ $$$$={A}_{\Delta{ABC}} −{A}_{{AD}'{BD}} \\ $$$$={A}_{\Delta{ABC}} −\left({A}_{\Delta{AD}'{D}} +{A}_{\Delta{DD}'{B}} \right) \\ $$$$=\frac{\mathrm{5}}{\mathrm{2}}−\left(\frac{\sqrt{\mathrm{2}}×\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\mathrm{1}×\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{5}}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$=\mathrm{1} \\ $$

Commented by mr W last updated on 31/Dec/20

or ∠BDC=∠BD′A=90° ∠ADC=360°−90°−45°−90°=135° A_(ΔADC) =(((√2)×2×sin 135°)/2)=1

$${or} \\ $$$$\angle{BDC}=\angle{BD}'{A}=\mathrm{90}° \\ $$$$\angle{ADC}=\mathrm{360}°−\mathrm{90}°−\mathrm{45}°−\mathrm{90}°=\mathrm{135}° \\ $$$${A}_{\Delta{ADC}} =\frac{\sqrt{\mathrm{2}}×\mathrm{2}×\mathrm{sin}\:\mathrm{135}°}{\mathrm{2}}=\mathrm{1} \\ $$

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