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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

$$\mathrm{\Delta }{AD}^{\prime }B\equiv \mathrm{\Delta }CDB$$$$AB=BC=\sqrt{1^2+2^2}=\sqrt{5}$$$$A_{\mathrm{\Delta }ABC}=\frac{\sqrt{5}\times \sqrt{5}}{2}=\frac{5}{2}$$$$A_{\mathrm{\Delta }ADC}=A_{\mathrm{\Delta }ABC}-(A_{\mathrm{\Delta }CDB}+A_{\mathrm{\Delta }ABD})$$$$=A_{\mathrm{\Delta }ABC}-A_{{AD}^{\prime }BD}$$$$=A_{\mathrm{\Delta }ABC}-(A_{\mathrm{\Delta }{AD}^{\prime }D}+A_{\mathrm{\Delta }{DD}^{\prime }B})$$$$=\frac{5}{2}-(\frac{\sqrt{2}\times \sqrt{2}}{2}+\frac{1\times 1}{2})$$$$=\frac{5}{2}-\frac{3}{2}$$$$=1$$

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

$$or$$$$\mathrm{\angle }BDC=\mathrm{\angle }{BD}^{\prime }A=90°$$$$\mathrm{\angle }ADC=360°-90°-45°-90°=135°$$$$A_{\mathrm{\Delta }ADC}=\frac{\sqrt{2}\times 2\times \sin 135°}{2}=1$$

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