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Question Number 18004 by ajfour last updated on 13/Jul/17

Commented by ajfour last updated on 13/Jul/17

$$\mathrm{The}\mathrm{angular}\mathrm{bisector}\mathrm{of}\mathrm{\angle }\mathrm{ABC}$$$$\mathrm{intersects}\mathrm{side}\mathrm{AD}\mathrm{at}\mathrm{point}\mathrm{M},\mathrm{and}$$$$\mathrm{the}\mathrm{perpendicular}\mathrm{dropped}\mathrm{from}$$$$\mathrm{vertex}\mathrm{A}\mathrm{to}\mathrm{side}\mathrm{BC}\mathrm{cuts}\mathrm{BC}\mathrm{at}$$$$\mathrm{point}\mathrm{N}\mathrm{so}\mathrm{that}\mathrm{BN}=\mathrm{NC}\mathrm{and}$$$$\mathrm{AM}=2\mathrm{MD}.\mathrm{Find}\mathrm{the}\mathrm{sides}\mathrm{and}\mathrm{area}$$$$\mathrm{S}\mathrm{of}\mathrm{the}\mathrm{convex}\mathrm{quadrilateral}$$$$\mathrm{ABCD}\mathrm{if}\mathrm{the}\mathrm{perimeter}\mathrm{is}\mathrm{equal}\mathrm{to}$$$$5+\sqrt{3},\mathrm{\angle }\mathrm{BAD}=90°,\mathrm{\angle }\mathrm{ABC}=60°.$$

Answered by mrW1 last updated on 14/Jul/17

$$\mathrm{let}\mathrm{x}=\mathrm{BN}=\mathrm{NC}$$$$\Rightarrow \mathrm{AB}=2\mathrm{x}$$$$\Rightarrow \mathrm{AM}=2\mathrm{x}/\sqrt{3}$$$$\Rightarrow \mathrm{MD}=\mathrm{AM}/2=\mathrm{x}/\sqrt{3}$$$$\mathrm{AD}=3\mathrm{x}/\sqrt{3}=\sqrt{3}\mathrm{x}$$$$\mathrm{AC}=\mathrm{AB}=2\mathrm{x}$$$$\mathrm{\angle }\mathrm{CAD}=30°$$$$\mathrm{DC}=\sqrt{{\left(\sqrt{3}\mathrm{x}\right)}^2+{\left(2\mathrm{x}\right)}^2-2\times \sqrt{3}\mathrm{x}\times 2\mathrm{x}\times \cos 30°}$$$$=\mathrm{x}\sqrt{3+4-2\times \sqrt{3}\times 2\times \frac{\sqrt{3}}{2}}=\mathrm{x}$$$${\mathrm{P}}_{\mathrm{ABCD}}=(5+\sqrt{3})\mathrm{x}=5+\sqrt{3}$$$$\Rightarrow \mathrm{x}=1$$$${\mathrm{S}}_{\mathrm{ABCD}}={\mathrm{S}}_{\mathrm{ABC}}+{\mathrm{S}}_{\mathrm{ADC}}=\frac{1}{2}\times 2\times 2\times \frac{\sqrt{3}}{2}+\frac{1}{2}\times \sqrt{3}\times 1=\frac{3}{2}\sqrt{3}$$

Commented by Tinkutara last updated on 15/Jul/17

$$\mathrm{Thanks}\mathrm{Sir}!$$

Commented by ajfour last updated on 14/Jul/17

$${\mathrm{S}}_{\mathrm{ADC}}=\frac{1}{2}\mathrm{AD}\times \mathrm{CD}=\frac{1}{2}\times \sqrt{3}\times 1$$$$\mathrm{so}{\mathrm{S}}_{\mathrm{ABCD}}=\frac{3\sqrt{3}}{2}.$$$$\mathrm{thanks}\mathrm{sir},\mathrm{little}\mathrm{error}\mathrm{of}\mathrm{yours}\mathrm{is}$$$$\mathrm{due}\mathrm{to}\mathrm{diagram},\mathrm{my}\mathrm{mistake};\mathrm{\angle }\mathrm{ADC}$$$$\mathrm{dont}\mathrm{appear}90°\mathrm{in}\mathrm{the}\mathrm{diagram}.$$

Commented by mrW1 last updated on 14/Jul/17

$$\mathrm{you}\mathrm{are}\mathrm{right}\mathrm{sir}.$$$$\mathrm{i}\mathrm{made}\mathrm{a}\mathrm{mistake}.$$

Commented by Tinkutara last updated on 14/Jul/17

$$\mathrm{Is}\mathrm{\angle }ADC=90°\mathrm{is}\mathrm{to}\mathrm{be}\mathrm{given}\mathrm{or}\mathrm{to}\mathrm{be}$$$$\mathrm{calculated}?$$

Commented by mrW1 last updated on 14/Jul/17

$$\mathrm{\angle }ADC=90°\mathrm{is}\mathrm{a}\mathrm{calculated}\mathrm{result},$$$$\mathrm{since}\mathrm{AC}=2,\mathrm{AD}=\sqrt{3}\mathrm{and}\mathrm{\angle }\mathrm{CAD}=30°.$$

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