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Question Number 209314 by SonGoku last updated on 06/Jul/24

Commented by SonGoku last updated on 06/Jul/24

$$\mathrm{How}\mathrm{to}\mathrm{find}\mathrm{the}\mathrm{of}\mathrm{this}\mathrm{polygon}?$$

Commented by mr W last updated on 07/Jul/24

$$youcanonlyfindtheperimeterof$$$$thepolygon,nothingelseofit.$$

Commented by SonGoku last updated on 07/Jul/24

$$\mathrm{So}\mathrm{the}\mathrm{only}\mathrm{way}\mathrm{to}\mathrm{determine}\mathrm{the}\mathrm{diagonal}\mathrm{of}\mathrm{ang}$$$$\mathrm{irreular}\mathrm{polygon},\mathrm{like}\mathrm{the}\mathrm{one}\mathrm{in}\mathrm{the}\mathrm{image},\mathrm{is}\mathrm{onlyo}$$$$\mathrm{thrugh}\mathrm{practice}?\mathrm{In}\mathrm{other}\mathrm{words},\mathrm{in}\mathrm{the}\mathrm{field}?$$

Commented by Frix last updated on 07/Jul/24

$$\mathrm{Diagonal}=d$$$$$$$$18<d<35$$$$$$$$\frac{\sqrt{197319}}{4}<\mathrm{area}⩽\frac{\sqrt{1278519}}{4}$$$$(\min \mathrm{at}d=35,\max \mathrm{at}d=\frac{\sqrt{638290}}{29})$$

Commented by Frix last updated on 07/Jul/24

$$\mathrm{For}1\mathrm{triangle}\mathrm{you}\mathrm{need}\mathrm{at}\mathrm{least}3\mathrm{measurements}.$$$$\mathrm{In}\mathrm{this}\mathrm{case},\mathrm{you}\mathrm{need}1\mathrm{additional}\mathrm{measurment}$$$$\mathrm{for}\mathrm{one}\mathrm{of}\mathrm{the}\mathrm{triangles},\mathrm{the}\mathrm{rest}\mathrm{follows}.$$

Commented by SonGoku last updated on 08/Jul/24

$$\mathrm{But},\mathrm{how}\mathrm{did}\mathrm{you}\mathrm{at}\mathrm{these}\mathrm{calculations}?$$

Commented by Frix last updated on 08/Jul/24

$$1^{\mathrm{st}}\mathrm{triangle}10,28,d\Rightarrow 18<d<38$$$$2^{\mathrm{nd}}\mathrm{triangle}15,20,d\Rightarrow 5<d<35$$$$\Rightarrow 18<d<35$$$$\mathrm{Area}\mathrm{of}\mathrm{triangle}a,b,c$$$$=\frac{\sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}}{4}$$$$\mathrm{Area}\mathrm{of}{1}^{\mathrm{st}}\mathrm{triangle}$$$$A_1=\frac{\sqrt{-d^4+1768d^2-467856}}{4}$$$$\mathrm{Area}\mathrm{of}{2}^{\mathrm{nd}}\mathrm{triangle}$$$$A_2=\frac{\sqrt{-d^4+1250d^2-30625}}{4}$$$$\mathrm{For}\mathrm{the}\mathrm{minimum}\mathrm{obviously}d=18\Rightarrow A_1=0$$$$\mathrm{For}\mathrm{the}\mathrm{maximum}\mathrm{I}\mathrm{used}\mathrm{differentiation}$$

Commented by mr W last updated on 08/Jul/24

$$maximumareaiswhenthe$$$$quadilateraliscyclic.$$$$s=\frac{a+b+c+d}{2}=\frac{10+28+15+20}{2}=36.5$$$$A_{max}=\sqrt{(s-a)(s-b)(s-c)(s-d)}$$$$=\sqrt{(36.5-10)(36.5-28)(36.5-15)(36.5-20)}$$$$=\frac{\sqrt{1278519}}{4}$$$$seealsoQ30233$$

Commented by SonGoku last updated on 09/Jul/24

$$\mathrm{Very}\mathrm{sophisticated}.$$$$\mathrm{Congratulations}$$$$\mathrm{I}\mathrm{willa}\mathrm{study}.$$$$\mathrm{Thank}\mathrm{you}.$$

Commented by SonGoku last updated on 09/Jul/24

$$\mathrm{So},\mathrm{can}\mathrm{I}\mathrm{use}\mathrm{this}\mathrm{formula}\mathrm{to}\mathrm{determine}\mathrm{the}\mathrm{area}\mathrm{of}$$$$\mathrm{any}\mathrm{quadrilateral}?$$

Commented by mr W last updated on 09/Jul/24

$$no!{it}^{\prime }sonlyforso-calledcyclic$$$$quadrilaterals.$$$$generallyaquadrilateralisnot$$$$uniquelydefinedwhenonlyitsfour$$$$sidesaregiven.youneedan$$$$additionalcondition.$$

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