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Question Number 204433 by emilagazade last updated on 17/Feb/24

Commented by emilagazade last updated on 17/Feb/24

$$findmaxvaluefor\mid DB\mid +\mid BC\mid$$

Answered by deleteduser1 last updated on 17/Feb/24

$$Let\mathrm{\angle }CAD=\theta \Rightarrow {CD}^2=3^2+64\times 2-2\times 3\times 8\sqrt{2}cos\left(\theta \right)$$$$\Rightarrow CD=\sqrt{137-48\sqrt{2}cos\left(\theta \right)}$$$${AC}^2={AB}^2+{BC}^2\Rightarrow 128={(3+BD)}^2+{BC}^2$$$$\Rightarrow 119={BD}^2+{BC}^2+6BD$$$${CD}^2={BD}^2+{BC}^2\Rightarrow 137-48\sqrt{2}cos\left(\theta \right)=119-6BD$$$$\Rightarrow BD=\frac{48\sqrt{2}cos\left(\theta \right)-18}{6}=8\sqrt{2}cos\left(\theta \right)-3$$$$\Rightarrow BC=\sqrt{128-128{cos}^2\theta }=8\sqrt{2}sin\left(\theta \right)$$$$\Rightarrow BD+DC=8\sqrt{2}(sin\theta +cos\theta )-3$$$$⩽8\sqrt{2}\left(\sqrt{1+1}\right)({sin}^2\theta +{cos}^2\theta )-3=13$$$$Equalitywhensin\left(\theta \right)=cos\left(\theta \right)\Rightarrow \theta =45°.$$

Answered by mr W last updated on 17/Feb/24

Commented by mr W last updated on 17/Feb/24

$${(\mid DB\mid +\mid BC\mid )}_{max}meansalso$$$${(\mid AB\mid +\mid BC\mid )}_{max},whichis$$$$\mid {AB}^{\prime }\mid +\mid B^{\prime }C\mid =2\mid {AB}^{\prime }\mid =2\times \frac{8\sqrt{2}}{\sqrt{2}}=16.$$$$\Rightarrow {(\mid DB\mid +\mid BC\mid )}_{max}=16-3=13✓$$

Commented by mr W last updated on 17/Feb/24

$$MethodII$$$$say\mathrm{\angle }CAB=\theta$$$$BC=8\sqrt{2}\sin \theta$$$$DB=8\sqrt{2}\cos \theta -3$$$$S=DB+BC=8\sqrt{2}(\cos \theta +\sin \theta )-3$$$$=16\sin (\frac{\pi }{4}+\theta )-3$$$$S_{max}=16-3=13at\theta =\frac{\pi }{4}.$$

Commented by emilagazade last updated on 17/Feb/24

$$nicethankyoualot$$

Answered by cortano12 last updated on 18/Feb/24

$$\underbrace{}$$

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