Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 67770 by TawaTawa last updated on 31/Aug/19

Answered by MJS last updated on 31/Aug/19

$$AD+DE=AE=6$$$$\Rightarrow AD=q;DE=6-q[\Rightarrow q<6]$$$$AB=DE-AD=6-2q[\Rightarrow q<3]$$$$A=\left(\begin{array}{c}0\\ 0\end{array}\right)D=\left(\begin{array}{c}q\\ 0\end{array}\right)E=\left(\begin{array}{c}6\\ 0\end{array}\right)$$$$\mathrm{line}AB:y=x\tan 60°\iff y=\sqrt{3}x\left[1\right]$$$$B\mathrm{on}\mathrm{this}\mathrm{circle}:x^2+y^2={(6-2q)}^2\left[2\right]$$$$\left[1\right]\mathrm{in}\left[2\right]$$$$x=3-q$$$$\Rightarrow B=\left(\begin{array}{c}3-q\\ \sqrt{3}(3-q)\end{array}\right)$$$$\mathrm{line}BD:y=\sqrt{3}\frac{q-3}{2q-3}(x-q)$$$$C_{\in BD}=\left(\begin{array}{c}p\\ \sqrt{3}\frac{q-3}{2q-3}(p-q)\end{array}\right)$$$$\mid CD{\mid }^2=\frac{{(p-q)}^2}{{(2q-3)}^2}(7q^2-30q+36)$$$$\mid {BD}^2\mid =7q^2-30q+36=\mid CD{\mid }^2$$$$\Rightarrow \frac{{(p-q)}^2}{{(2q-3)}^2}=1\Rightarrow p=3q-3$$$$c=\mid AB\mid =6-2q$$$$b=\mid AC\mid =2\sqrt{3q^2-9q+9}$$$$a=\mid BC\mid =2\sqrt{7q^2-30q+36}$$$$\mathrm{radius}\mathrm{of}\mathrm{circumcircle}\mathrm{of}\mathrm{\Delta }ABC$$$$R=\frac{abc}{\delta }=\frac{2}{q}\sqrt{(q^2-3q+3)(7q^2-30q+36)}$$$$\mathrm{center}$$$$O=\left(\begin{array}{c}\frac{1}{q}(4q^2-15q+18)\\ -\frac{\sqrt{3}}{q}(q-2)(2q-3)\end{array}\right)$$$$\mid OE{\mid }^2=R^2$$$$\Rightarrow q=\frac{3}{2}$$$$\Rightarrow$$$$A=\left(\begin{array}{c}0\\ 0\end{array}\right)B=\left(\begin{array}{c}\frac{3}{2}\\ \frac{3\sqrt{3}}{2}\end{array}\right)C=\left(\begin{array}{c}\frac{3}{2}\\ -\frac{3\sqrt{3}}{2}\end{array}\right)$$$$D=\left(\begin{array}{c}\frac{3}{2}\\ 0\end{array}\right)E=\left(\begin{array}{c}6\\ 0\end{array}\right)O=\left(\begin{array}{c}3\\ 0\end{array}\right)$$$$S_{ABC}=6+3\sqrt{3}$$

Commented by TawaTawa last updated on 31/Aug/19

$$\mathrm{Wow},\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{Thanks}\mathrm{for}\mathrm{your}\mathrm{time}.$$

Commented by TawaTawa last updated on 31/Aug/19

$$\mathrm{Sir},\mathrm{i}\mathrm{learnt}\mathrm{the}\mathrm{answer}\mathrm{should}\mathrm{be}:\frac{9\sqrt{3}}{4}=3.8971$$$$\mathrm{or}\frac{4.5\sqrt{3}}{2}=3.8971$$

Commented by MJS last updated on 31/Aug/19

$$\mathrm{just}\mathrm{draw}\mathrm{the}\mathrm{points}\mathrm{and}\mathrm{the}\mathrm{circle}\mathrm{of}\mathrm{my}$$$$\mathrm{solution},\mathrm{you}\mathrm{can}\mathrm{see}{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{right}$$

Commented by TawaTawa last updated on 01/Sep/19

$$\mathrm{I}\mathrm{will}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Commented by TawaTawa last updated on 01/Sep/19

$$\mathrm{Sir}{\mathrm{S}}_{\mathrm{ABC}}\mathrm{or}\mathrm{S}\left(\mathrm{ABC}\right)\mathrm{means}\mathrm{what}\mathrm{sir}?$$

Commented by TawaTawa last updated on 01/Sep/19

$$\mathrm{Is}\mathrm{it}\mathrm{side}\mathrm{ABC}?$$

Commented by TawaTawa last updated on 01/Sep/19

$$\mathrm{Or},\mathrm{it}\mathrm{also}\mathrm{means}\mathrm{Area}?$$

Commented by MJS last updated on 01/Sep/19

$$\mathrm{I}\mathrm{thought}\mathrm{it}\mathrm{means}\mathrm{perimeter}$$

Commented by MJS last updated on 01/Sep/19

$$\mathrm{the}\mathrm{area}\mathrm{is}\mathrm{indeed}\frac{9\sqrt{3}}{4}$$

Commented by TawaTawa last updated on 01/Sep/19

$$\mathrm{Me}\mathrm{too}\mathrm{sir}.\mathrm{I}\mathrm{thought}\mathrm{it}\mathrm{is}\mathrm{perimeter}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com