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 71410 by TawaTawa last updated on 15/Oct/19

Answered by MJS last updated on 15/Oct/19

$$A=\left(\begin{array}{c}0\\ 0\end{array}\right)B=\left(\begin{array}{c}p\\ 0\end{array}\right)C=\left(\begin{array}{c}p\\ q\end{array}\right)D=\left(\begin{array}{c}0\\ q\end{array}\right)$$$$DC:y=q$$$$AC:y=\frac{q}{p}x$$$$BE:y=-\frac{p}{q}x+\frac{p^2}{q}$$$$E=BE\cap DC=\left(\begin{array}{c}\frac{p^2-q^2}{p}\\ q\end{array}\right)$$$$F=AC\cap BE=\left(\begin{array}{c}\frac{p^3}{p^2+q^2}\\ \frac{p^{2}q}{p^2+q^2}\end{array}\right)$$$$\mathrm{area}\left(ABF\right)=\frac{p^{3}q}{2(p^2+q^2)}$$$$\mathrm{area}\left(BCF\right)=\frac{{pq}^3}{2(p^2+q^2)}$$$$\mathrm{area}\left(CEF\right)=\frac{q^5}{2p(p^2+q^2)}$$$$\mathrm{area}\left(ADEF\right)=\frac{q(p^4+p^2{q}^2-q^4)}{2p(p^2+q^2)}$$$$$$$$\frac{{pq}^3}{2(p^2+q^2)}=5\Rightarrow p^2+q^2=\frac{{pq}^3}{10}$$$$\frac{q^5}{2p(p^2+q^2)}=4\Rightarrow p^2+q^2=\frac{q^5}{8p}$$$$\frac{{pq}^3}{10}=\frac{q^5}{8p}\Rightarrow q=\frac{2\sqrt{5}}{5}p\Rightarrow p=\frac{3}{2}\sqrt[4]{125}\wedge q=3\sqrt[4]{5}$$$$\Rightarrow \mathrm{area}\left(ABF\right)=\frac{p^{3}q}{2(p^2+q^2)}=\frac{25}{4}=6.25$$$$\mathrm{area}\left(ADEF\right)=\frac{q(p^4+p^2{q}^2-q^4)}{2p(p^2+q^2)}=\frac{29}{4}=7.25$$

Commented by TawaTawa last updated on 15/Oct/19

$$\mathrm{I}\mathrm{appreciate}\mathrm{your}\mathrm{time}\mathrm{sir}$$

Commented by TawaTawa last updated on 15/Oct/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com