Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 212291 by hardmath last updated on 08/Oct/24

Commented by hardmath last updated on 08/Oct/24

$$\mathrm{m}\left(\mathrm{\angle }\mathrm{BCD}\right)=90°$$$$\mid \mathrm{AB}\mid =\mid \mathrm{AC}\mid$$$$\mathrm{AC}\cap \mathrm{BD}=\mathrm{K}$$$$\mathrm{Area}\left(\mathrm{ABCD}\right)=?$$

Commented by Frix last updated on 13/Oct/24

$${\mathrm{It}}^{\prime }\mathrm{s}\mathrm{either}90\mathrm{or}\frac{490}{9}$$

Answered by Frix last updated on 13/Oct/24

$$B=\left(\begin{array}{c}0\\ 0\end{array}\right)C=\left(\begin{array}{c}p\\ 0\end{array}\right)A=\left(\begin{array}{c}\frac{p}{2}\\ q\end{array}\right)D=\left(\begin{array}{c}p\\ r\end{array}\right)$$$$p,q,r>0$$$$\mathrm{Using}\mathrm{simple}\mathrm{geometry}$$$$l_{BD}:y=\frac{r}{p}x$$$$l_{AC}:-\frac{2q}{p}x+2q$$$$\Rightarrow K=\left(\begin{array}{c}\frac{2pq}{2q+r}\\ \frac{2qr}{2q+r}\end{array}\right)$$$$\mathrm{The}\mathrm{areas}\mathrm{require}\mathrm{the}\mathrm{sidelengths}\mathrm{to}\mathrm{use}$$$${\mathrm{Heron}}^{\prime }\mathrm{s}\mathrm{Formula}$$$$A=\frac{\sqrt{(a+b+c)(a+b-c)(a+c-b)(b+c-a)}}{4}$$$$\mathrm{We}\mathrm{get}$$$$\left(1\right)\frac{pqr}{2q+r}=25$$$$\left(2\right)\frac{p\mid 2q-r\mid r}{4(2q+r)}=10$$$$==========$$$$\left(1\right)p=\frac{25(2q+r)}{qr}\Rightarrow$$$$\left(2\right)\frac{25\mid 2q-r\mid }{4q}=10\Rightarrow q=\frac{5r}{2}\vee q=\frac{5r}{18}$$$$\mathrm{Inserting}\mathrm{and}\mathrm{using}{\mathrm{Heron}}^{\prime }\mathrm{s}\mathrm{Formula}\mathrm{again}$$$$\Rightarrow$$$$\mathrm{Area}\mathrm{of}ABCD=\frac{490}{9}\vee 90$$$$$$$$[\mathrm{We}\mathrm{are}\mathrm{free}\mathrm{to}\mathrm{choose}r>0]$$

Commented by Ghisom last updated on 13/Oct/24

$$\mathrm{but}\mathrm{the}\mathrm{one}\mathrm{with}\mathrm{area}490/9\mathrm{is}\mathrm{concave}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com