Question Number 212291 by hardmath last updated on 08/Oct/24
Commented by hardmath last updated on 08/Oct/24
$$\mathrm{m}\left(\angle\mathrm{BCD}\right)\:=\:\mathrm{90}° \\ $$$$\mid\mathrm{AB}\mid\:=\:\mid\mathrm{AC}\mid \\ $$$$\mathrm{AC}\:\cap\:\mathrm{BD}\:=\:\mathrm{K} \\ $$$$\mathrm{Area}\left(\mathrm{ABCD}\right)\:=\:? \\ $$
Commented by Frix last updated on 13/Oct/24
$$\mathrm{It}'\mathrm{s}\:\mathrm{either}\:\mathrm{90}\:\mathrm{or}\:\frac{\mathrm{490}}{\mathrm{9}} \\ $$
Answered by Frix last updated on 13/Oct/24
![B= ((0),(0) ) C= ((p),(0) ) A= (((p/2)),(q) ) D= ((p),(r) ) p, q, r >0 Using simple geometry l_(BD) : y=(r/p)x l_(AC) : −((2q)/p)x+2q ⇒ K= ((((2pq)/(2q+r))),(((2qr)/(2q+r))) ) The areas require the sidelengths to use Heron′s Formula A=((√((a+b+c)(a+b−c)(a+c−b)(b+c−a)))/4) We get (1) ((pqr)/(2q+r))=25 (2) ((p∣2q−r∣r)/(4(2q+r)))=10 ========== (1) p=((25(2q+r))/(qr)) ⇒ (2) ((25∣2q−r∣)/(4q))=10 ⇒ q=((5r)/2)∨q=((5r)/(18)) Inserting and using Heron′s Formula again ⇒ Area of ABCD =((490)/9)∨90 [We are free to choose r>0]](Q212406.png)
$${B}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\:\:\:\:{C}=\begin{pmatrix}{{p}}\\{\mathrm{0}}\end{pmatrix}\:\:\:\:\:{A}=\begin{pmatrix}{\frac{{p}}{\mathrm{2}}}\\{{q}}\end{pmatrix}\:\:\:\:\:{D}=\begin{pmatrix}{{p}}\\{{r}}\end{pmatrix} \\ $$$${p},\:{q},\:{r}\:>\mathrm{0} \\ $$$$\mathrm{Using}\:\mathrm{simple}\:\mathrm{geometry} \\ $$$${l}_{{BD}} :\:{y}=\frac{{r}}{{p}}{x} \\ $$$${l}_{{AC}} :\:−\frac{\mathrm{2}{q}}{{p}}{x}+\mathrm{2}{q} \\ $$$$\Rightarrow\:{K}=\begin{pmatrix}{\frac{\mathrm{2}{pq}}{\mathrm{2}{q}+{r}}}\\{\frac{\mathrm{2}{qr}}{\mathrm{2}{q}+{r}}}\end{pmatrix} \\ $$$$\mathrm{The}\:\mathrm{areas}\:\mathrm{require}\:\mathrm{the}\:\mathrm{sidelengths}\:\mathrm{to}\:\mathrm{use} \\ $$$$\mathrm{Heron}'\mathrm{s}\:\mathrm{Formula} \\ $$$${A}=\frac{\sqrt{\left({a}+{b}+{c}\right)\left({a}+{b}−{c}\right)\left({a}+{c}−{b}\right)\left({b}+{c}−{a}\right)}}{\mathrm{4}} \\ $$$$\mathrm{We}\:\mathrm{get} \\ $$$$\left(\mathrm{1}\right)\:\frac{{pqr}}{\mathrm{2}{q}+{r}}=\mathrm{25} \\ $$$$\left(\mathrm{2}\right)\:\frac{{p}\mid\mathrm{2}{q}−{r}\mid{r}}{\mathrm{4}\left(\mathrm{2}{q}+{r}\right)}=\mathrm{10} \\ $$$$========== \\ $$$$\left(\mathrm{1}\right)\:{p}=\frac{\mathrm{25}\left(\mathrm{2}{q}+{r}\right)}{{qr}}\:\Rightarrow \\ $$$$\left(\mathrm{2}\right)\:\frac{\mathrm{25}\mid\mathrm{2}{q}−{r}\mid}{\mathrm{4}{q}}=\mathrm{10}\:\Rightarrow\:{q}=\frac{\mathrm{5}{r}}{\mathrm{2}}\vee{q}=\frac{\mathrm{5}{r}}{\mathrm{18}} \\ $$$$\mathrm{Inserting}\:\mathrm{and}\:\mathrm{using}\:\mathrm{Heron}'\mathrm{s}\:\mathrm{Formula}\:\mathrm{again} \\ $$$$\Rightarrow \\ $$$$\mathrm{Area}\:\mathrm{of}\:{ABCD}\:=\frac{\mathrm{490}}{\mathrm{9}}\vee\mathrm{90} \\ $$$$ \\ $$$$\left[\mathrm{We}\:\mathrm{are}\:\mathrm{free}\:\mathrm{to}\:\mathrm{choose}\:{r}>\mathrm{0}\right] \\ $$
Commented by Ghisom last updated on 13/Oct/24
$$\mathrm{but}\:\mathrm{the}\:\mathrm{one}\:\mathrm{with}\:\mathrm{area}\:\mathrm{490}/\mathrm{9}\:\mathrm{is}\:\mathrm{concave} \\ $$