Question Number 81734 by jagoll last updated on 15/Feb/20
$${The}\:{vertices}\:{of}\:{quadrilateral} \\ $$$${lie}\:{on}\:{the}\:{graph}\:{of}\:{y}\:=\:{lnx}\:{and}\: \\ $$$${the}\:{x}−{coordinates}\:{of}\:{these}\:{vertices} \\ $$$${are}\:{consecutive}\:{positive}\:{integer} \\ $$$$.\:{The}\:{area}\:{of}\:{the}\:{quadrilateral} \\ $$$${is}\:{ln}\:\left(\frac{\mathrm{91}}{\mathrm{90}}\right).\:{what}\:{is}\:{the}\:{x}−{coordinate} \\ $$$${of}\:{the}\:{leftmost}\:{vertex} \\ $$
Commented by john santu last updated on 15/Feb/20
$$\mathrm{12}?\: \\ $$
Commented by jagoll last updated on 15/Feb/20
$${solution}\:{pls} \\ $$
Commented by mr W last updated on 15/Feb/20
$${A}=−\mathrm{ln}\:{n}+\mathrm{ln}\:\left({n}+\mathrm{1}\right)+\mathrm{ln}\:\left({n}+\mathrm{2}\right)−\mathrm{ln}\:\left({n}+\mathrm{3}\right) \\ $$$${A}=\mathrm{ln}\:\frac{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}{{n}\left({n}+\mathrm{3}\right)}=\mathrm{ln}\:\frac{\mathrm{91}}{\mathrm{90}} \\ $$$$\Rightarrow{n}^{\mathrm{2}} +\mathrm{3}{n}−\mathrm{180}=\mathrm{0} \\ $$$$\Rightarrow\left({n}+\mathrm{15}\right)\left({n}−\mathrm{12}\right)=\mathrm{0} \\ $$$$\Rightarrow{n}=\mathrm{12} \\ $$
Commented by jagoll last updated on 15/Feb/20
$${thank}\:{you}\:{sir} \\ $$