Question Number 19940 by ajfour last updated on 18/Aug/17
$${A}\:{circle}\:{is}\:{inscribed}\:{in}\:{an} \\ $$$${isosceles}\:{trapezium}.\:{Prove}\:{that} \\ $$$${the}\:{ratio}\:{of}\:{the}\:{area}\:{of}\:{the}\:{circle} \\ $$$${to}\:{the}\:{area}\:{of}\:{the}\:{trapezium}\:{is} \\ $$$${equal}\:{to}\:{the}\:{ratio}\:{of}\:{the}\:{circum}- \\ $$$${ference}\:{of}\:{the}\:{circle}\:{to}\:{the}\: \\ $$$${perimeter}\:{of}\:{the}\:{trapezium}. \\ $$
Commented by ajfour last updated on 18/Aug/17
Commented by ajfour last updated on 18/Aug/17
$${Area}\:{of}\:{trapezium}=\mathrm{4}×\frac{\mathrm{1}}{\mathrm{2}}×{R}\left({b}+{a}\right) \\ $$$$\:\:\:\:={R}\left({a}+{b}\right) \\ $$$$\frac{{Area}\:{of}\:{circle}}{{Area}\:{of}\:{trapezium}}=\frac{\pi{R}^{\mathrm{2}} }{\mathrm{2}{R}\left({a}+{b}\right)} \\ $$$$\:\:\:=\frac{\mathrm{2}\pi{R}}{\mathrm{4}\left({a}+{b}\right)}\:=\frac{{Perimeter}\:{of}\:{circle}}{{Perimeter}\:{of}\:{trapezium}}\:. \\ $$