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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-perimete




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

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