Applications Of Derivatives Question 15
Question: The largest area of a trapezium inscribed in a semi-circle of radius R, if the lower base is on the diameter, is
Options:
A) $ \frac{3\sqrt{3}}{4}R^{2} $
B) $ \frac{\sqrt{3}}{2}R^{2} $
C) $ \frac{3\sqrt{3}}{8}R^{2} $
D) $ R^{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ AD=AB\cos \theta =2R\cos \theta ,AE=AD\cos \theta $
$ =2R{{\cos }^{2}}\theta $ or $ EF=AB-2AE=2R-4R{{\cos }^{2}}\theta $
$ DE=AD\sin \theta =2R\sin \theta \cos \theta $ Thus, area of trapezium, $ S=\frac{1}{2}(AB+CD)\times DE $
$ =\frac{1}{2}(2R+2R-4R{{\cos }^{2}}\theta )\times 2R\sin \theta \cos \theta $
$ =4R^{2}{{\sin }^{3}}\theta \cos \theta $
$ \frac{dS}{d\theta }=12R^{2}{{\sin }^{2}}\theta {{\cos }^{2}}\theta -4R^{2}{{\sin }^{4}}\theta $
$ =4R^{2}{{\sin }^{2}}\theta (3cos^{2}\theta -sin^{2}\theta ) $ For maximum area, $ \frac{dS}{d\theta }=0 $ or $ {{\tan }^{2}}\theta =3 $ or $ \tan \theta =\sqrt{3} $ ( $ \theta $ is acute) or $ {S_{\max }}=\frac{3\sqrt{3}}{4}R^{2} $
BETA
