The largest area of the trapezium inscribed in a semi-circle or radius `R ,` if the lower base is on the diameter, is `(3sqrt(3))/4R^2` (b) `(sqrt(3))/2R^2` `(3sqrt(3))/8R^2` (d) `R^2`

Options

A. `((3sqrt(3))/(4)R^(2))`
B. `((sqrt(3))/(2)R^(2))`
C. `((3sqrt(3))/(8)R^(2))`
D. `R^(2)`

Answer

1

Explanation

`AD=AB cos theta =2R cos theta`
`AE=AD cos theta = 2R cos^(2) theta`
`EF=AB-2AE=2R-4R cos^(2)theta`
`DE=Ad sin theta =2R sin theta cos theta`
Thus area of trapezilum
`S=1/2 (AB+CD)xxDE`
`=1/2(2R+2R-4R cos^(2) theta)xx2R sin theta cos theta`
`=4R^(2)sin^(3)theta cos theta`
`(ds)/(d theta)=12R^(2) sin^(2) theta cos^(2) theta -4 R^(2) sin^(4) theta`
`=4R^(2)sin^(2)theta(3 cos^(2) theta - sin ^(2) theta)`
For maximum are `(ds)/(d theta)=0 or tan^(2) theta =3`
or `tan theta = sqrt(3)(theta is acute) or S_(max)=3sqrt(3)/(4)R^(2)`