A rectangle of the greatest area is inscribed in a trapezium `A B C D` , one of whose non-parallel sides `A B` is perpendicular to the base, so that one of the rectangles die lies on the larger base of the trapezium. The base of trapezium are 6cm and 10 cm and `A B` is 8 cm long. Then the maximum area of the rectangle is `24c m^2` (b) `48c m^2` `36c m^2` (d) none of these

Options

A. `24 cm^(2)`
B. `48 cm^(2)`
C. `36 cm^(2)`
D. none of these

Answer

2

Explanation

Let rectangle BEFG is inscribed
Its area A=xy
Now ,`triangle` FEC and `triangle` DHC are similer i.e
`(x)/(8)=(10-y)/(4)or y=10-(x)/(2)`
or `A=x(10-(x)/(2)) where x in (0,8)`
or Now `(dA)/(dx)=10 -x` .For `x in (0,8) , (dA)/(dx)gt0` i.e A increases
Hence `A_(max)` occurs when x =8
Hence max area =`A_(max)=8(10-(8))/(2)=48 cm^(2)`