In the given figure, ABCD is a rectangle inscribed in a quadrant of a circle of radius 10 cm. If `AD = sqrt25` cm then area of the rectangle is

Options

A. `32 cm^(2)`
B. `40 cm^(2)`
C. `44 cm^(2)`
D. `48 cm^(2)`

Answer

B

Explanation

`AB^(2)=AC^(2)-BC^(2)=(AC^(2)-AD^(2))={(10^(2))-2(sqrt5)^(2)}=(100-20)=80`
`rArr AB=sqrt80 = sqrt(16xx5) = 4sqrt5 cm`.
`therefore ar("rect. ABCD")=(ABxxAD)=(4sqrt5xx2sqrt5)cm^(2) = 40 cm^(2)`