ABCD is a parallelogram and X is the mid-point of AB. `(AXCD)= 24 cm^(2)`, then `ar (ABC) = 24 cm^(2)`.

Answer

`36 cm^(2)`

Explanation

`ar(triangleACD)=ar(triangleABC)=(1)/(2)ar("||gm ABCD")`,
`ar(triangleAMC)=ar(triangleMBC)=(1)/(2)ar(triangleABC)`.
`therefore ar(AMCD)=3ar(triangleAMC)rArr 24=3ar(triangleAMC)`
`rArr (triangleAMC) = 8 cm^(2)`.
And so, `ar(triangleABC)=2ar(triangleAMC)=16 cm^(2)`.