In trapezoid ABCD with legs AB and CD, ACAD=23 dm2. Find AABD.

It would be 23 dm2 as triangle CAD is equal to triangle ABD.

Given that ABCDis a trapezoid and legs are AB and CD.

Also given that area of triangle CAD = 23 dm square.

We know that in a trapezium we got two parallel lines AD and BC.

The legs are AB and CD

The triangle CAD liesbetween two parallellines AD and BC with distance say h.

Then area of triangle CAD = 1/2 (AD) (h) (using triangle formula)

To find ARea of ABD.

Triangle ABD has base as AD and height = distance between two parallel lines AD and BC = h

(By properties of parallel lines)

Hence area of Triangle ABD = 1/2 (AD) (h) = 23 dm square.

Answer = 23 dmsquare.