Question

In trapezoid ABCD with legs AB and CD , ACAD=23 dm^2. Find AABD.

Answer 1

In the figure, ABCD is a trapezoid with legs AB and CD.

Join AC and BD.

For the triangle CAD, from the vertex C, draw an altitude CE and for the triangle ABD, from the vertex B, draw an altitude BF.

Clearly, CE = BF = h (say) --- (1)

Note that the base of the triangles CAD and ABD are the same and is AD.

It is given that ar(CAD) = $23dm^{2}$.

Now, ar(ABD) =  $\frac{1}{2}(AD)(BF)$

= $\frac{1}{2}(AD)(CE)$ from (1)

= ar(CAD)

= $23dm^{2}$.

Hence, area of Δ ABD =  $23dm^{2}$..