How to find the side of this Parallelogram?

Question

Here, AB=10 cm and altitudes corresponding to the sides AB and AD are 6cm and 8cm respectively.

How can I find AD ? Or data is inadequate?

Answer 1

Let $DL$ be the height with respect to $AB$ and $BK$ the height with respect to $AD$. If it is a parallelogram its area is given by: $AB\times DL= AD\times BK$ , so you can find AD.

If it is not a parallelogram, but a trapezoid, note that the triangles $ADL$ and $AKB$ are similar so you can find anyway $AD:AB=DL:BK$. Note that in the first case also $BC$ is fixed, but in the second case it is not.

Answer 2

Let $BE$ be the perpendicular to $AD$.

In triangle $ADB$,

area of ADB considering AD as base is $1/2 * AD *8$

area of ADB considering AB as base is $1/2 * 10 *6$

Since both correspond to the same triangle,

$1/2 * AD *8$ = $1/2 * 10 *6$

which means that AD = 7.5

Answer 3

The triangle which can be constructed by the points $A$, $B$ and the orthogonal projection of point $B$ on to line $AD$, lets call that $B'$, is completely defined; namely $AB$ has length 10 cm, $BB'$ has length 8 cm and the angle $\angle AB'B$ is 90°. From this information you can derive the remaining lengths and angles of that triangle.

Now you can use the triangle $ALD$ to derive the length of $AD$. This can be done because the angle $\angle DAL$ is equal to $\angle B'AB$, which can be found using the previous triangle. This allows you to define the triangle $ALD$, because you know that $DL$ is 6 cm, the angle $\angle ALD$ is 90° and how to find $\angle DAL$.