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

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How can I find AD ? Or data is inadequate?


closed as off-topic by Daniel W. Farlow, Thomas, Mike Pierce, anomaly, Daniel Jul 9 '15 at 4:18

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  • 3
    $\begingroup$ Hint: Let $E$ be the point where $8$cm line hits $AD$. I claim that $ADL$ is similar to $ABE$. Further hint: they are right triangles and share an angle. Last hint: you know the length of $DL$ and $AE$. $\endgroup$ – Michael Burr Jul 8 '15 at 19:54
  • $\begingroup$ @MichaelBurr Is it 7.5? As If the are similar, then 8 is to 6, the corresponding 10 is to 7.5? $\endgroup$ – joey rohan Jul 8 '15 at 20:01
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    $\begingroup$ you can find from formula of square of parallelogram, |BE| * |AD|=|DL|*|AB|, 8*x=6*10,x=7.5 $\endgroup$ – haqnatural Jul 8 '15 at 20:12
  • $\begingroup$ Is it really a parallelogram? this means AD=BC and AB=DC $\endgroup$ – Emilio Novati Jul 8 '15 at 20:15

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.


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


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 $\angle DAL$.


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