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One of the angles of the parallelogram is of $150^\circ$. Altitudes are drawn from the vertex of this angle. If these altitudes measures $6 cm$ and $8 cm$,

then find the perimeter of parallelogram?

In this question, I am not getting how you can draw two altitudes from the vertex in parallelogram. I am not able to make make a diagram. I need a hint so that I can try further. Thanks in advance!

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Let $ABCD$ be the parallelogram with angle $D$ equal to $150^o$.

One altitude is drawn from the vertex $D$ on the side $AB$ (let this is $6$ and meets $AB$ in $E$) another from the same vertex $D$ on side $BC$ of the parallelogram.

Now triangle $DEA$ is a $30^o{-}60^o{-}90^o$ triangle. So side $AD=12=BC$. Similarly considering the $30^o{-}60^o{-}90^o$ triangle that corresponds to the $8$cm altitude, we find side $CD=16\text{cm}=AB$.

Now perimeter is $16+12+16+12=56$cm.

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Draw one side from $O=(0,0)$ to $A=(a,0)$, and draw another side from $O$ to $B=(b\cos 150^{\circ},b\sin 150^{\circ})$. Then $\overline{OA}=a$, $\overline{OB}=b$ and $\angle AOB=150^{\circ}$.

Note that the last vertex of the parallelogram, call it $C$, can be found at the terminus of the vector sum $\overrightarrow{OA}+\overrightarrow{OB}$.

So much for the visualization. If you can work out the altitudes as vector projections, you can solve the problem. @Pedja's answer doest that. I call the altitudes $r,s$ in the diagram below, where $$\eqalign{r=h_1&=a\,\sin 30{^\circ}\\s=h_2&=b\,\sin 30^{\circ}}$$

enter image description here

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Let's denote sides of parallelogram as $a$ and $b$ , and corresponding heights as $h_1$ and $h_2$ , then :

$h_1=a \cdot \sin 30^{\circ}$

$h_2=b \cdot \sin 30^{\circ}$

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I think you mean $$\eqalign{h_1&=a\,\sin 30^\circ\\h_2&=b\,\sin 30^\circ}$$ – bgins Apr 5 '12 at 13:02
@bgins Of course...thanks.. – pedja Apr 5 '12 at 13:06

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