Finding the perimeter of a parallelogram.
By James Williams
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ABCD is a parallellogram in which $$\angle D = 120^\circ .$$
The bisector of angle D bisects the side AB. If the length of the bisector is 5 cm, then what is the perimeter of the parallelogram.
Can this be solved using the sin rule or cosine rule?
How i can achieve this? Thanks in advance.
$\endgroup$ 03 Answers
$\begingroup$Hint: Let $A$ be to top left vertex, $B$ the top right, $C$ the bottom right and $D$ the bottom left. Let $E$ be the point of intersection of the angle bisector with $AB$.
Draw a picture and note that $DE$, $AE$, and $DA$ are the sides of an equilateral triangle. Each side of this triangle has length 5. From this you can find the perimeter of $ABCD$.
Let's denote perimeter as $p$ and length of bisector as $x$ (see picture below) , then :
$p=2\cdot(2x+x)=6 \cdot x=30$ cm
$30$, because $\angle A=120^\circ$ we have $\angle B=\angle C= \frac{360^\circ-2\cdot 120^\circ}{2}=60^\circ$. Therefore $BDS$ is a eqilateral triangle, so $\overline{BD}=\overline{SD}=5\text{cm}$ ($S$ is the point, where the bisector hits $\overline{AB}$).
Then you have $\overline{AB}=2\overline{BS}=2\overline{BD}=10\text{cm}$.
Summing it up, you get $\overline{AB}+\overline{BD}+\overline{DC}+\overline{AC}=10\text{cm}+5\text{cm}+10\text{cm}+5\text{cm}=30\text{cm}$.
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