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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.

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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 $ABCD$.


enter image description here

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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

enter image description here

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$30$, 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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