Question
Construct a quadrilateral $ABCD$ in which $AB = BC = 6\ cm, AD = DC = 4.5\ cm$ and $\angle\text{B}=120^\circ.$
✓
Answer
Steps of construction:
Step $I$: Draw $AB = 6\ cm$.
Step $II$: Construct $\angle\text{ABC}=120^\circ.$
Step $III$: With B as the centre and radius $6\ cm$, cut off $BC = 6\ cm$. Now, we can see that $AC$ is about $10.3\ cm$ which is greater than $AD + CD = 4.5 + 4.5 = 9\ cm$.
We know that sum of the lengths of two sides of triangle is always greater than the third side but here, the sum of $AD$ and $CD$ is less than $AC$.
So, construction of the given quadrilateral is not possible.
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