What quadrilateral has both diagonals bisecting each other but they are not congruent or perpendicular?

Mathematics

1 answer:

Bond [772]3 years ago

8 0

Answer:

A parallelogram with different side lengths.

Step-by-step explanation:

If we consider a parallelogram (not rectangle, not square, and not a rhombus), then its diagonals will bisect each other but would be not congruent or perpendicular. See attached image.

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I really don’t understand this and just need an explanation of how to do it.

sertanlavr [38]

$\bf ~\hspace{12em}\left( \cfrac{2n}{-3n\cdot -2n^2} \right)^4
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
\cfrac{2n}{-3n\cdot -2n^2}\implies \cfrac{1}{-3n}\cdot \cfrac{2n}{-2n^2}\implies \cfrac{1}{-3n}\cdot \cfrac{2n}{2n\cdot -n}\implies \cfrac{1}{-3n}\cdot \cfrac{2n}{2n}\cdot \cfrac{1}{-n}$

$\bf \cfrac{1}{-3n}\cdot \boxed{1}\cdot \cfrac{1}{-n}\implies \cfrac{1}{-3n\cdot -n}\implies \cfrac{1}{3n^2}
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
\left( \cfrac{2n}{-3n\cdot -2n^2} \right)^4\implies \left( \cfrac{1}{3n^2} \right)^4\implies \stackrel{\textit{distributing the exponent}}{\cfrac{1^4}{3^4n^{2\cdot 4}}}\implies \cfrac{1}{81n^8}$