Geometry - Can a trapezoid ever be a kite, and vice versa?

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A trapezoid cannot be a kite because it has only one pair of parallel sides, while a kite has two pairs of adjacent congruent sides, which would imply a parallelogram if it were a trapezoid. Conversely, a kite cannot be a trapezoid since it does not possess the necessary angle properties, specifically having two pairs of opposite congruent angles. The definitions of trapezoids and kites allow for squares to fit both categories, as they meet the criteria for each shape. The discussion emphasizes the importance of understanding the definitions and properties of quadrilaterals. Overall, the conclusion is that trapezoids and kites are distinct categories in geometry.
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Homework Statement



Can a trapezoid ever be a kite?
Can a kite be a trapezoid?

2. The attempt at a solution

I believe that the answer to both is no.

1. If a trapezoid was a kite, wouldn't the pair of parallel sides, along with the congruent consecutive sides, imply that it was a parallelogram? But a trapezoid only has one pair of parallel sides?

2. If a kite was a trapezoid, then wouldn't it have two pairs of opposite congruent angles, rather than exactly one?

I feel like my answers are correct, it's just that they don't seem mathematically sound. Any help is appreciated. :)
 
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Read http://www.mathsisfun.com/quadrilaterals.html, and see the picture:

quadrilateral-class.gif


The definition of trapezoid does not say that the other sides can not be also parallel.

Is square a trapezoid?

The definition of kite does not say that the the other pair of opposite angles can not be also equal.

Is square a kite?

Are both quadrilaterals?

ehild
 
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I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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