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March 11th, 2010, 06:32 PM   #1
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Isosceles Trapezoid

Another question I'm confused on while preparing for state;

An isosceles trapezoid has bases of respective lengths 10 and 20. A quadrilateral is formed by joining the midpoints of the consecutive sides of the isosceles trapezoid. If one of the sides of that quadrilateral has a length of 12, the sum of the lengths of the diagonals are...?

Thanks in advance, you guys are great
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March 11th, 2010, 06:46 PM   #2
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Re: Isosceles Trapezoid

Looking at the information, I've recognized that the quadrilateral is a kite, but I'm not sure where to go from there
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March 11th, 2010, 07:33 PM   #3
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The length of one of the diagonals of the quadrilateral may be deduced easily from the length information given for the bases of the trapezoid.
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March 11th, 2010, 07:52 PM   #4
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Re: Isosceles Trapezoid

Ahh, so I think I've recognized it. The answer I got fits the correct answer as well
Tell me if my approach here doesn't break any mathematical rules:

From the information given I have determined that the figure was a rhombus (why did I do that?), and every side must then be equal to 12 units.

Since one of the diagonals of the rhombus went through the median of the trapezoid, it must be the average of the two bases ((20+10)/2 = 15) and since diagonals in a rhombus are perpendicular bisectors, one of four right triangles formed from the intersection of the two diagonals has a length of 15/2

Upon Pythagorean Theorem, the other leg must equal 3 root 39 / 2

Multiplying by two gives the measure of the second diagonal

Most of it makes sense to me now, but my final question is how can I be positive that it was in fact a rhombus?
At state I don't want to make wrong assumptions and perform calculations leading to the wrong answer
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March 11th, 2010, 08:32 PM   #5
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Use the midpoint theorem to prove it's a parallelogram, then the symmetry of the diagram implies it's a rhombus.
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