Question 267266
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A rhombus and its diagonals form four congruent right triangles with legs that are each one-half the measure of one of the diagonals and the hypotenuse equal in measure to one of the sides of the rhombus.


So, half of 12 is 6 and half of 16 is 8, so the measure of the side of the rhombus must be:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ s\ =\ \sqrt{6^2\ +\ 8^2}]


And, like a square, the perimeter of a rhombus is 4 times the measure of one side, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P\ =\ 4\sqrt{6^2\ +\ 8^2}]


Yep, you get to do your very own arithmetic.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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