Destiny designs necklaces.
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Check the picture below.
so the rhombus has the diagonals of AC and BD, now keeping in mind that the diagonals bisect each, namely they cut each other in two equal halves, let's find the length of each.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{-4}~,~\stackrel{y_1}{-2})\qquad C(\stackrel{x_2}{6}~,~\stackrel{y_2}{8})\qquad \qquad % distance value d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ AC=\sqrt{[6-(-4)]^2+[8-(-2)]^2}\implies AC=\sqrt{(6+4)^2+(8+2)^2} \\\\\\ AC=\sqrt{10^2+10^2}\implies AC=\sqrt{10^2(2)}\implies \boxed{AC=10\sqrt{2}}\\\\ -------------------------------](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%0A%5C%5C%5C%5C%0AA%28%5Cstackrel%7Bx_1%7D%7B-4%7D~%2C~%5Cstackrel%7By_1%7D%7B-2%7D%29%5Cqquad%20%0AC%28%5Cstackrel%7Bx_2%7D%7B6%7D~%2C~%5Cstackrel%7By_2%7D%7B8%7D%29%5Cqquad%20%5Cqquad%20%0A%25%20%20distance%20value%0Ad%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0AAC%3D%5Csqrt%7B%5B6-%28-4%29%5D%5E2%2B%5B8-%28-2%29%5D%5E2%7D%5Cimplies%20AC%3D%5Csqrt%7B%286%2B4%29%5E2%2B%288%2B2%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0AAC%3D%5Csqrt%7B10%5E2%2B10%5E2%7D%5Cimplies%20AC%3D%5Csqrt%7B10%5E2%282%29%7D%5Cimplies%20%5Cboxed%7BAC%3D10%5Csqrt%7B2%7D%7D%5C%5C%5C%5C%0A-------------------------------)
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ B(\stackrel{x_1}{-2}~,~\stackrel{y_1}{6})\qquad D(\stackrel{x_2}{4}~,~\stackrel{y_2}{0})\qquad \qquad BD=\sqrt{[4-(-2)]^2+[0-6]^2} \\\\\\ BD=\sqrt{(4+2)^2+(-6)^2}\implies BD=\sqrt{6^2+6^2} \\\\\\ BD=\sqrt{6^2(2)}\implies \boxed{BD=6\sqrt{2}}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%0A%5C%5C%5C%5C%0AB%28%5Cstackrel%7Bx_1%7D%7B-2%7D~%2C~%5Cstackrel%7By_1%7D%7B6%7D%29%5Cqquad%20%0AD%28%5Cstackrel%7Bx_2%7D%7B4%7D~%2C~%5Cstackrel%7By_2%7D%7B0%7D%29%5Cqquad%20%5Cqquad%20BD%3D%5Csqrt%7B%5B4-%28-2%29%5D%5E2%2B%5B0-6%5D%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0ABD%3D%5Csqrt%7B%284%2B2%29%5E2%2B%28-6%29%5E2%7D%5Cimplies%20BD%3D%5Csqrt%7B6%5E2%2B6%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0ABD%3D%5Csqrt%7B6%5E2%282%29%7D%5Cimplies%20%5Cboxed%7BBD%3D6%5Csqrt%7B2%7D%7D)
that simply means that each triangle has a side that is half of 10√2 and another side that's half of 6√2.
namely, each triangle has a "base" of 3√2, and a "height" of 5√2, keeping in mind that all triangles are congruent, then their area is,
![\bf \stackrel{\textit{area of the four congruent triangles}}{4\left[ \cfrac{1}{2}(3\sqrt{2})(5\sqrt{2}) \right]\implies 4\left[ \cfrac{1}{2}(15\cdot (\sqrt{2})^2) \right]}\implies 4\left[ \cfrac{1}{2}(15\cdot 2) \right] \\\\\\ 4[15]\implies 60](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20four%20congruent%20triangles%7D%7D%7B4%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%283%5Csqrt%7B2%7D%29%285%5Csqrt%7B2%7D%29%20%5Cright%5D%5Cimplies%204%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%2815%5Ccdot%20%28%5Csqrt%7B2%7D%29%5E2%29%20%5Cright%5D%7D%5Cimplies%204%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%2815%5Ccdot%202%29%20%5Cright%5D%0A%5C%5C%5C%5C%5C%5C%0A4%5B15%5D%5Cimplies%2060)
so the rhombus has the diagonals of AC and BD, now keeping in mind that the diagonals bisect each, namely they cut each other in two equal halves, let's find the length of each.
that simply means that each triangle has a side that is half of 10√2 and another side that's half of 6√2.
namely, each triangle has a "base" of 3√2, and a "height" of 5√2, keeping in mind that all triangles are congruent, then their area is,
So for this, you'll be using the pythagorean theorem, which is . In this case, 10 and 24 are the legs, and AB is the hypotenuse. Our equation will be
Firstly, solve the exponents:
Next, combine 100 and 576:
And lastly, square root both sides to get
And in short, Line AB is 26 ft, or the first option.