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The best answer I can give to this problem is $2\times\sqrt{130}$, using the law of cosine to find the angle between the diagonals $[\cos\frac{\sqrt{14}}{12} ] ,$ then using cosine law to find the obtuse angle of the parallelogram $(\cos \frac{-11}{24}),$ then converting that angle in sin $[ \frac{\sqrt{455}}{24} ] ,$ then using this formula $S = a\times b\times \sin$ (angle between those sides) to find the surface area to be $2\times \sqrt{455},$ and then using this formula $S = \frac12 \times$ first diagonal $\times$ second diagonal $\times \cos$ (angle between diagonals) to find the second diagonal, which I calculate $2 \times \sqrt{130}$

Is this answer correct? Is there a more elegant solution?

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  • $\begingroup$ You may consider using MathJax for better formatting. $\endgroup$ – Cave Johnson Sep 26 '16 at 15:44
  • $\begingroup$ how can you utilize the law of cosines? $\endgroup$ – Saketh Malyala Dec 22 '16 at 17:52
  • $\begingroup$ 6^2 + 8^2 - 2(6)(8)cos(a) = 144 100 - 96cos(a) = 144 96cos(a) = -44 cos(a) = -11/24 cos(π-a) = cos(b) cos(π)cos(a) - sin(π)sin(a) 11/24 6^2 + 8^2 - 2(6)(8)(11/24) = 100 - 96 * 11/24 = 100-44 = 56 The length of the other diagonal is sqrt(56) $\endgroup$ – Saketh Malyala Dec 22 '16 at 17:57
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Use the concept that area of triangles lying between same parallel lines and with same base are equal.

Then use Heron's formula find the area of one triangle and equate it with the second one.

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area of $\triangle ACD=$ area of$\triangle BDC$

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$$\|A-B\|^2 + \|A+B\|^2 = 2(\|A\|^2+\|B\|^2) $$ hence in a parallelogram the sum of the squared lengths of the diagonals equals the sum of the squared lengths of the sides. So in our case the answer is given by $$ \sqrt{6^2+6^2+8^2+8^2-12^2}=\color{red}{2\sqrt{14}}.$$

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An elegant way to proceed is by making use of the vector calculus. Let the vectors $\mathbf u$ and $\mathbf v$ be the sides of the parallelogram. Then $\mathbf u$-$\mathbf v$ and $\mathbf u+\mathbf v$ represent the diagonals. By using the formula of the length of a vector in terms of a scalar product, namely $$ |\mathbf w|^2=\mathbf w\cdot \mathbf w, $$ one obtains $$ |\mathbf u- \mathbf v|^2=|\mathbf u|^2-2\mathbf u\cdot \mathbf v+|\mathbf v|^2 $$ and $$ |\mathbf u+ \mathbf v|^2=|\mathbf u|^2+2\mathbf u\cdot \mathbf v+|\mathbf v|^2 $$ By summing up the two equations, $$ |\mathbf u- \mathbf v|^2+|\mathbf u+ \mathbf v|^2=2\left(|\mathbf u|^2+|\mathbf v|^2\right) $$ By substituting the data of the problem, you have $$ 12^2+|\mathbf u+ \mathbf v|^2=2(6^2+8^2)=200 $$ Therefore $ |\mathbf u+ \mathbf v|^2=56$ and the result is $\sqrt{56}$.

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If we can use standard formula for median length that bisects $a,$

$$\frac 12 \sqrt{2b^2 + 2c^2 - a^2} $$

From property of diagonal bisection we take double this length

$$ b=6,c-8, a=12, \rightarrow 2 \sqrt{14}. $$

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