Length of Parallelogram Diagonals

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Homework Statement


Find the length of the two diagonals of a paralellogram with the sides (-2,-2) and (-10,-2)


Homework Equations





The Attempt at a Solution


What I have tried is, I drew out the 2 vectors, and then drew out their components and used trig to find the angles where the connect. I also found the magnitude of both the vectors.
Am i on the right track?
 
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Larrytsai said:

Homework Statement


Find the length of the two diagonals of a paralellogram with the sides (-2,-2) and (-10,-2)


Homework Equations





The Attempt at a Solution


What I have tried is, I drew out the 2 vectors, and then drew out their components and used trig to find the angles where the connect. I also found the magnitude of both the vectors.
Am i on the right track?
Not really. If u = <-2, -2> and v = <-2, -10> are your two vectors, a vector that represents the main diagonal of the parallelogram is u + v. The other diagonal is given by u - v.
 
OMG I totally missed that oh man I fail! Thanks so much, I over complicated it!
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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