How Do You Solve the Rhombus Side Length in Rectangle ABCD?

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Homework Statement



In Rectangle ABCD, AB=4 and BC=3. Find the side length of Rhombus AXYZ, where X is on AB, Y is on BC and Z is on BD.

2. Relevant Questions:


The Attempt at a Solution



Hi, so here's my picture for the problem...I tried to draw the exact picture with exact value of the problem (using rulers, compass, etc.)
https://mail.google.com/mail/u/0/?ui=2&ik=65096f80a0&view=fimg&th=142e05c217ab4a7e&attid=0.1&disp=inline&safe=1&attbid=ANGjdJ8VlUlSyOcZxwQH7UU8SWsWZZS5AQMIGuuyqaGtZEzxSajqjBH3pw_2NeEwrZCyqh-ea_Lq9g6IEgwbZ80bNLtI2nwqrzmgJlmwZALZiJEnTpNHIJZJno8zncw&ats=1386743610684&rm=142e05c217ab4a7e&zw&sz=w1488-h627
I was able to find a pair of similar triangles and figured out that: YZ = (4/3)BY
I didn't get really far after that...
Please help me !
 
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ThienNguyen said:

Homework Statement



In Rectangle ABCD, AB=4 and BC=3. Find the side length of Rhombus AXYZ, where X is on AB, Y is on BC and Z is on BD.

2. Relevant Questions:

The Attempt at a Solution



Hi, so here's my picture for the problem...I tried to draw the exact picture with exact value of the problem (using rulers, compass, etc.)
https://mail.google.com/mail/u/0/?ui=2&ik=65096f80a0&view=fimg&th=142e05c217ab4a7e&attid=0.1&disp=inline&safe=1&attbid=ANGjdJ8VlUlSyOcZxwQH7UU8SWsWZZS5AQMIGuuyqaGtZEzxSajqjBH3pw_2NeEwrZCyqh-ea_Lq9g6IEgwbZ80bNLtI2nwqrzmgJlmwZALZiJEnTpNHIJZJno8zncw&ats=1386743610684&rm=142e05c217ab4a7e&zw&sz=w1488-h627
I was able to find a pair of similar triangles and figured out that: YZ = (4/3)BY
I didn't get really far after that...
Please help me !

YZ = (4/3)BY is correct. Use the other property of the rhombus, that all sides are of equal length.ehild
 
40700d803214069171fe87bec1811471c283e285.png
is a rhombus implies
d1dd0ff0c99be4bb0ed589e9fea29bb8f4a5ecb4.png
. In
1a7a0327109ede54bf3abf9eb325bb58a503afc2.png
by similarity we get
6d282a64f3a730b4fadbba8d53d29a28e9003bcd.png


And then we get
081b45483abfce53e98405dcacc80c9801fec8f0.png
. Thus, we result in the equation
a0ccde74f902f48e407c67b4dd90cf551630994d.png
.

Sorry if the LaTeX or images I put in my reply don't come out properly

ET ag
 
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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