Finding the diagonals of a rhombus with side length $13$, where the sum of the diagonals is $34$ 
How can we find diagonals of the rhombus with side length $a=13$ cm and sum of diagonals $d_1+d_2=34$ cm?

Anything doesn't seem to work...
I would really appreciate it, if anyone could help me / solve it :)
Since I have only basic knowledge in geometry, please post your EASIEST answer possible.
 A: Hint: The diagonals of a rhombus are perpendicular. Set up a system of equations using the given information and the Pythagorean theorem.
A: You can also use the theorem of cosines:
$$d_1^2=13^2+13^2-2\cdot 13^2\cos(180^{\circ}-\alpha)$$
$$d_2^2=13^2+13^2-2\cdot 13^2\cos(\alpha)$$ and use that $$d_1+d_2=34$$ and note that $$\cos(\pi-x)=-\cos(x)$$
A: Results on quadratic equations make  this problem easy to solve:


*

*As the diagonals in a rhombus are perpendicular, applying Pythagoras results in
$$d_1^2+d_2^2=4\cdot13^2. $$

*On the other, the hypothesis $\;d_1+d_2=34$ implies
$$4\cdot 17^2=(d_1+d_2)^2=d_1^2+d_2^2+2d_1d_2=4\cdot13^2+2d_1d_2,$$
so that $\;d_1d_2=240$.
Therefore, it comes down to the standard problem of finding two numbers , given their sum $s$ and their product $p$. We know they're the roots (if any) of the quadratic equation
$$x^2-sx+p=0.$$
A: Here's how to do this problem the wrong way. Be suspicious of the number $13$ and remember that "$5$-$12$-$13$" forms a right triangle. Guess that the diagonals split the rhombus into $4$ seperate "$5$-$12$-$13$" right triangles, which implies diagonals of lengths $10$ and $24$, which do actually add to $34$, so that must be the correct answer.
A: Perpendicular lengths of rhombus semi diagonals $(x,y)$ can be mentally solved
$$ 2x+2y=34\quad x+y=17 \tag1$$
Perimeter 
$$ 4 \sqrt{x^2+y^2} = 4 (13)\quad \rightarrow {x^2+y^2}=13^2 $$
Using identity
$$ 2xy= (x+y)^2-(x^2+y^2) =17^2-13^2= 30(4)=120 \rightarrow xy=60 \tag2 $$
Since the sum and product of $x$ and $y$ are known we can factorize
$$ (x-5)(y-12)=0 \rightarrow d_1= 2x =10, d_2=2y= 24. \tag3 $$
