The sum of the diagonals of a rhombus is 5√2.The area is 4 cm²What's the perimeter? The result is 2√34Please, show your work!Thanks :)

Greetings!
Let ABCD be a rhombus
AC + BD = 5√2 cm

Area of ABCD = Ar△ABD + Ar △BCD
[tex] = \frac{1}{2} \times BD \times AO + \frac{1}{2} \times BD \times OC[/tex]
[tex] = \frac{1}{2} BD (AO + OC)[/tex]
[tex] \frac{1}{2} BD \times AC[/tex]

[tex]So, \frac{ BD \times AC}{2} = 4 \: cm {}^{2} [/tex]
[tex]BD \times AC = 8[/tex]
[tex]Now, AC + \: BD = 5 \sqrt{2} [/tex]

Squaring both sides, we get
[tex]AC {}^{2} + BD {}^{2} + 2 AC.BD =50[/tex]
[tex]AC {}^{2} + BD {}^{2} + 2 \times 8 = 50[/tex]
[tex]AC {}^{2} + BD {}^{2} = 50 - 16[/tex]
[tex]AC {}^{2} + BD {}^{2} = 34[/tex]

In △AOB, we have
[tex]OA {}^{2} + OB {}^{2} =AB {}^{2} [/tex]
[tex]( \frac{ AC }{2} ) {}^{2} + ( \frac{ BD}{2} ) {}^{2} = AB {}^{2} [/tex]
[tex] \frac{ AC {}^{2} } {4} + \frac{ BD {}^{2} }{4} = AB {}^{2} [/tex]
[tex]AC {}^{2} + BD {}^{2} = 4 AB {}^{2} [/tex]
[tex]34 = 4 AB {}^{2} [/tex]

Square rooting both sides
[tex] \sqrt{34} = 2 AB[/tex]
[tex]Perimeter = 4 \: AB \\ = 2 \times 2 \: AB \\ = 2 \: \times \sqrt{34 } \\ = 2 \sqrt{34 \: } units[/tex].

Hope it helps!