I Don't know how to do this plz help

\(\text{well we can get a few equations using the Pythagorean theorem}\\ \text{Let the left hand unlabeled leg be $a$ and the right hand one be $b$}\\ x^2+4^2=a^2\\ x^2+16=a^2\\~\\ a^2+b^2=(10+4)^2= 14^2=196\\~\\ x^2+10^2=b^2\\ x^2+100=b^2\)

\(\text{We can add the first and third equations}\\ x^2+16+x^2+100=a^2+b^2\\ 2x^2 + 116 = a^2+b^2\\~\\ \text{and note this equals the left hand side of the second equation}\\ 2x^2+116 = 196\\~\\ 2x^2=80\\ x^2=40\\ x=\sqrt{40}=2\sqrt{10}\\~\\ \text{choice B}\)

Can you help me with another one like this one

m∠ABD + m∠CBD  =  90°

m∠ABD + m∠BAD  =  90°

So  m∠CBD =  m∠BAD

And  m∠BDA  =  m∠CDB  because they are both right angles.

So by the AA similarity theorem,  △ABD  ~ △BCD

BD / CD  =  AD / BD

x / 10  =  4 / x

x² / 10  =  4

x²  =  40

x  =  sqrt( 40 )

x  =  2 sqrt( 10 )

Tan BCD = x / 10 =Cotan BAD =4 / x
Or: x / 10 = 4 / x    Cross multiply
x^2 = 40
x = 2 Sqrt(10)