The ladder is moving away from the wall at the bottom, namely dx/dt
at a rate of 0.5 m/s, now... when it's 6m away from the wall, it has been moving for 12seconds already then, at 0.5ms, you need 12s to get 6meters away, thus x = 6
the ladder is moving downwards, namely dy/dt
at a rate of 0.375 m/s, in 12seconds, it has moved then 4.5meters, thus y = 4.5
so hmmm, notice, the ladder is sliding down and away from the wall, but the ladder itself itself elongating or shrinking, so is a constant
let us use the pythagorean theorem for that, notice picture below
thus
![\bf r^2=x^2+y^2\\\\ -----------------------------\\\\ 2r=2x\frac{dx}{dt}+2y\frac{dy}{dt}\impliedby \textit{taking common factor }2 \\\\\\ r=x\frac{dx}{dt}+y\frac{dy}{dt}\qquad \begin{cases} \frac{dy}{dt}=0.375\\\\ \frac{dx}{dt}=0.5\\\\ x=6\\ y=4.5\\\\ r=ladder \end{cases} \\\\\\ \left.\cfrac{}{} r \right|_{x=6}\implies ?](https://tex.z-dn.net/?f=%5Cbf%20r%5E2%3Dx%5E2%2By%5E2%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0A2r%3D2x%5Cfrac%7Bdx%7D%7Bdt%7D%2B2y%5Cfrac%7Bdy%7D%7Bdt%7D%5Cimpliedby%20%5Ctextit%7Btaking%20common%20factor%20%7D2%0A%5C%5C%5C%5C%5C%5C%0Ar%3Dx%5Cfrac%7Bdx%7D%7Bdt%7D%2By%5Cfrac%7Bdy%7D%7Bdt%7D%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0A%5Cfrac%7Bdy%7D%7Bdt%7D%3D0.375%5C%5C%5C%5C%0A%5Cfrac%7Bdx%7D%7Bdt%7D%3D0.5%5C%5C%5C%5C%0Ax%3D6%5C%5C%0Ay%3D4.5%5C%5C%5C%5C%0Ar%3Dladder%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Cleft.%5Ccfrac%7B%7D%7B%7D%20r%20%5Cright%7C_%7Bx%3D6%7D%5Cimplies%20%3F)
pretty sure is very obvious :)