Related Rates Ladder Question A ladder 25 feet in length creates a right triangle with the wall it's leaning against. If the base of the ladder is being pulled horizontally away from the wall at a rate of 2ft/s, what is the rate that the area of the triangle changes when the base of the ladder is 11 feet away from the wall. 
I've tried solving this question and am unsure whether or not my methodology is correct. I solved for height using the Pythagorean theorem (22.45 ft) and then found the rate that the height changes using the following:
$$
2h\frac{dh}{dt} = -2b\frac{db}{dt}
$$
I then surmised that the rate of change in area was:
$$
\frac{dA}{dt} = .5b\frac{dh}{dt} + .5h\frac{db}{dt}
$$
Plugging things in, I got 17.06. Am I correct?
 A: Let, $h$ & $a$ be the height & the base of the right triangle formed at any instance then 
In right triangle, using Pythagorean theorem, $$25^2=h^2+a^2$$$$\implies h=\sqrt{625-a^2}$$ Now, the area of the right triangle at any instance $$A=\frac{1}{2}(\text{base})(\text{height})=\frac{1}{2}(a)(h)$$
$$A=\frac{1}{2}a\sqrt{625-a^2}$$
Hence, the rate of change of the area of right triangle, $$\frac{dA}{dt}=\frac{1}{2}\frac{d}{dt}(a\sqrt{625-a^2})$$ 
$$=\frac{1}{2}\left(\sqrt{625-a^2}-\frac{a^2}{\sqrt{625-a^2}}\right)\frac{da}{dt}$$ 
$$=\frac{1}{2}\left(\frac{625-2a^2}{\sqrt{625-a^2}}\right)\frac{da}{dt}$$ 
Now, setting the corresponding values, length of base, $a=11$ ft & speed of ladder base, $\frac{da}{dt}=2$ ft/sec, we get 
$$\frac{dA}{dt}=\frac{1}{2}\left(\frac{625-2\cdot11^2}{\sqrt{625-11^2}}\right)(2)$$
$$\color{red}{\frac{dA}{dt}}=\color{blue}{\frac{383}{2\sqrt{126}}\approx 17.06  ft^2/sec}$$
A: Basically you are correct. It is convenient and also provides some physical insight when handled symbolically till the last step of inputting/plugging in.  
Let $  h,b $ be height and base length at floor contact point of ladder. Ladder length $L$
Area $$ A= b \, h/2,\,  h^2 +b^2 = L^2,  $$
$$ 2 A = b \cdot \sqrt{L^2 - b^2}  $$
Differentiate with respect to time $t$ and simplify. 
$$ 2 \dot A = \dfrac{\dot b( L^2 -2 b^2) }{ \sqrt{L^2 - b^2}} $$
You can see that when $ b/L = \dfrac{1}{\sqrt2}$ , or when ladder makes $45^0 $to ground, the  area $A$ remains constant. That part of it is convincing already.
Only now  plug in given values: $ L=25, b=112,\dot b = 2 ; \, \dot A = 17.0602 $ square ft per sec.  
