a right triangle has sides a, b, and c (where c is the hypotenuse) and its sides are increasing such that…

Explanation:

Step1: Find side $a$

The area of a right - triangle is $A=\frac{1}{2}ab$. Given $A = 6$ and $b = 4$, we have $6=\frac{1}{2}a\times4$. Solving for $a$: $6 = 2a$, so $a = 3$.

Step2: Use the Pythagorean theorem

By the Pythagorean theorem, $a^{2}+b^{2}=c^{2}$. Differentiating both sides with respect to $t$ using the chain - rule, we get $2a\frac{da}{dt}+2b\frac{db}{dt}=2c\frac{dc}{dt}$, which simplifies to $a\frac{da}{dt}+b\frac{db}{dt}=c\frac{dc}{dt}$.

Step3: Find side $c$

Since $a = 3$ and $b = 4$, by the Pythagorean theorem $c=\sqrt{a^{2}+b^{2}}=\sqrt{3^{2}+4^{2}} = 5$.

Step4: Substitute and solve

We know that $\frac{db}{dt}=3\frac{da}{dt}$, so $\frac{da}{dt}=\frac{1}{3}\frac{db}{dt}$. Also, $\frac{dc}{dt}=1$. Substitute $a = 3$, $b = 4$, $c = 5$, $\frac{da}{dt}=\frac{1}{3}\frac{db}{dt}$, and $\frac{dc}{dt}=1$ into $a\frac{da}{dt}+b\frac{db}{dt}=c\frac{dc}{dt}$: $3\times\frac{1}{3}\frac{db}{dt}+4\frac{db}{dt}=5\times1$. $ \frac{db}{dt}+4\frac{db}{dt}=5$. $5\frac{db}{dt}=5$. So, $\frac{db}{dt}=1$.

Answer:

$1$