A slight deformation

First, we find the inradius, combining two versions of the area of the initial right-angled triangle:

$$

$\left. \begin{matrix} \text{Base-height formula: } Area =\frac{1}{2}base \times height = \frac{1}{2} \times 3 \times 4 = 6 \\ \text{ } \\ \text{Semiperimeter-inradius formula: } Area = s \cdot r = \frac{{3 + 4 + 5}}{2} \cdot r = 6r \\ \end{matrix} \right\}\Rightarrow r=1$

Now we move to the acute $\triangle ABC$ .
Denote $I$ , $D$ , $E$ , $F$ , the incenter and the contact points of the incircle with the triangle, as seen on figure.
Let $AB=3$ , $AC=4$ , $AD=AF=x$ , $\angle IAD=\angle IAF=\theta$ and area of $\triangle ABC=A$ .

The semiperimeter of $\triangle ABC$ is $s' = \frac{{AB + AC + BC}}{2} = \frac{{3 + 4 + \left( {7 - 2x} \right)}}{2} = 7 - x$

Hence, $A = s' \cdot r = \left( {7 - x} \right) \cdot 1 \Rightarrow A = 7 - x\ \ \ \ \ (1)$

Again, $A = \frac{1}{2}AB \cdot AC \cdot \sin \left( {2\theta } \right) = \frac{1}{2}\cdot 3 \cdot 4 \cdot \sin \left( {2\theta } \right) \Rightarrow A = 6\sin \left( {2\theta } \right)\ \ \ \ \ (2)$

$\left( 1 \right),\left( 2 \right) \Rightarrow 6\sin \left( {2\theta } \right) = 7 - x \Rightarrow \sin \left( {2\theta } \right) = \frac{{7 - x}}{6}\ \ \ \ \ (3)$

From $\triangle AID$ we see that $x = \cot \theta\ \ \ \ \ (4)$

Alongside, $\sin \left( {2\theta } \right) = \frac{{2\cot \theta }}{{1 + {{\cot }^2}\theta }}\ \ \ \ \ (5)$

$\begin{gathered} \left( 3 \right),\left( 4 \right),\left( 5 \right) \Rightarrow \frac{{7 - x}}{6} = \frac{{2x}}{{1 + {x^2}}} \\ \Leftrightarrow {x^3} - 7{x^2} + 13x - 7 = 0 \\ \Leftrightarrow \left( {x - 1} \right)\left( {{x^2} - 6x + 7} \right) = 0 \\ \Leftrightarrow x = 1{\text{ or }}x = 3 \pm \sqrt 2 \\ \end{gathered}$

Since $x = \left| {\overline {AD} } \right| < \left| {\overline {AB} } \right| = 3$ , the value $3 + \sqrt 2$ is rejected.
$x=1$ leads to the initial right triangle. Hence, for the acute $\triangle ABC$ we have $x = 3 - \sqrt 2.$

From $(1)$ , $A = 7 - \left( {3 - \sqrt 2 } \right) = 4 + \sqrt 2$

Taking the difference of the areas we find the loss: $6 - \left( {4 + \sqrt 2 } \right) = 2 - \sqrt 2$

For the answer, $a=2$ , $b=2$ , $a+b=\boxed{4}$ .

The inradius $r$ of a triangle with area $\Delta$ and side lengths $a$ , $b$ , and $c$ is given by $r = \frac \Delta s$ , where $s = \frac {a+b+c}2$ is the semi-perimeter. From the 3-4-5 right triangle, we have $r = \frac {3\times 4/2}{(3+4+5)/2} = 1$ .

Another formula for inradius is as follows:

$r = \frac 12 \sqrt{\frac {(b+c-a)(c+a-b)(a+b-c)}{a+b+c}}$

Assuming $a$ is the unknown side length, $b=3$ , and $c=4$ , we have:

$\begin{aligned} \frac 12 \sqrt{\frac {(7-a)(a+1)(a-1)}{a+7}} & = 1 \\ (7-a)(a^2-1) & = 4 (a+7) \\ \implies a^3 -7a^2 + 3a + 35 & = 0 & \small \blue{\text{Obviously }x=5 \text{ is a root.}} \\ (a-5)(a^2 - 2a-7) & = 0 \\ \implies a & = 1+2\sqrt 2 & \small \blue{\text{The other }a.} \end{aligned}$

The area of the new triangle is $\Delta_1 = rs_1 = \frac {1+2\sqrt 2+3+4}2 = 4 + \sqrt 2$ . The loss in area is $\frac {3\times 4}2 - 4 - \sqrt 2 = 2 - \sqrt 2$ . Therefore, the required answer is $\boxed 4$ .

Label the triangle as usual with vertices $A,B,C$ and the sides opposite them $a,b,c$ ; let $a=3,b=4$ .

When $c=5$ , the inradius is given by $r=\frac12 (a+b-c)=1$ .

Now consider the deformed triangle. Let the points of tangency of the incircle be $A'$ on side $BC$ , $B'$ on side $CA$ and $C'$ on side $AB$ .

Let $CA'=x$ . By the two tangent theorem, we also have $CB'=x$ . Working around the triangle, we get $\begin{aligned} CA'&=CB'=x \\ AB' &= AC' = 4-x \\ BA' &= BC' = 3-x \end{aligned}$

so the triangle's sides are $a=3,b=4,c=7-2x$ . Its semiperimeter is $s=7-x$ .

Now we can work out its area in two different ways to get an equation. Using $\Delta=sr$ , we have $\Delta=7-x$ . Plugging this in to Hero's formula, we get $\begin{aligned} \Delta^2 &=s(s-a)(s-b)(s-c) \\ (7-x)^2 &= x(7-x)(3-x)(4-x) \\ 7-x &= x(3-x)(4-x) \\ x^3-7x^2+13x-7 &=0 (x-1)(x^2-6x+7)&=0 \end{aligned}$

with solutions $x=1$ , $x=3\pm\sqrt2$ . The root $x=1$ corresponds to the original $3-4-5$ triangle. $x=3+\sqrt2$ is too big; so the solution we're after is $x=3-\sqrt2$ .

We find $\Delta=4+\sqrt2$ , which is a loss of area from the original triangle (whose area is $6$ ) of $\boxed{2-\sqrt2}$ square units.