Area of a right triangle if $a/b$ is $1.05$ and the difference between the radii of the circumscribed and inscribed circles is $17$? What's the area of a right triangle if the quotient of its legs is $1.05$ and the difference between the radii of the inscribed and circumscribed circles is $17$?
I've been trying to solve this and I've got:
($R$ - radius of circumscribed circle, $r$ - radius of inscribed circle)
$1.$ $ \frac{a}{b}=1.05$
$2.$ $c^2=a^2+b^2$ 
$3.$ $a + b - 2r = c$ 
$4.$ $c-2r=34$
$5.$ $ab=(a+b+c)r$
Using the first four equations, I can substitute for one of the legs from $1.$ and for $r$ through $4.$ which leaves me with
$b(2.05)-2c=34$ 
$c=b\sqrt{1.05^2+1}$
However, solving this simply evades me, as I don't find myself getting rid of the square root which I don't know how to calculate.
I do know my equations give the right answer so I'm probably missing a simpler way to solve the system of equations.
Help much appreciated.
 A: You are missing that $2R=c$, so
$$(2R)^2=a^2+b^2=(1+1.05^2)b^2$$
but $R=17+r$, so
$$4(17+r)^2=(1+1.05^2)b^2 \quad (1)$$
We also have:
$$a+b-2r=2R\to(1+1.05)b=2(r+R)=2(17+2r)\quad (2)$$
Combining $(1)$ and $(2)$ you get:
$$\frac{b\sqrt{1+1.05^2}}{2}-17=\frac{1}{2}\left(\frac{b(1+1.05)}{2}-17\right)$$
$$\frac{1.45b}{2}-17=\frac{1}{2}\left(1.025b-17\right)\to1.45b-34=1.025b-17\\
0,425b=17\to b=40$$
Now you can calculate $a=40\cdot 1.05=42$.
The area will be $$A=\frac{40\cdot 42}{2}=840$$.
A: You have correctly arrived at the $2$ equations, namely
$$2.05b - 2c = 34$$ and
$$c = b \sqrt{1 + 1.05^{2}}$$
Substitute $c$ from second equation into first equation to get 
$$2.05b - 34 = 2c = 2b\sqrt{1 + 1.05^{2}}$$
Now square both the sides and simplify to get the following quadratic
$$4.2075b^2 + 139.4b - 1156 = 0 \implies (b+40)(4.2075b - \frac{1156}{40}) = 0$$
Taking the positive root of $b$ gives $b = 6.\overline{86}$
Now the area could be calculated as $A = \frac{ab}{2} = 0.525 b^2$
A: The inradius of a right triangle in terms of its sides is given by the relationship $$r \frac{a+b+c}{2} = rs =|\triangle ABC| = \frac{ab}{2},$$ hence $$r = \frac{ab}{a+b+c}.$$  The circumradius is trivially $$R = \frac{c}{2}.$$  Thus the given conditions may be summarized as $$17 = R - r = \frac{c}{2} - \frac{ab}{a+b+c} = \frac{(a+b)c + c^2 - 2ab}{2(a+b+c)} = \frac{(a+b)c + (a-b)^2}{2(a+b+c)}, \\ \frac{a}{b} = \frac{21}{20}, \\ a^2 + b^2 = c^2.$$  We note that because $21^2 + 20^2 = 29^2$, we have $$a : b  : c \equiv 21 : 20 : 29,$$ thus the computation is greatly simplified by letting $a = 21k$, $b = 20k$, $c = 29k$, to obtain from the first equation $$17(2)(70)k = (41)(29)k^2 + (21-20)^2 k^2,$$ or $$2380k = 1190k^2,$$ or $k = 2$.  Thus the desired triangle is $(a,b,c) = (42, 40, 58)$ and its area is $$|\triangle ABC| = 840.$$
A: Here is a rather simple solution using analytic geometry.

Let us call $AOB$ the right triangle, taking the main vertex $O$ as the origin, and axes resp. directed by OA and OB.
By assumption, the coordinates of $A$ and $B$ are resp. $(21s,0)$ and $(0,20s)$ for a certain $s$. Hypotenuse (H) has thus length $\sqrt{(20s)^2+(21s)^2}=29s$. Therefore the radius of the circumscribed circle is the half of the hypotenuse: $R:=\tfrac{29}{2}s$.
The incenter (center of the inscribed circle), being on the angle bissector of right angle $A$, has coordinates $(t,t)$, where $t$ is the radius of the incircle.
The constraint on the radii becomes:
$$\tag{1}\tfrac{29}{2}s-t=17$$
The equation of line AB is $\tfrac{x}{21s}+\tfrac{y}{20s}=1 \ \iff \ 20x+21y-420s$. 
Using the classical formula of the distance of a point to a straight line, we have: $d(I,(H))=\tfrac{|20t+21t-420s|}{\sqrt{20^2+21^2}}$ which must be equal to $t$, yielding finally the following constraint:
$$\tag{2}\tfrac{420s-41 t}{29}=t.$$
The linear system (1)+(2) gives instantly : $s=2$ and $t=12$.
Thus $OA=21s=42$ and $OB=20s=40$ and the area is $\frak{A}$$=\tfrac12 \times 42 \times 40=840.$ 
Remark: the problem is based on the pythagorean triangle $(20,21,29)$.
