Three sides of a $\triangle$ are known. If a circle with it's center on base of $\triangle$ touches the other two sides , find the radius of circle.

Question

In $\triangle ABC$, $AB = 10, AC = 12$ and $BC = 18$. A circle is drawn such that its center is on side $ BC$ and it touches lines $AC$ and $AB$. Find the radius of the circle.

By pythagoras theorem ,
$r^2 + (10-x)^2=(18-y)^2$
$(12-x)^2+r^2=y^2$
So there are three variable and 2 equations. Also I think my algebraic approach is quite cumbersome. So can anyone find an elegant solution? Any help is apreciated.

Answer 1

$(ABC)=(ADB)+(ADC)=5r+6r=11r$. So, by Heron's formula, $$11r=\sqrt{20\cdot10\cdot8\cdot2}=40\sqrt2$$So, $r=\frac{40}{11}\sqrt2$.

Answer 2

Use Heron's formula to compute the area $$S=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{20\cdot 10\cdot 8\cdot 2}=40\sqrt 2. $$ We also have $$S=\frac12bh_B=\frac12ch_C$$ where $h_B$ and $h_B$ are the respective heights of $B$ and $C$. Then $$r:(h_C-r)=(a-y):y = (h_C-r):r$$ from the intercept theorem (once with centre $B$, once with centre $C$), so $$r^2=(h_C-r)(h_B-r)=h_Bh_C-(h_B+h_C)r+r^2,$$ which gives us $$ r=\frac{h_Bh_C}{h_B+h_C}=\frac{2S}{b+c}=\frac{40\sqrt 2}{12+10}=\frac{20}{11}\sqrt 2.$$