Radius of an Inscribed Semicircle

Let $O$ be the center of the semicircle and let $D$ and $E$ be the points of tangency on $AB$ and $BC$ , respectively. From circle properties, we have that $OD$ is perpendicular to $AB$ , $OE$ is perpendicular to $BC$ and $BD = BE$ . Thus $\angle DOE = 90^\circ$ and $BDOE$ is a square, where the sidelength is the radius of the semicircle. Let $BD = r$ , thus $DA =36 - r$ and $EC = 27 - r$ .

Solution 1: Notice that triangles $ADO$ and $ABC$ are similar, hence $\frac {36-r}{r} = \frac {36}{27}$ , which implies that $\frac {36}{r} = \frac {36 + 27}{27}$ . Thus $r = \frac {36 \times 27}{36 + 27} =\frac {108}{7}$ . Hence $a+b = 108+7=115$ .

Solution 2: The area of $ABC$ is equal to $\begin{aligned} [ABC] &= [ADO] + [BDOE] + [OEC] \\ &= \frac{(36 - r)(r)}{2} + r^2 + \frac{(27-r)(r)}{2} \\ &= \frac{63r}{2} \\ \end{aligned}$

We also have that $[ABC] = \frac{36 \cdot 27}{2}$ . Equating the two, we get $r = \frac{108}{7}$ . Hence $a + b = 108 + 7 = 115$ .

The semicircle has its diameter on the hypotenuse.

Call the center $R$

Call the radius $r$

. This splits the right triangle into 2 triangles: $ABR$ and $BRC$

Area of $ABC=ABR+BRC$ , which is $\frac{36(27)}{2}=\frac{36r}{2}+\frac{27r}{2}$

Since all the denominators are equal, equate the numerator: $36(27)=36r+27r$ and $r=\frac{972}{63}=\frac{108}{7}=\frac{a}{b}$

So the answer is $\boxed{115}$

I solved the area of a rectangle (assuming that its dimensions were the aforementioned) with 36 27=972.. Given that the semi-perimeter of the rectangle is (36 2+27*2)/2=63, dividing it to the area of the rectangle will yield the radius of the largest circle that can be inscribed in a rectangle = 972/63=108/7, which I get 108+7=115..

if we take off the square from the right triangle . we have 2 similar right triangles. so x/(27-x)=(36-x)/x

by solving the equation x = 108/7

given diagram we get the eqn square root [(36-x)^2+x^2]+square root[(27-x)^2+x^2]=45.when we solve the eqn we get x=108/7.then a+b=108+7=115.