Minimario's triangle

Let O be the center of the semicircle. The area of triangle ABC can be represented as the area of triangle OBC + triangle AOC. The area of these triangles is base*height or $\dfrac{BC\cdot R}{2}$ and $\dfrac{AC\cdot R}{2}$ where R is the radius of the semicircle. Thus $[ABC]=\dfrac{(25+17)R}{2}=21R$

Using Heron's Formula, $[ABC]=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{35\cdot7\cdot10\cdot18}=5\cdot6\cdot7=21R\\\Rightarrow R=10$ Therefore the diameter is $\boxed{20}$

We find the area of $\Delta ABC$ , denoted $[\Delta ABC]$ in two ways.

Method 1: Since a $28-25-17$ triangle is a $8-15-17$ right triangle glued to a $15-20-25$ right triangle along the $15$ edge, we have

$[\Delta ABC] = \dfrac{1}{2} \cdot 15 \cdot 8 + \dfrac{1}{2} \cdot 15 \cdot 20 = 210$ .

Method 2: Draw a picture of $\Delta ABC$ along with the semicircle. Let $X$ be the center of the semicircle. Let $r$ be the radius of the semicircle. Draw in two radii from $X$ to the tangency points on $AC$ and $BC$ .

Since $\Delta ACX$ has a base of $AC = 17$ and a height of $r$ , $[\Delta ACX] = \dfrac{1}{2} \cdot 17 \cdot r = \dfrac{17}{2}r$ .

Since $\Delta BCX$ has a base of $BC = 25$ and a height of $r$ , $[\Delta BCX] = \dfrac{1}{2} \cdot 25 \cdot r = \dfrac{25}{2}r$ .

Since $\Delta ACX$ and $\Delta BCX$ join to form $\Delta ABC$ , we have $[\Delta ABC] = [\Delta ACX] + [\Delta BCX] = \dfrac{17}{2}r + \dfrac{25}{2}r = 21r$ .

However, both methods must yield the same value of $[\Delta ABC]$ .

Therefore, $21r = 210 \leadsto r = 10$ , and so, the diameter is $2r = \boxed{20}$ .

Let O be the center of the semicircle with radius r, and D and E the points of tangency at BC and AC, respectively. Applying Heron's Formula, with semiperimeter $s = 35,$

$A_{ABC} = \sqrt{35(35-28)(35-25)(35-17)} = 210.$

$A_{AOC} =\frac12AC\times r = \frac{17r}{2}$

$A_{BOC} =\frac12BC\times r = \frac{25r}{2}$

$A_{ABC} = A_{AOC} + A_{BOC}$

$210 = \frac{17r}{2} + \frac{25r}{2} = 21r$

$r = 10$

Hence, $d = 20.$

let O be the center of circle and let the radius is 2r.

the circle tangent BC at D, AC at E.

Sum of area triangles AOC and BOC equal area ABC

17r+25r=sqrt(35x10x7x18)

42r=210

r=5

the diameter = 4r=4*5=20

Let $M$ be the center of the inscribed semicircle and $r$ its radius (note that $M$ lies on $AB$ ). Let $P$ be the point on $AC$ where the semicircle touches $AC$ . Then $MP$ is perpendicular to $AC$ . Therefore, the area of the triangle $AMC$ equals

$T_1 = \frac12 |MP| |AC| = \frac{17}2 r$ .

By similar means, we find that the area of the triangle $BMC$ equals

$T_2 = \frac{25}2 r$ .

Since these two triangles together make up the triangle $ABC$ , that area equals $T = T_1 + T_2 = 21r$ .

We can also compute the area of the triangle $ABC$ using Heron's formula:

$T = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{35 \cdot 10 \cdot 18 \cdot 7} = 210$ .

Comparing our two formulas for $T$ , we thus find $r = 10$ , hence the diameter is $2r = \boxed{20}$ .

Let O be the centre of the semicircle. Let E be the foot of the perpendicular drawn from O to AC and F be the foot of the perpendicular drawn from O to BC. Then let $r = OE = OF$ be the radius of the semicircle. By Heron's Formula, area of triangle ABC is $\sqrt{35(35 - 17)(35-25)(35-28)} = 210$ where 35 is the semiperimeter.

However,

Area of triangle ABC = Area of triangle OAC + Area of triangle OBC = 210.

$\triangle OBC = \frac12 \times BC \times OF = \frac12 \times 25 \times r$

$\triangle OAC = \frac12 \times AC \times OE = \frac12 \times 17 \times r$

From the above two equations, we can deduce that $r = 10$ and therefore the diameter of the circle is $20$ .

Mirror the triangle about AB to get a quadrilateral. The circle is now inscribed in this quadrilateral. For such a tangential quadrilateral , the area is the product of the radius of the circle and the semiperimeter of the quadrilateral. Since the area of the triangle is 210 (and hence the area of the quadrilateral is 420) and the semiperimeter of the quadrilateral is 42, we obtain d=2*420/42=20.

We represent the radius of the semicircle as $r$ . Draw the radii from the center of the semicircle (call this center $D$ ) to the tangent lines. Call the points on $\overline{AC},\overline{BC}$ the names $R, S$ , respectively. Note that $\Delta DRC\cong \Delta DSC$ because of the SSS congruence. We represent $\overline{RC}=\overline{SC}$ as $x$ . Since $\angle RDC=\angle SDC$ , we can use the angle bisector theorem to obtain that $\overline{AD}=\dfrac{34}{3}$ and $\overline{BD}=\dfrac{50}{3}$ . Note that both $\Delta ADR$ and $\Delta BDS$ are right triangles. Therefore we can set up these equations by Pythagorean Theorem: $\left(\dfrac{34}{3}\right)^2-(17-x)^2=r^2$ $\left(\dfrac{50}{3}\right)^2-(25-x)^2=r^2$ Using elimination to cancel off the $r^2$ we obtain that $x=\dfrac{35}{3}$ . Plugging $x$ back in and solving for $r$ gives $r=10$ . Therefore the answer is $d=2r=\boxed{20}$

Let the center of the semicircle be $O$ . Join $O$ and $C$ .

Let $P$ and $Q$ be the feet of perpendiculars from $O$ to $BC$ and $AC$ respectively.

It is clear that $OP=OQ=$ the radius of the semicircle. We're going to call it $r$ from this point on.

Now consider the right triangles $\triangle OPC$ and $\triangle OQC$ . Notice that they are congruent.

So we can say $\angle OCP=\angle OCQ$ .

Now we apply the angle bisector theorem on $\triangle ABC$ which tells us:

$\frac{AC}{BC}=\frac{AO}{BO}$

$\Rightarrow \frac{17}{25}=\frac{AO}{BO}$

$\Rightarrow \frac{17}{25}+1=\frac{AO}{BO}+1$

$\Rightarrow \frac{17+25}{25}=\frac{AO+BO}{BO}$

$\Rightarrow \frac{42}{25}=\frac{AB}{BO}$

$\Rightarrow \frac{42}{25}=\frac{28}{BO}$

$\Rightarrow BO=\frac{50}{3}$ $\cdots (1)$ .

Now let's use the cosine law on $\triangle ABC$ to find $\cos \angle ABC$ .

$\cos \angle ABC=\frac{AB^2+BC^2-AC^2}{2\times AB\times BC}$ .

Plugging in all the values, we get $\cos \angle ABC=\frac{4}{5}$ .

That means $\sin \angle ABC=\sqrt{1-(\frac{4}{5})^2}=\frac{3}{5}$ $\cdots (2)$ .

Now consider the right triangle $OPB$ .

$OP=r=OB\times \sin \angle OBP=OB\times \sin \angle ABC$ .

Plugging in the values from $(1)$ and $(2)$ we get $r=\frac{50}{3}\times \frac{3}{5}=10$ .

We want the diameter. So just multiply it by $2$ to get $2r=20$ .

Notice that $\triangle ABC$ can be split into two right triangles, a 15-20-25 triangle and a 8-15-17 triangle that share a height of length 15. Let $O$ be the center of the semicircle, and then the distances from $O$ to $\overline{AC}$ and $\overline{BC}$ are equal. Drawing in those perpendiculars, we see that similar right triangles are formed. Since $AO=x$ , the distance from $O$ to $\overline{AC}$ is $\dfrac{15}{17}x$ , and since $OB=28-x$ , the distance from $O$ to $\overline{BC}$ is $\dfrac{3}{5}(28-x)$ . We have $\dfrac{15}{17}x=\dfrac{3}{5}(28-x)$ $75x=51(28-x)$ $126x=28\cdot 51$ $18x=204$ $x=\dfrac{34}{3}$ Now, the answer is twice the distance from $O$ to $\overline{AC}$ , which is $\dfrac{15}{17}x$ , so the answer is $2\cdot\dfrac{15}{17}\cdot\dfrac{34}{3}=\boxed{20}$ .

By Heron's formula, we know that the area of $\bigtriangleup ABC=210$ .

We know that side $AC$ is perpendicular to the radius, and side $BC$ as well. This means that $\frac{17r}{2}+\frac{25r}{2}=210$ , solving we get $r=10$ . Therefore, the diameter is $\boxed{20}$ .

We use Heron's formula on $\bigtriangleup ABC$ to find its area: $s = \frac{28 + 25 + 17}{2} = 35 \to$ $A_{\bigtriangleup{ABC}} = \sqrt{35\cdot7\cdot10\cdot18} =$ $\sqrt{7^2\cdot5^2\cdot2^2\cdot3^2} = 210$ .

Reflecting the entire figure over segment $\overline{AB}$ gives us a quadrilateral $ACBC'$ which has the now full-circle (the semicircle became a full circle when we reflected it over its diameter) as an incircle.

Clearly, the area of this quadrilateral is twice that of $\bigtriangleup ABC$ , which is $420$ . By the equality $A = s \cdot R$ , where $A,s,R$ are the area, semiperimeter, and radius of the incircle of a polygon, respectively, we have $420 = \frac{25 + 17 + 25 + 17}{2} \cdot R \to$ $R = \frac{420}{42} = 10$ . Thus, the diameter of the circle is $2 \cdot 10 = \fbox{20}$ .

Draw another triangle attached to side AB and congruent to triangle ABC. The vertex directly opposite C can be labeled D. There exists a quadrilateral ADBC with AC length 17, AD length 17, BC and BD length 25. The semicircle can be completed to form a quadrilateral with a circle inscribed inside. A quadrilateral's area with a circle inscribed can be found by (semi-perimeter)(inradius). The semi-perimeter is 42. So the area of the quadrilateral is equal to 42r. The area of the quadrilateral is also equal to 2 times the area of triangle ABC. Area of triangle ABC can be found using Herons Formula: \sqrt{s(s-a)(s-b)(s-c)}. s = semi-perimeter and a, b, and c, are the three sides. That gives us the area to be 210. That's area of triangle ABC, quadrilateral ADBC is twice that, giving 420. 42r = 420. r = 10. The diameter is twice that, 20.

Let $A=(0,0)$ and $B=(28,0)$ .

Finding C: $\sqrt{Cx^2+Cy^2}=17\\ \sqrt{(Cx-28)^2+Cy^2}=25\\ Cx=8\\Cy=\pm 15$

C can be on both sides of line AB and the triangle still looks the same, so we can simply pick out $C=(8,15)$ without loss of generality.

Let X be the center of the circle. X must lie on AB.

$0\leq \text{Xx}\leq 28\\\text{Xy}=0$

Let D be the point where the circle touches AC. D must lie on AC. DX must also be perpendicular to AC.

$\text{Dy}=\frac{15}{8}\text{Dx}\\ \frac{\text{Xy}-\text{Dy}}{\text{Xx}-\text{Dx}}=-\frac{8}{15}$

Let E be the point where the circle touches BC. E must lie on BC. EX must also be perpendicular to BC.

$\text{Ey}=\frac{15}{8-28} (\text{Ex}-28)\\ \frac{\text{Xy}-\text{Ey}}{\text{Xx}-\text{Ex}}=-\frac{(8-28)}{15}$

DX and EX are both radii to the same circle, hence they must have the same length.

$r=\sqrt{(\text{Xx}-\text{Dx})^2+(\text{Xy}-\text{Dy})^2}=\sqrt{(\text{Xx}-\text{Ex})^2+(\text{Xy}-\text{Ey})^2}$

Solving this big mess gives us:

$D=\left(\frac{128}{51},\frac{80}{17}\right)\\ E=\left(\frac{52}{3},8\right)\\ X=\left(\frac{34}{3},0\right)\\ r=10$

Diameter of circle: $2r=20$

Let $I$ be the center of the semicircle inscribed in $\triangle ABC$ tangent to $AC$ and $BC$ .

Since $(I)$ tangent to $AC$ and $BC$ , the distances from $AC$ and $BC$ to $I$ must equal to each other. Hence, $I$ lies on the bisector $d$ of $\widehat{ACB}$ .On the other hand, since a diameter of $(I)$ lies on $AB$ , $I$ lies on $AB$ . Therefore, $I$ must be the intersection of $d$ and $AB$ .

Let $r$ be the radius of $(I)$ , then $r$ is the distance from $I$ to $AC$ and $BC$ as well. By using Heron's formula, we can find that $S_{ABC} = 210$ . On the other hand, we have $2S_{ABC} = 2S_{IBC} + 2S_{IAC} = r \cdot BC + r \cdot CA.$ Then $r = \frac{2S_{ABC}}{BC+CA} = \frac{2 \times 210}{25+17} = 10$ , and $d = 2r = 20$ .

Let $r$ be the radius of the semicircle. Call the center of the semicircle $O$ , and let $D$ and $E$ be the points of tangency of the semicircle on the sides $AC$ and $BC$ , respectively. $OD$ and $OE$ both have lengths of $r$ and are both perpendicular to their respective sides. Thus, the areas of $\triangle AOC$ and $\triangle BOC$ can be written as $\frac{17r}{2}$ and $\frac{25r}{2}$ , and their sum is easily seen to be the area of $\triangle ABC$ .

The area of $\triangle ABC$ can also be calculated using Heron's Formula, which yields $\sqrt{35(35-20)(35-17)(35-28)} = 210$ .

We thus have the equation $\frac{r}{2}(25+17) = 210$

$\therefore r = 10$

Thus the diameter is 20.

Let $D$ be the center of the semicircle of line $AB$ . $CD$ splits the triangle into two parts. Furthermore, we know that dropping $D$ onto lines $AC$ and $BC$ will create perpendiculars, with the length of the radius. Call this length $r$

Using Hero's Formula, $[ABC]=210$ . So, $.5 \times AC \times r + .5 \times BC \times r=210$ . Making Substitutions for AC and BC gives $r=10$ . SO the answer is $20$

Let $P$ , $Q$ be the points where the semicircle touches $AC$ and $BC$ , respectively, and $M$ its center. Note that $MP \perp AC$ and $MQ \perp BC$ . If $(ABC)$ denotes the area of $\bigtriangleup ABC$ and $r$ the radius of the semicircle, we have

$(ABC)$

$= (ACM) + (BCM)$

$= \frac{1}{2} (AC \cdot PM + BC \cdot QM)$

$= \frac{r}{2} (AC + BC) = 21 r$ .

On the other hand, using Heron's formula, we can deduce $(ABC) = \sqrt{35 \cdot 10 \cdot 7 \cdot 18} = 210$ .

Hence, $r = 10$ and the diameter of the semicircle is $20$ .

Look at the triangle ABC. Let points D, E, and O are lies in sides BC, AC, and AB respectively, so that O is the center of the semicircle, OD is perpendicular with BC, and OE is perpendicular with AC. so we get CE = CD, OD = OE = r , where r is the radius of the semicircle. the triangle ODC is similar with triangle OEC, and angle OCE and angle OCD both have same value of degree. Base on Angle Bisector Theorem, we get

$\frac{AC}{BC}=\frac{AO}{BO}$

$\frac{17}{25}=\frac{AO}{28-AO}$

$AO=\frac{34}{3}$

Then, by using the law of cosinus base on triangle ABC

$25^{2}=17^{2}+28^{2}-2.17.28.\cos \alpha$

$\cos \alpha = \frac{8}{17}$ , and $\sin \alpha = \frac{15}{17}$

Look at the triangle EAO. we know that

$\sin \alpha = \frac{OE}{AO}$

$OE=AO \sin \alpha$

$OE=\frac{34}{3} . \frac{15}{7}$

$OE = 10$

So, the diameter of the semicircle is $2r = 2.10=20$

We can find the area of ABC by Heron's formula: $A=210$ . Let D be the centre of the inscribed semi-circle of radius $r$ . The area of ACD is $\frac{25r}{2}$ and of BCD is $\frac{17r}{2}$ since the radii from D are altitudes. So the area of ABC is $21r$ and hence the radius is 10 and the diameter is 20.

Denote I is the center of the semicircle. Put △ABC in Oxy with A(0,0) , B(28,0) C(20,15)

I in the bisector of ACB, so $\frac{AI}{AC}$ = $\frac{BI}{BC}$ AI+BI=28 => AI = $\frac{50}{3}$ => I( $\frac{50}{3},0)$

Equaltion of the line AC is -3x+4y = 0

Distance from I to AC is 10 => diameter of the semicircle is 20

half of perimeter of $\triangle ABC = s = \frac{28+25+17}{2} = 35$

Area $= \sqrt{35*(35-28)*(35-25)*(35-17)} = 210$

Let $r$ be the radius and $O$ be the center of the semi-circle. In $\triangle AOC$ and $\triangle BOC$ , $r$ is the height.

Area of $\triangle AOC = \frac{1}{2}*17*r = \frac{17*r}{2}$

Area of $\triangle BOC = \frac{1}{2}*25*r = \frac{25*r}{2}$

Now,

$\frac{17*r}{2} + \frac{25*r}{2} = 210$ $\Rightarrow r = 10$

So, the diameter is $(10*2) = \boxed{20}$

Let $O$ be the centre of the semicircle, which has radius $r$ .

Since the circle is tangent to $OB$ , the radius at the tangent point makes a right angle with $OB$ . So $[BCO]$ is $\frac{1}{2} BC \cdot r$ .

Similarly, $[ACO]$ is $\frac{1}{2} AC \cdot r$ . Adding these two together gives $[ABC] = \frac{1}{2}r(25 + 17)$ so $[ABC] = 21r$

Now, using Heron's formula, with $s = \frac{1}{2}(28 + 25 + 17) = 35$ : ${[ABC]}^{2} = 35 \cdot 28 \cdot 25 \cdot 17$ which leads to $[ABC] = 210$

But we already worked out that $[ABC] = 21r$ , therefore $r = 10$ , and the diameter is twice that.

Assume that O is the point on AB where the tangents to AC and BC meet. If we extend a line from C to O we now have triangle BCO and ACO. The ratio of which BC/BO = AC/AO. Therefore, if BO = x:

25/x = 17/(28-x)

x = 16.67

Using the law of Cosines, let the degree at B = theta:

17^2 = 25^2 + 28^2 - 2(25)(28)Cos (theta)

Cos (theta) = 0.8

theta = 36.87 degrees

Then using the law of sin, let the radius at tangent BC = y:

Sin(36.87) = y/x

y = 16.67 * 0.6 = 10

Therefore, the diameter of the semicircle is 10 * 2 = 20

Let the center of the semicircle be denoted by "P", which lies on AB. Drawing the radii "r" of the semicircle from P to the points of tangency on AC and BC, we get right angles. Dividing triangle ABC along the line from point C to point P, we see that [ABC] = [APC] + [BPC] = (AC)r/2 + (BC)r/2 = (17+25)r/2 = 21r. Using Heron's formula, we find that [ABC] = 210. Thus 210 = 21r, so r = 10, so the diameter is 2r = 20 .

Using Heron's formula, we find $s = (28+25+17)/2 = 35$ , and $[ABC] = \sqrt{35(35-28)(35-25)(35-17)} = 210.$ Consider $A'$ on $AC$ extended and $B'$ on $BC$ extended such that $A'B' || AB$ and $A'B'$ is tangent to the incircle of which the semicircle is part. Thus the distance between $A'B'$ and $AB$ is $r$ , the inradius, and we wish to find $2r$ . The altitude from $C$ to $AB$ is $h = 2[ABC]/(AB) = 2(210)/28 = 15$ , so by similarity of triangles, the semiperimeter of $\triangle A'B'C$ is $s' = \frac{h+r}{h} s = 35(1+r/15)$ , and the area is $[A'B'C] = \left(\frac{h+r}{h}\right)^2 [ABC] = 210(1+r/15)^2$ . But we also have $[A'B'C] = rs'$ , hence $210(1+r/15)^2 = 35(1+r/15)r,$ and since $r > 0$ , we can cancel the common factor and obtain the solution $r = 10$ , so the diameter of the semicircle is $2r = \boxed{20}$ .

Indeed, the radius of the semicircle in the general case can be expressed as $r = \frac{[ABC]}{AC+BC}.$

We draw $\triangle$ $ABC$ with $AB$ as base $\big($ to facilitate clarity $\big)$ . Let the semicircle with diameter on $AB$ touch $AC$ and $BC$ at $D$ and $E$ respectively. Let the centre of semicircle be $O$ and radius be $r$ . From geometry we have $\angle$ $OEB=\:$ $\angle$ $ODA=\:$ $90^{\circ}$ .

If $AD=\:x$ then $DC=\:CE=\:17 - x$ and $EB=\:8 + x\,.$

$\;\;\;\;\;\;\;\;$ $\big($ Lengths of tangents from an external point to a circle are equal $\big)$

Now $[ABC]=\:[ADO] + \big([ODC] + [OEC]\big) + [OEB]$

$\;\;\;\;\;\;\;\;$ $\big($ where $[PQRS]$ denotes the area of quadrilateral $PQRS\big)$

$\implies\:$ $[ABC]=\:\displaystyle{\frac{rx}{2} + r.(17 - x) + \frac{r.(8 + x)}{2}}$

$\implies\:$ $[ABC]=\:21.r$ $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ -- $(i)$

Again $[ABC]=\:\displaystyle{\frac{1}{2} \times AB \times BC \times \sin \angle ABC=\:\frac{1}{2} \times28 \times 25 \times \frac{3}{5}}$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ -- $(ii)$

$\;\;\;\;\;\;\;\;$ $\big($ Using cosine-rule in $\triangle ABC$ $\big)$

From $(i)$ and $(ii)$ we get $,$

$21.r=\:\displaystyle{\frac{1}{2} \times28 \times 25 \times \frac{3}{5}}$

$\implies\:$ $r=\:10$

So diameter = $2.r=\:\boxed{20} .$

First note that the length of the altitude from $C$ to $AB$ is $15$ because $\triangle ABC$ is composed of an $8-15-17$ and a $15-20-25$ triangle. Call the center of the semicircle $O$ . Now call the altitude from $O$ to $AC$ and $O$ to $BC$ $A'$ and $B'$ respectively. Note that $OA'=OB'=$ the radius of the semicircle. Now take $\triangle AOA'$ and $\triangle BOB'$ and join them together such that $A'$ and $B'$ are the same point. This new triangle will be called $\triangle ABO$ . Because of how $\triangle ABO$ was constructed, $\triangle ABO \sim \triangle ABC$ and the length of the altitude from $O$ to $AB$ is the length of the radius of the semicircle. Now we need to find the factor $k$ that $\triangle ABC$ was scaled by to get $\triangle ABO$ . The sum of the lengths $AC$ and $BC$ is $17+25=42$ . The sum of the lengths $AO$ and $BO$ is $28$ because it is equal to $AB$ of triangle $ABC$ . This means the scale factor $k$ is $\frac{28}{42}=\frac{2}{3}$ . Thus the radius of the semicircle is $15\times\frac{2}{3}=10$ and the diameter of the semicircle is $20$ .

Call the origin of the semicircle $O$ and the points of tangency to $BC$ and $AC$ , $D$ and $E$ respectively. Therefore the right triangles $ODB$ and $OEA$ are formed.

Given the two sides $17$ and $25$ we notice the nice Pythagorean triples $8,15,17$ and $15,20,25$ or we can drop an altitude and solve for this if it isn't apparent.

Therefore we have the $\sin A = \frac{15}{17}$ and $\sin B = \frac{3}{5}$ .

We use this fact on the right triangles $ODB$ and $OEA$ . Now, $OD = OE$ since they are both radii. Calling $OD = OE = r$ , we get $\frac{r}{AO} = \frac{15}{17}$ and $\frac{r}{OB} = \frac{3}{5}$ .

Solving the system with $AO + OB = 28$ we get $r$ to be $10$ . Doubling this for the diameter we get $\boxed{20}$

Let the points of tangency of AC and BC to the semicircle, respectively, be D and E. Also, let the center of the circle be O. Thus, OD is perpendicular to AC and OE is perpendicular to CB.

We can find the area of AOC and BOC in terms of r, the radius, since we know the height is r and the side lengths are AC = 17 and BC = 25. Adding the two areas together, we get [AOC] + [BOC] = 17r/2 + 25r/2 = 21r = [ABC].

We can also find the area of ABC using Heron's formula with side lengths 17, 25, and 28 and semiperimeter 35. We get that [ABC] = 210, so using our expression for [ABC] from earlier, 21r = 210, so r = 10 and d = 20.

Let center of the semicircle be O and let the foot of the altitude drawn from C to AB be H. Note that [ABC] = 14 CH = 21 r where r is the radius of the semicircle. Solving for CH in terms of r yields CH = $\frac{3r}{2}$ . Also we can now express AB in terms of r using Pythagorean theorem on triangles ACH and BCH. Let $\frac{3r}{2} = x$ . , $28 = \sqrt{17^2-x^2} + \sqrt{25^2-x^2}$ . After solving for x, we get that x = 15 so r = 10. Thus, the diameter of the circle or 2r = 20.