Half Circle in a Triangle

Join A and O.Drop a perpendicular from O to AB and to AC,they will be equal to r.Area of ABO = $\frac{AB.r}{2}$ similarly for AOC $\frac{AC.r}{2}$ .Adding this 2 areas give the area of whole triangle which is equal to $24\sqrt{6}$ by Heron's formula.This gives area of full circle =24pi so half area=12pi.

I tried to make this solution as detailed as I could.

Let trace the line that goes from $O$ to the $\overline { BC }$ side of the triangle, forming 90º with the same. Automatically, this segment takes on the character of semicircle's radius.

Now, let trace the segment $\overline { AO }$ .

Therefore, the area of the triangle $ABC$ is the sum of the areas of triangles $AOB$ , $AOD$ and $OCD$ . Then, denoting the area a triangle $ABC$ by $\left[ ABC \right]$ , we have:

$(1) \left[ ABC \right]=\left[ AOB \right]+\left[ AOD \right]+\left[ COD \right]$

Note that the triangles $AOB$ and $AOD$ are congruent by the case of three equal sides. So, we get:

$(2) \left[ ABC \right]= 2\times\left[ AOD \right]+\left[ COD \right]$ .

Calculating the area of the triangle $ABC$ by the Hieron's formula,

$\left[ ABC \right]=\sqrt { p(p-a)(p-b)(p-c) }$ , where p is the semiperimeter of the triangle $ABC$ and $a,b,c$ the sides. We have:

= $\sqrt { 18(18-12)(18-14)(18-10) }$

So, $\left[ ABC \right]=24\sqrt{6}$ .

Denoting the radius of the semicircle by (\r), we have that he areas of the triangles (\AOD and COD) are, as a function of $r$ :

$(3) \left[ AOD \right]=\frac { r\times 10 }{ 2 }=5r$ ,and

$(4) \left[ COD \right]=\frac { r\times 4 }{ 2 }=2r.$

Comparing the equations $(2), (3)$ and $(4)$ , we get:

$\left[ ABC \right]=24\sqrt{6}=2\times5r+2r$

$24\sqrt{6}=12r$

$r = 2\sqrt{6}$

Finally, calculating the area of the circle and comparing it to $a\pi$ , we get:

$Area=\frac { \pi \times { r }^{ 2 } }{ 2 } =\frac { \pi \times { (2\sqrt { 6 } ) }^{ 2 } }{ 2 } =\frac { \pi \times 24 }{ 2 } =12\pi$

...Comparing,

$a\pi=12\pi$

So, $a=$ $\boxed { 12 }$

$\quad Q.E.D$

We can reflect Triangle $ABC$ over line $BC$ to form Quadrilateral $ABDC$ . Now label the points where the circle touches the quadrilateral, $E$ , $F$ , $B$ , and $H$ where $E$ is between $A$ and $B$ , $F$ is between $B$ and $D$ , $G$ is between $C$ and $D$ and $H$ is between $A$ and $C$ . Now since $AE$ = $AH$ , we have that $AO$ an angle bisector. After applying angle bisector theorem, we have that $BO$ = 5 and $CO$ = 7. Now we apply Stewart's theorem to find that $AO$ = $\sqrt{105}$ . Now let $AE$ be $x$ and and $BE$ be $y$ . Now we apply Pythagorean theorem on triangles $AEO$ and $BEO$ . Since both share a side length of $r$ , we can set the two equations equal to each other. Thus, we have $105 - x^{2} = 25 - y^{2}$ . This can be rearranged into $x^{2} - y^{2} = 80$ . However, we know that $AE + BE = AB$ , so $x + y = 10$ . Divided the first equation by the second yields $x - y = 8$ . Now that we have the two equations $x + y = 10$ and $x - y = 8$ , we can easily find that $x = 9$ and $y = 1$ .

Using Pythagorean theorem again, we find that $r^{2} = 24$ . The area of our semicircle is $\frac{\pi r^{2}}{2}$ so our answer is $12$ .

Thanks to Saloni Arora . It is his solution that leads to mine.

1. Let radius of semicircle R.
2. Using Cosine Law: cos B = 1/5 or sin B = (2/5)sqrt(6) ; cos C = 5/7 or sin C = (2/7)sqrt(6).
3. From the figure, cos (90 -B) = sin B = R / BO, sin C = R / OC.
4. Using algebra: BO + OC = 12 or (R / sin B) + (R / sin C) = 12, or R = 12 / sqrt(6).
5. Area of semicircle = (1/2)(pi)(R^2), A = (1/2)(R^2) = 12. (Q.E.D.)

using cosine rule i.e a^2=b^2+c^2-2(b)(c)(cos@) where @=angle opposite to line a. taking a to be the line i.e 10 we solve this equation and find @ @=Cos^-1((144+196-100)/2(14)(12) so @=44.415 than as we know that when a tangent touches a circle it makes an angle of 90 with the line made from the centre of the circle. which happens when the line with lenght 14 touches the semicircle at an imaginary point x. we form a right angle triangle OXY where Yis the point where line 14 and 12 meet. than using sine rule i.e sina/a=sinb/b like this sin(44.415)/r=sin(90)/(12-r) where r=radius of circle we simplify the equation to 8.4-0.7r=r and get r to be equal to 4.94 than using formula area of semicirle= pi r^2/2 we equate Api=r^2/2 pi and we get A=12.2.

1/2 πr² =πa r =√2a 10/14=(√2a+x)/(12-(√2a+x)), where x is difference of length of radius and gap of semicircle. Solve above equation considering x as negligibly small and calculate, a =12.5 if you consider value of x now we get ”a” less than 12.5 we get a=12 from option.

Drop the radius on AB and AC at points X and y respectively such that ∠BXO = ∠CYO= 90 degree. Let r be the radius of the circle. Now, in both these triangles BOX and CYO,

By cosine rule, cos B= (a²+c²-b²)/2ac

cos B= 1/5

sin B= 2√6/5

Also, by Pythagoras theorem, r/(12-y)= 2√6/5 (Here y is CO, thus BO= 12- y) .....eqn 1

Now in triangle CYO, by same cosine rule and pythagoras theorem,

r/y = 2√6/7 ....eqn 2

Solving eqn 1 and 2-

y=5 and r= 2√6

∏r²/2= a∏

and a= 12 (substitute r)

The solution looks much more easier if accompanied by a diagram alongwith, unfortunately we can't do that here

my way isn't very accurate but it worked

by heron's formula: area of the triangle is sqrt{18 * 8 * 4 * 6}= 24 sqrt{6}

Let's drop a perpendicular BD of height h from B on AC

drop another perpendicular OE from O on AC, it's obvious that OE= r

now triangles COE and CBD are similar then

CO/CB= OE/BD

let's assume that CO= 12-r "small part of BO on the left of the semi-circle is neglected"

CB=12

OE=r

BD= h=24/7 sqrt{6}

by substitution you get slightly larger value of r and the SA of semi-circle appears to be 12.204*pi

the closest answer is 12

Draw perpendiculars from O on side AB & AC. Say, they will intersect AB at D & AC at E. Evidently they will be the tangent points. So OD=OE=radius of the half circle.Now join A to O. Two triangles formed, AOB and AOC. Area of AOB=AB r/2 and for AOC=AC r/2. Sum of these two will be the area of ABC which is equal to root square of S (S-10)(S-12)(S-14).S=half perimetre of ABC. From this we can make an equation and find r. Now its easy to find the value of a. Ans is 12.