Not your average shuriken

$AB=2x=\frac{14}{\cos{(2\theta})}=\frac{14}{1-2\sin^2\theta}=\frac{14}{1-\frac{2\times{7^2}}{11^2+7^2}}=\frac{595}{18}$

solve for x and y.

Just for sake of variety ....

Let $O$ be the center of the circle and $T$ be the point of tangency of the "top" right triangle with the circle. Also, let the lower-left vertex of this triangle be $Q$ and its uppermost vertex be $P.$

Then $\Delta PQB$ and $\Delta OTB$ are similar triangles. Now let $|OB| = x$ and $|TB| = y.$ Since $|PT| = |PQ| = 11$ and $|QO| = 7,$ we have by similarity that

$\dfrac{|PQ|}{|QB|} = \dfrac{|OT|}{|TB|} \Longrightarrow \dfrac{11}{7 + x} = \dfrac{7}{y} \Longrightarrow 11y - 7x = 49.$ (i),

and also that

$\dfrac{|PQ|}{|PB|} = \dfrac{|OT|}{|OB|} \Longrightarrow \dfrac{11}{11 + y} = \dfrac{7}{x} \Longrightarrow 11x - 7y = 77,$ (ii).

Multiplying (i) by $7$ and (ii) by $11$ and adding the resulting equations gives us that

$-49x + 121x = 7*49 + 11*77 \Longrightarrow x = \dfrac{1190}{72} = \dfrac{595}{36}.$

Now $|AB| = 2x = \dfrac{595}{18},$ so $ab = 595*18 = \boxed{10710}.$

Let point $C,D$ be the intersections of line $\overline{AB}$ through the circle from left to right. Let point $E$ be the other vertex of the right triangle on the top semicircle as shown. $\overline{BE}$ is the tangent line that intersects with the circle at point $F$ .

Since both $\overline{EC}$ and $\overline{EF}$ are tangents to the circle, $EC=EF=11.$ Let $BD = x$ and $BF=y$ , by the power of the point theorem, we have

$y^2=x(x+14) \implies y^2-x^2=14x.$ (1)

According the Pythagorean Theorem, $11^2+(x+14)^2 = (y+11)^2.$

Regroup the equation and substitute (1) we have $121 + 22y = 14x+317.$

Therefore, $y=\dfrac{7(x+14)}{11}.$ (2)

Substitute (2) into (1),

$121(x^2+14x)=49(x^2+28x+196), \implies x=\dfrac{343}{36}.$

Thus, $AB= 2\times \dfrac{343}{36} + 14 = \dfrac{595}{18} = \dfrac ab.$

We have $ab = 595 \times 18 = \boxed{10710}.$

Since the triangles are similar you can look at the figure like this.

AB = 14 + 2AC. We can use the formula r = Area of triangle (A)/semi-perimeter of triangle (s). Since you're given the radius (7) and the base (22), you can get the value of h. h = 935/36.

Using Pythagorean theorem. You can get the value of the other leg. The value of the other leg is 847/36.

AC = 847/36 - 14

AC = 343/36

AB = 14 + 2(343/36)

AB = 595/18

ab = 595*18 = 10710

$AB=2*(Long~ leg)~-~Diameter=2*11*Tan\{2*Tan^{-1}(7/11)\}~-~14=22*\dfrac{2*\frac 7{11} }{1-(\frac 7{11})^2}~-~14=\dfrac{847}{18}~-~14=\dfrac{595}{18}=\dfrac a b.\\ \therefore~a*b=10710.$

Firstly, $BD=y$ and $BG=x$

Since $CE$ and $DE$ are tangents, $CE=ED=11$

$CF=FG=\frac{14}{2}=7$ since $BD$ is a tangent.

Now applying Pythagorean Theorem to the right triangles $FDB$ and $ECB$ , we have the following pair of equations:

$\boxed{7^{2}+y^{2} = (7+x)^{2}}$ and $\boxed{11^{2} + (14+x)^{2} = (11+y)^{2}}$

Solving for $x$ , we come across a quadratic equation $36x^{2}+161x-4802=0$ . The only legitimate value of $x$ happens to be $\frac{343}{36}$ .

Now, we have $AB=14+2x=\boxed{\frac{595}{18}}$

Multiplying their numerator and denominator, we have $595 \times 18 = \boxed{10710}$

I hope it helped. :)

Connect center of circle $O$ to a point $C$ on the point of tangency of the upper right triangle.

Connect these to points to get $OC$ , the radius, which is $7$ .

This forms 2 right triangles that are similar by $AA$ .

Call $CB=x$

We set up a proportion: $\frac{hyp}{hyp}=\frac{leg}{leg}$

The hypotenuse of the smaller right triangle is $\sqrt{x^2+7^2}$

Substitute the known values into the proportion: $\frac{\sqrt{x^2+7^2}}{11+x}=\frac{7}{11}$

Solve this equation to get $x=\frac{539}{36}$

Thus $AB=2(\sqrt{(\frac{539}{36})^2+7^2}-7)+14$ (from the Pythagorean theorem for the smaller triangle)

$AB=\boxed{\frac{595}{18}}$

Firstly, notice that $CE$ and $BE$ are both tangents to the circle. This means that angles $\angle BEO$ and $\angle OEC$ are the same. $\tan(\angle OEC) = 7/11$ so $tan(\angle BEC) = \tan(2\angle OEC) \\ = \frac{2\frac{7}{11}}{1 - (\frac{7}{11})^2 } \\ = \frac{77}{36}$

From $\tan(\angle BEC) = \frac{CB}{11}$ , $CB = 11.\frac{77}{36} = \frac{847}{36}$ As the diagram is symmetrical and $DB = CB - CD= \frac{847}{36} - 14 = \frac{343}{36}$ $AB = AC + CD + DB = \frac{343}{36} + 14+ \frac{343}{36} \\ = \boxed{\frac{595}{18}}$

Multiplying out gives the answer.