Free Test

We are given that $AC=CD=6$ and $BD=14$ , and we infer from the semicircle that $AO=BO=CO=DO=r$ .

Let F be the midpoint of BD. Then, since $\triangle BOD$ is isosceles, $\angle OFD$ is right. Similarly, E is the midpoint of AD by symmetry and $\triangle AOD$ is isosceles, so $\angle OED$ is right. Thirdly, $\angle ADB$ is right, since it is subtended by the diameter of a circle.

Hence, $OFDE$ is a rectangle, which implies $EO=DF=\frac{14}2=7$ .

Now, consider the right triangle $\triangle ODE$ . By the Pythagorean theorem, $\begin{aligned}ED^2+EO^2&=DO^2\\ \implies ED^2+7^2&=r^2\\ \implies ED^2&=r^2-7^2\end{aligned}$ Now consider the right triangle $\triangle CDE$ . Notice that $CE=CO-EO=r-7$ . By the Pythagorean theorem again, $\begin{aligned}ED^2+CE^2&=CD^2\\ \implies r^2-7^2+(r-7)^2&=6^2\\ \implies r^2-49+r^2-14r+49&=36\\ \implies 2r^2-14r-36&=0\\ \implies 2(r-9)(r+2)&=0\\ \implies r=9&,r=-2\end{aligned}$ Since $r$ cannot be negative, the solution is $r=9$ .

Let the center of the circle be $O$ and the four vertices of the given quadrilateral be $A, B, C, D$ respectively with $\overline {AD}$ as diameter. Let the angle $\angle {AOB}$ be $α$ . Then $\angle {COD}$ is $π-2α$ . Using Cosine Rules to the triangles $\triangle {AOB}$ and $\triangle {COD}$ we get $\cos α=1-\dfrac{18}{r^2}=\dfrac{7}{r}$ , where $r$ is the radius of the circle. Solving we get $r=\boxed 9$

Let the angle extended at the center of the semicircle by the side of quadrilateral with length 6 be $\theta$ . By cosine rule we have:

$\begin{cases} r^2 + r^2 - 2r^2 \cos \theta = 6^2 & \implies r^2(1-\cos \theta) = 18 & ...(1) \\ r^2 + r^2 - 2r^2 \cos (\pi - 2\theta)) = 14^2 & \implies r^2(1-\cos (\pi - 2\theta) = 98 & ...(2) \end{cases}$

$\begin{aligned} \frac {(1)}{(2)}: \quad \frac {1-\cos \theta}{1\blue{-\cos (\pi-2\theta)}} & = \frac 9{49} & \small \blue{\text{Note that }\cos (\pi - \phi) = - \cos \phi} \\ \frac {1-\cos \theta}{1\blue{+\cos 2\theta}} & = \frac 9{49} & \small \blue{\text{and }\cos 2\phi = 2\cos^2 \phi-1} \\ \frac {1-\cos \theta}{\blue{2\cos^2 \theta}} & = \frac 9{49} & \small \blue{\text{Rearrange}} \\ 18\cos^2 \theta + 49\cos \theta - 49 & = 0 \\ (9\cos \theta - 7)(2\cos \theta + 7) & = 0 \\ \implies \cos \theta & = \frac 79 & \small \blue{\text{Since }\cos \theta > 0} \end{aligned}$

From $(1): \ r^2\left(1-\frac 79\right) = 18 \implies r = \boxed 9$ .