9) A quadrilateral $A B C D$ has perpendicular diagonals and is inscribed in a circle $c$ with diameter $A C$. The area and perimeter of the quadrilateral are $48 \mathrm{~cm}^{2}$ and $28 \mathrm{~cm}$, respectively. What is the radius of the circle $c$?
(A) $4 \mathrm{~cm}$,
(B) $5 \mathrm{~cm}$,
(C) $6 \mathrm{~cm}$,
(D) $7 \mathrm{~cm}$,
(E) $8 \mathrm{~cm}$.
9. The answer is $(\mathbf{B})$.

We observe that triangles $A B C$ and $A C D$ are right-angled because they are inscribed in a semicircle. Let $a$ and $b$ be the lengths of sides $A B$ and $B C$ respectively. The area of quadrilateral $A B C D$ is the sum of the areas of triangles $A B C$ and $A C D$, so

$$
\frac{a b}{2}+\frac{a b}{2}=a b=48 \mathrm{~cm}^{2}
$$

![](https://cdn.mathpix.com/cropped/2024_04_17_1da74e1307acd0f8cf0eg-3.jpg?height=520&width=582&top_left_y=794&top_left_x=794)

Furthermore,

$$
2 a+2 b=28 \mathrm{~cm} \Rightarrow a+b=14 \mathrm{~cm} .
$$

Thus, we know the sum and product of $a$ and $b$, and we know that they are the solutions to the quadratic equation

$$
x^{2}-14 x+48=0,
$$

from which we obtain $a=8 \mathrm{~cm}$ and $b=6 \mathrm{~cm}$ or $a=6 \mathrm{~cm}$ and $b=8 \mathrm{~cm}$. In either of these two cases, we have

$$
\overline{A C}=\sqrt{6^{2}+8^{2}} \mathrm{~cm}=10 \mathrm{~cm}
$$

and thus the radius of $c$ measures $5 \mathrm{~cm}$.