, 2026

The area of $DMBN$ is equal to
the sum of the areas of triangles $DMB$ and $DNB$, and so we will first find these two
areas.

Each of $DA$, $DB$ and $DC$ is a radius of the circle, and so
$DA=DB=DC=5$.

Since $\triangle DAB$ is an isosceles
triangle and $M$ is the midpoint of
$AB$, then $DM$ is perpendicular to $AB$ ($DM$ is the height of $\triangle DAB$).

Using the Pythagorean Theorem in $\$\triangle$
DMB$, we get$DB^2=DM^2+MB^2$
and since $\$MB=\dfrac{AB}{2}=2\text{}$
cm}$, then$5^2=DM^2+2^2$.

Solving for $DM$, we get $DM^2=25-4=21$, and so $DM=\sqrt{21}\text{ cm}$ (since $DM>0$).

Therefore, the area of $\$\triangle$
DMB$is$$\dfrac12\times MB\times DM=\dfrac12\times 2\text{ cm}\times \sqrt{21}\text{ cm}=\sqrt{21}\text{}$
cm}^2$.

We can similarly determine the area of $\triangle DNB$.

Since $\$NB=\dfrac12\times BC=3\text{}$
cm}$, then$5^2=DN^2+3^2$.

Solving for $DN$, we get $DN^2=25-9=16$, and so $DN=\sqrt{16}=4\text{ cm}$ (since $DN>0$).

Therefore, the area of $\$\triangle$
DNB$is$$\dfrac12\times NB\times DN=\dfrac12\times 3\text{ cm}\times 4\text{}$ cm}=6 $\text{}$
cm}^2$.

Adding the two areas together, the area of $DMBN$ is $\sqrt{21}\text{ cm}^2+6\text{ cm}^2$,
which is $10.6\text{ cm}^2$ when
rounded to one decimal place.