, 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.