1. **Problem statement:** Find the perimeter of polygon ABCD where $m(\angle B) = m(\angle ACD) = 90^\circ$, $AB = 4$ cm, $BC = 3$ cm, and $AD = 13$ cm. 2. **Understanding the problem:** The polygon ABCD has two right angles at vertices B and C. We know three sides: $AB$, $BC$, and $AD$. We need to find the perimeter, which is the sum of all sides: $AB + BC + CD + DA$. 3. **Using the Pythagorean theorem:** Since $\angle B = 90^\circ$, triangle $ABC$ is right-angled at B. We can find $AC$: $$AC = \sqrt{AB^2 + BC^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ cm}$$ 4. **Finding side $CD$:** Since $\angle ACD = 90^\circ$, triangle $ACD$ is right-angled at C. We know $AD = 13$ cm and $AC = 5$ cm, so: $$CD = \sqrt{AD^2 - AC^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12 \text{ cm}$$ 5. **Calculating the perimeter:** Sum all sides: $$P = AB + BC + CD + DA = 4 + 3 + 12 + 13 = 32 \text{ cm}$$ **Final answer:** The perimeter of polygon ABCD is $32$ cm.