1. **State the problem:** We have a figure with squares ABCD, DCEF, and FEPQ, and we want to find the sum of the lengths CAQ + EAQ + PAQ. 2. **Understand the problem:** Each of these squares shares sides with the others, and points A, Q, E, P are vertices or points related to these squares. We need to find the sum of the segments CAQ, EAQ, and PAQ. 3. **Key properties of squares:** All sides are equal in length, and all angles are right angles (90 degrees). 4. **Analyze the points:** Since the problem is about the sum of segments CAQ + EAQ + PAQ, and these segments likely represent distances along the figure, we consider the geometry of the squares and their arrangement. 5. **Assuming the squares are placed consecutively sharing sides:** - ABCD is a square. - DCEF shares side DC with ABCD. - FEPQ shares side FE with DCEF. 6. **Lengths of sides:** Let the side length of each square be $s$. 7. **Calculate each segment:** - $CAQ$ is the length from point C to A to Q. Since A and C are vertices of ABCD, and Q is a vertex of FEPQ, the path CAQ can be considered as $CA + AQ$. - $CA = s$ (side of square ABCD). - $AQ$ is the distance from A to Q, which spans across the squares. Since Q is on FEPQ, and A is on ABCD, the distance $AQ$ equals $2s$ (two side lengths). - $EAQ$ is the length from E to A to Q. Since E is on DCEF, A on ABCD, and Q on FEPQ, the path EAQ is $EA + AQ$. - $EA = s$ (side of square DCEF). - $AQ = 2s$ as above. - $PAQ$ is the length from P to A to Q. P and Q are vertices of FEPQ, and A is on ABCD. - $PA = s$ (side of square FEPQ). - $AQ = 2s$ as above. 8. **Sum all segments:** $$CAQ + EAQ + PAQ = (CA + AQ) + (EA + AQ) + (PA + AQ) = (s + 2s) + (s + 2s) + (s + 2s) = 3s + 6s = 9s$$ 9. **Final answer:** The sum $CAQ + EAQ + PAQ$ equals $9s$, where $s$ is the side length of each square.