1. **Problem Statement:** Find the perimeter of trapezoid ABCD where AB is parallel to DC. Given lengths: AB = 5 cm, AD = 13 cm, DC = 35 cm, and BC is the hypotenuse of right triangle BCD. 2. **Understanding the problem:** The perimeter of polygon ABCD is the sum of all its sides: $$P = AB + BC + CD + DA$$ We know AB, DC, and AD, but need to find BC. 3. **Finding BC:** Since BC is the hypotenuse of right triangle BCD, we use the Pythagorean theorem: $$BC = \sqrt{BD^2 + CD^2}$$ We know CD = 35 cm, but BD is unknown. 4. **Finding BD:** Since AB is parallel to DC, trapezoid ABCD has right angles at B and C (assuming from the right triangle BCD). Given AD = 13 cm and AB = 5 cm, and the trapezoid shape, BD can be found by considering triangle ABD or using coordinate geometry. 5. **Using coordinates:** Place point D at origin (0,0), C at (35,0) since DC = 35 cm. Since AB is parallel to DC, AB is horizontal. Point A is at (x,y), point B at (x+5,y) because AB = 5 cm. AD = 13 cm means distance from A to D is 13: $$AD = \sqrt{x^2 + y^2} = 13$$ 6. **Finding coordinates of A and B:** Let A = (x,y), then B = (x+5,y). 7. **Finding BD:** Point B = (x+5,y), D = (0,0), so $$BD = \sqrt{(x+5)^2 + y^2}$$ 8. **Using right triangle BCD:** Triangle BCD is right angled at C, so: $$BC^2 = BD^2 + CD^2$$ 9. **Express BC:** $$BC = \sqrt{BD^2 + CD^2} = \sqrt{(x+5)^2 + y^2 + 35^2}$$ 10. **Using AD = 13:** $$x^2 + y^2 = 169$$ 11. **Express BC in terms of x and y:** $$BC = \sqrt{(x+5)^2 + y^2 + 1225}$$ 12. **Simplify BC:** $$BC = \sqrt{x^2 + 10x + 25 + y^2 + 1225} = \sqrt{(x^2 + y^2) + 10x + 1250}$$ 13. **Substitute $x^2 + y^2 = 169$:** $$BC = \sqrt{169 + 10x + 1250} = \sqrt{1419 + 10x}$$ 14. **Find x:** Since AB is parallel to DC and trapezoid, the height y is constant for A and B. Assuming B lies vertically above C (35,0), then B's x-coordinate is 35. But B's x-coordinate is $x+5$, so: $$x + 5 = 35 \Rightarrow x = 30$$ 15. **Calculate BC:** $$BC = \sqrt{1419 + 10 \times 30} = \sqrt{1419 + 300} = \sqrt{1719}$$ 16. **Calculate perimeter:** $$P = AB + BC + CD + DA = 5 + \sqrt{1719} + 35 + 13$$ 17. **Approximate $\sqrt{1719}$:** $$\sqrt{1719} \approx 41.47$$ 18. **Final perimeter:** $$P \approx 5 + 41.47 + 35 + 13 = 94.47 \text{ cm}$$ **Answer:** The perimeter of trapezoid ABCD is approximately 94.47 cm.