1. **Problem Statement:** Find the midsegment length of trapezoids and solve for $x$ using the midsegment theorem. 2. **Midsegment Theorem:** The midsegment length $m$ of a trapezoid is the average of the lengths of the two bases: $$m = \frac{\text{base}_1 + \text{base}_2}{2}$$ 3. **Step-by-step solutions for part 1:** **a)** Given bases $XY = 12$ m and $ZW = 18$ m, find $MN$: $$MN = \frac{12 + 18}{2} = \frac{30}{2} = 15 \text{ m}$$ **b)** Given bases $QR = 6.5$ cm and $UT = 12.5$ cm, find $VS$: $$VS = \frac{6.5 + 12.5}{2} = \frac{19}{2} = 9.5 \text{ cm}$$ **c)** Given bases $AB = 34$ yd and $ED = 48$ yd, find $FC$: $$FC = \frac{34 + 48}{2} = \frac{82}{2} = 41 \text{ yd}$$ **d)** Given bases $ZW = 8$ yd and $QR = 26$ yd, find $TS$: $$TS = \frac{8 + 26}{2} = \frac{34}{2} = 17 \text{ yd}$$ **e)** Given bases $MF = 18.8$ ft and $VB = 16.6$ ft, find $DC$: $$DC = \frac{18.8 + 16.6}{2} = \frac{35.4}{2} = 17.7 \text{ ft}$$ 4. **Step-by-step solutions for part 2 (solve for $x$):** **a)** Given top base $= 10$ yd, midsegment $= 2x + 10$ yd, bottom base $= 24$ yd: $$2x + 10 = \frac{10 + 24}{2}$$ $$2x + 10 = \frac{34}{2} = 17$$ Subtract 10 from both sides: $$2x + \cancel{10} - \cancel{10} = 17 - 10$$ $$2x = 7$$ Divide both sides by 2: $$\frac{\cancel{2}x}{\cancel{2}} = \frac{7}{2}$$ $$x = 3.5$$ **b)** Given top base $= 61$ ft, midsegment $= 5x + 11$ ft, bottom base $= 11$ ft: $$5x + 11 = \frac{61 + 11}{2}$$ $$5x + 11 = \frac{72}{2} = 36$$ Subtract 11 from both sides: $$5x + \cancel{11} - \cancel{11} = 36 - 11$$ $$5x = 25$$ Divide both sides by 5: $$\frac{\cancel{5}x}{\cancel{5}} = \frac{25}{5}$$ $$x = 5$$ **c)** Given top base $= 22$ m, midsegment $= x + 12$ m, bottom base $= 48$ m: $$x + 12 = \frac{22 + 48}{2}$$ $$x + 12 = \frac{70}{2} = 35$$ Subtract 12 from both sides: $$x + \cancel{12} - \cancel{12} = 35 - 12$$ $$x = 23$$ **Final answers:** 1a) $MN = 15$ m 1b) $VS = 9.5$ cm 1c) $FC = 41$ yd 1d) $TS = 17$ yd 1e) $DC = 17.7$ ft 2a) $x = 3.5$ 2b) $x = 5$ 2c) $x = 23$