Question

find xy in each trapezoid.
11.
12.
13.

Explanation:

Step1: Identify the Trapezoid Type (Problem 11)

This is a trapezoid with a midline (XY) connecting the midpoints of the non-parallel sides. The formula for the midline (or median) of a trapezoid is the average of the two parallel sides. Wait, but here the parallel sides (BF and CE? Wait, no, looking at the markings: BF is length 5, CE is length 2.5? Wait, no, actually, the sides: BF and CE? Wait, the figure has BF (length 5) and CE (length 2.5), and XY is the midline. Wait, the midline of a trapezoid (when it's a midsegment) should be the average of the two bases. Wait, but maybe it's a trapezoid with the two parallel sides being BF (length 5) and CE (length 2.5)? Wait, no, the midline (XY) in a trapezoid is the average of the two parallel sides. Wait, but the markings: BX = XC, FY = YE, so XY is the midline. So the formula is \( XY = \frac{BF + CE}{2} \). Wait, BF is 5, CE is 2.5? Wait, no, maybe BF and CE are the two bases? Wait, BF is 5, CE is 2.5? Then \( XY = \frac{5 + 2.5}{2} = 3.75 \)? Wait, no, maybe I got the sides wrong. Wait, the other sides: BF is 5, FE? Wait, no, the figure: B, X, C; F, Y, E. So BF is vertical (length 5), CE is vertical (length 2.5), and XY is horizontal. So the two parallel sides are BF (length 5) and CE (length 2.5), so the midline XY is the average. So \( XY = \frac{5 + 2.5}{2} = 3.75 \)? Wait, but maybe it's a different trapezoid. Wait, problem 11: Find XY in the trapezoid. The markings: BX = XC, FY = YE, so XY is the midline. The two parallel sides (bases) are BF (length 5) and CE (length 2.5). So midline formula: \( XY = \frac{BF + CE}{2} = \frac{5 + 2.5}{2} = 3.75 \). Wait, but maybe I made a mistake. Wait, another approach: in a trapezoid, the midsegment (XY) is the average of the two bases. So if the two bases are 5 and 2.5, then XY is (5 + 2.5)/2 = 3.75. Wait, but let's check problem 12: Trapezoid HILM, with HI = 16, ML = 8, and XY is the midline (since HX = XM, IY = YL). So midline formula: \( XY = \frac{HI + ML}{2} = \frac{16 + 8}{2} = 12 \). Ah, that makes sense. So problem 12: HI is 16, ML is 8, so XY = (16 + 8)/2 = 12. Problem 13: ST = 15, VU = 6, so XY is the midline, so \( XY = \frac{ST + VU}{2} = \frac{15 + 6}{2} = 10.5 \). Wait, but the question is problem 11? Wait, the user's question: "Find XY in each trapezoid." Let's take problem 11 first. Wait, problem 11: the figure has BF = 5, CE = 2.5, and XY is the midline. So using the midline formula: \( XY = \frac{BF + CE}{2} = \frac{5 + 2.5}{2} = 3.75 \). Wait, but maybe BF and CE are not the bases. Wait, the other sides: BF is 5, FE? Wait, the markings: BX = XC, FY = YE, so XY is the midline, so it should be the average of the two parallel sides. So if BF and CE are the two parallel sides (lengths 5 and 2.5), then XY is (5 + 2.5)/2 = 3.75. Alternatively, maybe BF and CE are the legs? No, the legs are the non-parallel sides. Wait, maybe the two bases are BF (length 5) and CE (length 2.5), so XY is the midline. So the answer for problem 11 is 3.75? Wait, but let's confirm with problem 12: HI = 16, ML = 8, midline XY = (16 + 8)/2 = 12. That's correct. Problem 13: ST = 15, VU = 6, midline XY = (15 + 6)/2 = 10.5. So for problem 11, the two bases are 5 and 2.5, so XY = (5 + 2.5)/2 = 3.75.

Step2: Calculate XY (Problem 11)

Using the midline formula for a trapezoid: \( XY = \frac{\text{Base}_1 + \text{Base}_2}{2} \). Here, Base₁ = 5, Base₂ = 2.5. So \( XY = \frac{5 + 2.5}{2} = \frac{7.5}{2} = 3.75 \).

Step3: Calculate XY (Problem 12)

For trapezoid HILM, HI = 16 (top base), ML = 8 (bottom base). Midline XY: \( XY = \frac{16 + 8}{2} = \frac{24}{2} = 12 \).

Step4: Calculate XY (Problem 13)

For trapezoid STUV, ST = 15 (top base), VU = 6 (bottom base). Midline XY: \( XY = \frac{15 + 6}{2} = \frac{21}{2} = 10.5 \).

Answer:

For problem 11: \( \boxed{3.75} \)
For problem 12: \( \boxed{12} \)
For problem 13: \( \boxed{10.5} \)