Question

points a, b, e, and f are the midpoints of xc, xd, \overline{yc}, and \overline{yd}. complete.

1. if cd = 24, then ab = ? and ef = ?.
2. if ab = k, then cd = ? and ef = ?.
3. if ab = 5x - 8 and ef = 3x, then x = ?.
4. if cd = 8x and ab = 3x + 2, then x = ?.

Explanation:

Step1: Recall Midline Theorem

In a triangle, the segment connecting the midpoints of two sides is parallel to the third side and half its length. For \(\triangle XCD\), \(A\) and \(B\) are midpoints of \(XC\) and \(XD\), so \(AB\) is the midline. Thus, \(AB=\frac{1}{2}CD\). For \(\triangle YCD\), \(E\) and \(F\) are midpoints of \(YC\) and \(YD\), so \(EF\) is the midline. Thus, \(EF = \frac{1}{2}CD\), so \(AB=EF\).

Step1 (Problem 1): Calculate \(AB\) and \(EF\)

Given \(CD = 24\). By midline theorem, \(AB=\frac{1}{2}CD=\frac{1}{2}\times24 = 12\), \(EF=\frac{1}{2}CD=\frac{1}{2}\times24=12\).

Step2 (Problem 2): Calculate \(CD\) and \(EF\)

Given \(AB = k\). Since \(AB=\frac{1}{2}CD\), then \(CD = 2AB=2k\). And \(EF = AB=k\) (as \(AB = EF\) from midline theorem).

Step3 (Problem 3): Solve for \(x\)

Given \(AB = 5x - 8\) and \(EF = 3x\). Since \(AB = EF\), set \(5x - 8=3x\). Subtract \(3x\) from both sides: \(2x - 8 = 0\). Add 8 to both sides: \(2x=8\). Divide by 2: \(x = 4\).

Step4 (Problem 4): Solve for \(x\)

Given \(CD = 8x\) and \(AB = 3x + 2\). Since \(AB=\frac{1}{2}CD\), substitute \(CD = 8x\) into \(AB=\frac{1}{2}CD\): \(3x + 2=\frac{1}{2}\times8x\). Simplify right side: \(\frac{1}{2}\times8x = 4x\). So equation is \(3x + 2=4x\). Subtract \(3x\) from both sides: \(2=x\), so \(x = 2\).

Answer:

1. \(AB = \boldsymbol{12}\), \(EF = \boldsymbol{12}\)
2. \(CD = \boldsymbol{2k}\), \(EF = \boldsymbol{k}\)
3. \(x = \boldsymbol{4}\)
4. \(x = \boldsymbol{2}\)