Question

e)
side lengths: $x + 6$, $x$, 30, 9
f)
side lengths: $y$, $x$, 6, 12, 10, 8

Explanation:

Step1: Solve for x (Part E)

By the triangle midsegment theorem, the ratio of the smaller triangle side to the larger triangle side equals the ratio of the segment lengths.
$$\frac{x}{x+6} = \frac{9}{30}$$
Simplify $\frac{9}{30}$ to $\frac{3}{10}$, so:
$$\frac{x}{x+6} = \frac{3}{10}$$
Cross-multiply:
$$10x = 3(x+6)$$
$$10x = 3x + 18$$
Subtract $3x$ from both sides:
$$7x = 18$$
$$x = \frac{18}{7}$$

Step2: Solve for x (Part F)

By the triangle midsegment theorem, set up the proportion:
$$\frac{12}{x} = \frac{6}{8}$$
Simplify $\frac{6}{8}$ to $\frac{3}{4}$, so:
$$\frac{12}{x} = \frac{3}{4}$$
Cross-multiply:
$$3x = 48$$
$$x = 16$$

Step3: Solve for y (Part F)

Set up the proportion for the other side:
$$\frac{y}{12} = \frac{10}{8}$$
Simplify $\frac{10}{8}$ to $\frac{5}{4}$, so:
$$\frac{y}{12} = \frac{5}{4}$$
Cross-multiply:
$$4y = 60$$
$$y = 15$$

Answer:

For part E: $x = \frac{18}{7}$
For part F: $x = 16$, $y = 15$