Question

$\overleftrightarrow{ae}$, $\overleftrightarrow{bf}$, $\overleftrightarrow{cg}$, and $\overleftrightarrow{dh}$ are parallel, with $ef = fg = gh$. complete.

1. if $ab = 5$, then $ad = \underline{\quad?\quad}$.
2. if $ac = 12$, then $cd = \underline{\quad?\quad}$.
3. if $ab = 5x$ and $bc = 2x + 12$, then $x = \underline{\quad?\quad}$.
4. if $ac = 22 - x$ and $bd = 3x - 22$, then $x = \underline{\quad?\quad}$.

b 14. if $ab = 15$, $bc = 2x - y$, and $cd = x + y$, then $x = \underline{\quad?\quad}$ and $y = \underline{\quad?\quad}$.

1. if $ab = 12$, $bc = 2x + 3y$, and $bd = 8x$, then $x = \underline{\quad?\quad}$ and $y = \underline{\quad?\quad}$.

Explanation:

Problem 10

Step1: Identify the theorem

Since \(\overrightarrow{AE}, \overrightarrow{BF}, \overrightarrow{CG}, \overrightarrow{DH}\) are parallel and \(EF = FG = GH\), by the Basic Proportionality Theorem (Thales' theorem) or the theorem of parallel lines cutting transversals proportionally, the segments on the transversal are equal. So \(AB = BC = CD\).

Step2: Calculate \(AD\)

Given \(AB = 5\), then \(BC = 5\) and \(CD = 5\). So \(AD=AB + BC+CD=5 + 5+5 = 15\).

Step1: Identify the equal segments

From the parallel lines, \(AB = BC = CD\). Let \(AB = BC = CD = x\). Then \(AC=AB + BC=2x\).

Step2: Solve for \(x\) and \(CD\)

Given \(AC = 12\), so \(2x=12\), then \(x = 6\). Since \(CD=x\), \(CD = 6\).

Step1: Set up the equation

Since \(AB = BC\) (from parallel lines), given \(AB = 5x\) and \(BC=2x + 12\), we set \(5x=2x + 12\).

Step2: Solve for \(x\)

Subtract \(2x\) from both sides: \(5x-2x=2x + 12-2x\), so \(3x=12\). Divide both sides by 3: \(x = 4\).

Answer:

\(15\)

Problem 11