Question

for the questions below, each pair of triangles is similar. first state how you know the triangles are similar, then solve for x (and y).

1. △efg~△abc
2. △def~△dvw
3. △jkl~△jbc
4. △fgh~△fvu

11)
12)

Explanation:

Problem 7: $\triangle EFG \sim \triangle ABC$

Step 1: Identify Similarity Criterion

$\triangle EFG \sim \triangle ABC$ by AA (Angle-Angle) similarity: $\angle A = \angle E$ (marked) and $\angle B = \angle F = 90^\circ$ (right angles).

Step 2: Set Up Proportion

Corresponding sides: $AB = 8$, $EF = 36$; $BC = 12$, $FG = x$.
Proportion: $\frac{AB}{EF} = \frac{BC}{FG}$ → $\frac{8}{36} = \frac{12}{x}$.

Step 3: Solve for $x$

Cross-multiply: $8x = 36 \times 12$ → $8x = 432$ → $x = \frac{432}{8} = 54$.

Problem 8: $\triangle DEF \sim \triangle DVW$

Step 1: Identify Similarity Criterion

$\triangle DEF \sim \triangle DVW$ by AA (vertical angles at $D$ and $\angle E = \angle V$? Wait, actually, vertical angles $\angle EDF = \angle VDW$, and $\angle E = \angle V$ (alternate interior? Wait, better: corresponding angles. Let’s check sides. $DV = 4$, $DE = 8$; $VW = 5$, $EF = 3x + 1$; $DW = 6$, $DF = 12$. Wait, ratio of $DV$ to $DE$: $\frac{DV}{DE} = \frac{4}{8} = \frac{1}{2}$. So scale factor is $\frac{1}{2}$ (smaller to larger? Wait, $DV = 4$, $DE = 8$: $DE = 2 \times DV$. So $\triangle DVW$ is smaller, $\triangle DEF$ is larger. So $\frac{VW}{EF} = \frac{1}{2}$ → $\frac{5}{3x + 1} = \frac{1}{2}$? Wait, no: $DV = 4$, $DE = 8$ (so $DE = 2 \times DV$), $DW = 6$, $DF = 12$ (so $DF = 2 \times DW$). So scale factor is 2 (from $\triangle DVW$ to $\triangle DEF$). Thus, $EF = 2 \times VW$? Wait, $VW = 5$, so $EF = 2 \times 5 = 10$? No, wait $EF = 3x + 1$. Wait, let’s set proportion: $\frac{DV}{DE} = \frac{VW}{EF} = \frac{DW}{DF}$.

$DV = 4$, $DE = 8$ → ratio $\frac{4}{8} = \frac{1}{2}$.
$VW = 5$, $EF = 3x + 1$ → $\frac{5}{3x + 1} = \frac{1}{2}$ → $3x + 1 = 10$ → $3x = 9$ → $x = 3$. Wait, but $DW = 6$, $DF = 12$: $\frac{6}{12} = \frac{1}{2}$, which matches. So yes, $x = 3$.

Problem 9: $\triangle JKL \sim \triangle JBC$

Step 1: Identify Similarity Criterion

$\triangle JKL \sim \triangle JBC$ by AA: $\angle J = 41^\circ$ (common angle) and $\angle K = 39^\circ$ (wait, $\triangle JBC$: angles are $41^\circ$, $100^\circ$ (since $180 - 41 - 39 = 100$? Wait, $\triangle JKL$: $\angle J = 41^\circ$, $\angle K = 39^\circ$, so $\angle L = 100^\circ$. $\triangle JBC$: $\angle J = 41^\circ$, $\angle C = 100^\circ$, so $\angle B = 39^\circ$. So AA: $\angle J$ common, $\angle K = \angle B = 39^\circ$.

Step 2: Set Up Proportion

$JC = 5$, $JL = 25$; $JB = 10$, $JK = 8x + 2$.
Ratio: $\frac{JC}{JL} = \frac{JB}{JK}$ → $\frac{5}{25} = \frac{10}{8x + 2}$.

Step 3: Solve for $x$

Simplify $\frac{5}{25} = \frac{1}{5}$ → $\frac{1}{5} = \frac{10}{8x + 2}$ → $8x + 2 = 50$ → $8x = 48$ → $x = 6$.

Problem 10: $\triangle FGH \sim \triangle FVU$

Answer:

s:

1. $x = \boldsymbol{54}$
2. $x = \boldsymbol{3}$
3. $x = \boldsymbol{6}$
4. $x = \boldsymbol{8}$
5. $x = \boldsymbol{8}$, $y = \boldsymbol{\frac{15}{4}}$ (or $3.75$)
6. $x = \boldsymbol{60}$, $y = \boldsymbol{\frac{8}{3}}$ (or $2.67$)