geometry unit 7 retake
1. what is the scale factor of figure a to figure b?
2 points
2. if the scale factor of figure a to figure b is 2:3, find the value of x.
1 point
3. if $ \triangle cde\sim\triangle fge $, find the value of x.
3 points
4. if $ \triangle hbn\sim\triangle lyr $, find the value of x.
4 points
5. if $ \triangle pml\sim\triangle trq $, find lm.
5 points
for questions 6 - 8, determine how (if possible) the triangles can be proved similar.
3 points
9. which of the follow triangles is similar to $ \triangle pqr $?
1 point

Question

Accepted Answer

### Question 1
# Explanation:
## Step1: Identify corresponding sides
Corresponding sides of similar figures are $10$ (Figure A) and $4$ (Figure B).
## Step2: Calculate scale factor
Scale factor $=\frac{\text{side of A}}{\text{side of B}}=\frac{10}{4}=\frac{5}{2}$

# Answer:
$5:2$

### Question 2
# Explanation:
## Step1: Set up proportion
Since scale factor $A:B = 2:3$, and corresponding sides $x$ (A) and $24$ (B). Proportion is $\frac{x}{24}=\frac{2}{3}$
## Step2: Solve for $x$
Cross - multiply: $3x = 2\times24$, $3x=48$, $x = 16$

# Answer:
$x = 16$

### Question 3
# Explanation:
## Step1: Use similarity of $\triangle CDE\sim\triangle FGE$
Corresponding sides: $\frac{CD}{FG}=\frac{DE}{GE}$. Here $CD = 27$, $FG = 40$, $DE = 25$, $GE=x$
Proportion: $\frac{27}{40}=\frac{25}{x}$
## Step2: Solve for $x$
Cross - multiply: $27x=25\times40$, $27x = 1000$, $x=\frac{1000}{27}\approx37.04$

# Answer:
$x=\frac{1000}{27}$

### Question 4
# Explanation:
## Step1: Use $\triangle HBN\sim\triangle LYR$
Corresponding sides: $\frac{HB}{LY}=\frac{HN}{LR}$. $HB = 28$, $LY=x + 11$, $HN=3x-7$, $LR = 21$
Proportion: $\frac{28}{x + 11}=\frac{3x-7}{21}$
## Step2: Cross - multiply
$28\times21=(3x - 7)(x + 11)$
$588=3x^{2}+33x-7x-77$
$3x^{2}+26x-665 = 0$
Using quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ where $a = 3$, $b = 26$, $c=-665$
$x=\frac{-26\pm\sqrt{26^{2}-4\times3\times(-665)}}{2\times3}=\frac{-26\pm\sqrt{676+7980}}{6}=\frac{-26\pm\sqrt{8656}}{6}=\frac{-26\pm93.04}{6}$
We take the positive root $x=\frac{-26 + 93.04}{6}=\frac{67.04}{6}\approx11.17$ (Another root is negative, so we discard it as side length cannot be negative)

# Answer:
$x\approx11.17$

### Question 5
# Explanation:
## Step1: Use $\triangle PML\sim\triangle TRQ$
Corresponding sides: $\frac{PM}{TR}=\frac{PL}{TQ}$. $PM = 35$, $TR = 14$, $PL=7x-9$, $TQ=2x + 2$
Proportion: $\frac{35}{14}=\frac{7x-9}{2x + 2}$
## Step2: Simplify $\frac{35}{14}=\frac{5}{2}$
So $\frac{5}{2}=\frac{7x-9}{2x + 2}$
Cross - multiply: $5(2x + 2)=2(7x-9)$
$10x+10 = 14x-18$
$14x-10x=10 + 18$
$4x=28$, $x = 7$
## Step3: Find $LM$
Since $PM = 35$ and $x = 7$, $LM$ is corresponding to $RQ$ (not needed as $PM$ is given. But if we assume $LM$ is related to $PM$ (assuming proper similarity ratios), but if we just consider $x = 7$ and $PM = 35$ (as per problem statement maybe a mis - label, but if we go back to proportion $\frac{PM}{TR}=\frac{LM}{QR}$ (assuming full similarity setup). But if we just consider the value of $x$ from $\frac{35}{14}=\frac{7x-9}{2x + 2}$ gives $x = 7$. Then $LM$ (if $PM = 35$ and scale factor $\frac{35}{14}=\frac{5}{2}$, but no $QR$ given. But if we assume $LM$ is related to $PM$ (maybe problem has typo, but if we go with $x = 7$)

# Answer:
$x = 7$, $LM$ (assuming proper setup with $x = 7$ and if $LM$ is $35$ (if it's a typo and $PM$ is $LM$) but more accurately $x = 7$

### Question 6
# Explanation:
## Step1: Check ratios
$\frac{8}{9}=\frac{20.8}{23.4}$ (since $8\times23.4=187.2$ and $9\times20.8 = 187.2$) and included angle is common. So $SAS\sim$

# Answer:
C. $SAS\sim$

### Question 7
# Explanation:
## Step1: Check ratios
$\frac{24}{36}=\frac{20}{30}
eq\frac{42}{...}$ (ratios not equal). So not similar

# Answer:
D. Not Similar

### Question 8
# Explanation:
## Step1: Vertical angles are equal
Two triangles have vertical angles (equal) and another pair of angles (by transversal or parallel lines assumption, but with vertical angles equal, $AA\sim$ (if two angles of one triangle equal to two angles of another triangle))

# Answer:
A. $AA\sim$

### Question 9
# Explanation:
## Step1: Find ratio for $\triangle PQR$
$\triangle PQR$ has sides $18$ and $30$. Check ratios for options.
For option A: $\frac{12}{18}=\frac{2}{3}$, $\frac{15}{30}=\frac{1}{2}$ (not equal)
For option B: $\frac{9}{18}=\frac{1}{2}$, $\frac{12}{30}=\frac{2}{5}$ (not equal)
For option C: $\frac{6}{18}=\frac{1}{3}$, $\frac{10}{30}=\frac{1}{3}$ (but need two pairs. If assume included angle (but no info). Wait, $\triangle PQR$ has sides $18$ (say $PR$) and $30$ (say $PQ$). If we check $\frac{6}{18}=\frac{10}{30}=\frac{1}{3}$ (if two sides in proportion and included angle (common angle in $\triangle PQR$ and option C (if angle is same). But more accurately, if we assume $SAS$ (if angle is same). But if we check $\frac{18}{6}=\frac{30}{10}=3$ (if sides are in proportion and included angle (the angle at $P$ in $\triangle PQR$ and corresponding angle in option C (if it's same)). So option C

# Answer:
C.

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QUESTION IMAGE

Question

Explanation:

Question 1

Step1: Identify corresponding sides

Step2: Calculate scale factor

Step1: Set up proportion

Step2: Solve for \(x\)

Step1: Use similarity of \(\triangle CDE\sim\triangle FGE\)

Step2: Solve for \(x\)

Answer:

Question 2