SOLUTION: 13.1 Similar Figures Practice Solutions - Studypool - Free Printable

Let's go through the practice problems on similar figures, step by step, and verify the answers with explanations.

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Problem 1: Similar Quadrilaterals

Given:
$ABCD \sim LMNO$

This means quadrilateral $ABCD$ is similar to quadrilateral $LMNO$, so their corresponding angles are equal and sides are proportional.

#### a. $\angle L$ corresponds to ___

Since the order of the vertices matters in similarity notation:

- $A \leftrightarrow L$
- $B \leftrightarrow M$
- $C \leftrightarrow N$
- $D \leftrightarrow O$

So, $\angle L$ corresponds to $\angle A$. ✔

> Answer: $\angle A$

#### b. $\overline{DC}$ corresponds to ___

$\overline{DC}$ connects $D$ to $C$. From the correspondence:
- $D \leftrightarrow O$
- $C \leftrightarrow N$

So $\overline{DC} \leftrightarrow \overline{ON}$

> Answer: $\overline{ON}$ ✔

#### c. $\angle C$ corresponds to ___

$C \leftrightarrow N$, so $\angle C \leftrightarrow \angle N$

> Answer: $\angle N$ ✔

#### d. $\overline{NM}$ corresponds to ___

$\overline{NM}$ connects $N$ to $M$. From correspondence:
- $N \leftrightarrow C$
- $M \leftrightarrow B$

So $\overline{NM} \leftrightarrow \overline{CB}$

> Answer: $\overline{CB}$ ✔

✔ All answers for #1 are correct.

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Problem 2: Similar Triangles

Given:
$ANT \sim FLY$

So:
- $A \leftrightarrow F$
- $N \leftrightarrow L$
- $T \leftrightarrow Y$

#### a. $\angle N$ corresponds to ___

$N \leftrightarrow L$, so $\angle N \leftrightarrow \angle L$

> Answer: $\angle L$ ✔

#### b. $\overline{NA}$ corresponds to ___

$\overline{NA}$ goes from $N$ to $A$. Corresponding points:
- $N \leftrightarrow L$
- $A \leftrightarrow F$

So $\overline{NA} \leftrightarrow \overline{LF}$

> Answer: $\overline{FL}$ (same segment, just written backwards — acceptable) ✔

Note: $\overline{FL}$ is the same as $\overline{LF}$ in terms of length, but direction doesn't matter for segments.

#### c. $\angle Y$ corresponds to ___

$Y \leftrightarrow T$, so $\angle Y \leftrightarrow \angle T$

> Answer: $\angle T$ ✔

#### d. $\overline{FY}$ corresponds to ___

$F \leftrightarrow A$, $Y \leftrightarrow T$, so $\overline{FY} \leftrightarrow \overline{AT}$

> Answer: $\overline{AT}$ ✔

✔ All answers for #2 are correct.

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Problem 3: Find Missing Side Using Scale Factor

We have two similar triangles:
- Larger triangle: side = 12 and 20
- Smaller triangle: side = 3 and $x$

The sides are proportional.

#### Step 1: Identify scale factor

Compare corresponding sides:
$$
\frac{12}{3} = 4 \quad \text{or} \quad \frac{20}{x} = 4
$$

So scale factor from small to large is 4.

Alternatively, from large to small: $\frac{1}{4}$

But we’re solving for $x$, the missing side.

Set up proportion:
$$
\frac{12}{3} = \frac{20}{x}
$$

$$
4 = \frac{20}{x} \Rightarrow x = \frac{20}{4} = 5
$$

> Answer: $x = 5$ ✔

Scale factor: $\frac{12}{3} = 4$ or $\frac{20}{5} = 4$

So scale factor = 4 (from small to large), or $\frac{1}{4}$ (from large to small)

✔ Correct.

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Problem 4: Another Similar Triangle

Given:
- Small triangle: side = 6
- Large triangle: side = 10, and another side = $x$

Wait — let’s look at the image again.

From the handwritten work:
- One triangle has a side of 6
- The other has a side of 10 and another side labeled $x$
- Proportion: $\frac{6}{16} = \frac{x}{8}$ → Wait, this seems off.

Wait, actually, looking closely:

There's a small triangle with side 6 and a large triangle with side 10 and side $x$.

But the proportion shown is:
$$
\frac{6}{16} = \frac{x}{8}
$$

That suggests:
- 6 corresponds to 16?
- But that can’t be — unless the labeling is different.

Wait — perhaps the smaller triangle has side 6, and the larger has side 10?

But then the proportion should be based on corresponding sides.

Looking at the student’s work:
$$
\frac{6}{16} = \frac{x}{8} \Rightarrow 16x = 48 \Rightarrow x = 3
$$

But also they wrote: “$x = 2$” in a box.

Wait — there's confusion.

Let me re-analyze.

Actually, the student wrote:
$$
\frac{6}{16} = \frac{x}{8}
$$
Then:
$$
16x = 48 \Rightarrow x = 3
$$

But they boxed $x = 2$. That’s incorrect.

Wait — maybe the figure shows:

- Left triangle: side = 6
- Right triangle: side = 10, and another side = $x$

But the student wrote:
$$
\frac{6}{16} = \frac{x}{8}
$$

Wait — 16? Where did 16 come from?

Ah! Possibly a typo or misreading.

Looking more carefully:

The left triangle has a side labeled 6, and the right triangle has a side labeled 10, and another side labeled x.

And the student wrote:
$$
\frac{6}{16} = \frac{x}{8}
$$

But 16 isn't in the diagram.

Wait — maybe the side of 6 corresponds to a side of 16? No, that’s not shown.

Alternatively, maybe the large triangle has a side of 16, and the small one has 6?

But the diagram shows:

- Small triangle: side = 6
- Large triangle: side = 10 and side = $x$

But the student wrote:
$$
\frac{6}{16} = \frac{x}{8}
$$

That doesn’t match the numbers.

Wait — maybe it's:
$$
\frac{6}{x} = \frac{16}{8}
$$

No — that would be $6/x = 2 \Rightarrow x = 3$

But the student says $x = 2$

Wait — let’s check the handwritten math:

They wrote:
$$
\frac{6}{16} = \frac{x}{8}
\Rightarrow 16x = 48 \Rightarrow x = 3
$$

But then they boxed $x = 2$ — which is wrong.

Wait — but in the next line, they wrote:
“$16x = 72$” — wait, no, it says “16x = 72”? Let’s see.

Actually, in the image, it says:
> $\frac{6}{16} = \frac{x}{8}$
> $16x = 48$ → $x = 3$
> Then boxed: $x = 2$

But this is inconsistent.

Wait — perhaps the actual problem is:

Left triangle: side = 6
Right triangle: side = 10, and side = $x$

But the student used 6 and 16 — which may be a mistake.

Alternatively, maybe the scale factor was calculated as:
$$
\frac{10}{6} = \frac{5}{3}
$$

But then if another side is 8, then $x = ?$

Wait — perhaps the student meant:
- Side 6 corresponds to side 16? No.

Wait — looking at the handwritten solution:

They wrote:
$$
\frac{6}{16} = \frac{x}{8}
\Rightarrow 16x = 48 \Rightarrow x = 3
$$

But then they wrote “$x = 2$” in a green box — that’s incorrect.

But earlier, they wrote:
> Scale Factor = $\frac{10}{16} = \frac{16}{4} = 4$ — that doesn’t make sense.

Wait — let’s read carefully:

They wrote:
> Scale Factor = $\frac{10}{16} = \frac{16}{4} = 4$ — that’s wrong

$\frac{10}{16} = \frac{5}{8}$, not 4.

Then they said:
> $\frac{6}{16} = \frac{x}{8}$ → $16x = 48$ → $x = 3$

But then boxed $x = 2$

So the final answer $x = 2$ is incorrect.

But the proportion setup seems to be wrong too.

Wait — maybe the correct setup is:

If the triangles are similar, and one side is 6 in small triangle, and corresponding side is 10 in large triangle, then scale factor = $10/6 = 5/3$

Then if another side in small triangle is 8, then large side = $8 \times \frac{5}{3} = \frac{40}{3} \approx 13.33$

But that doesn’t match.

Alternatively, if the large triangle has a side of 10, and the small triangle has a side of 6, and another side in small is 8, then:

$$
\frac{6}{10} = \frac{8}{x} \Rightarrow 6x = 80 \Rightarrow x = \frac{40}{3}
$$

Still not matching.

Wait — maybe the diagram shows:

- Small triangle: side = 6
- Large triangle: side = 10, and side = $x$
- And another side in small triangle is 8?

But the student wrote:
$$
\frac{6}{16} = \frac{x}{8}
$$

That suggests a side of 16 in the large triangle.

But only 10 is labeled.

I think there’s a mislabeling or error in the student’s work.

But in the green box, they wrote $x = 2$, and showed:

$$
\frac{6}{16} = \frac{x}{8} \Rightarrow 16x = 48 \Rightarrow x = 3
$$

But they boxed $x = 2$ — contradiction.

Also, they wrote:
> Scale Factor = $\frac{10}{16} = \frac{16}{4} = 4$ — wrong

$\frac{10}{16} = 0.625$, not 4.

And $\frac{16}{4} = 4$, but that’s unrelated.

So the student made errors in Problem 4.

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Correct Solution for Problem 4

Assuming:
- Two similar triangles
- One triangle has side 6
- Other triangle has corresponding side 10
- Another side in small triangle is 8, find corresponding side $x$ in large triangle

Then scale factor from small to large: $\frac{10}{6} = \frac{5}{3}$

So:
$$
x = 8 \times \frac{5}{3} = \frac{40}{3} \approx 13.33
$$

But that’s not matching.

Alternatively, suppose:
- Small triangle: side = 6
- Large triangle: side = 10
- Another side in large triangle is $x$, and corresponding side in small is 8

Then:
$$
\frac{6}{10} = \frac{8}{x} \Rightarrow 6x = 80 \Rightarrow x = \frac{40}{3}
$$

Still not 2.

But the student wrote:
$$
\frac{6}{16} = \frac{x}{8} \Rightarrow x = 3
$$

But then boxed $x = 2$

So either the diagram is misread, or the student made a mistake.

But in the image, it looks like the student intended:

- Small triangle: side = 6
- Large triangle: side = 16
- Then $x = ?$ for another side of 8

So:
$$
\frac{6}{16} = \frac{x}{8} \Rightarrow x = \frac{6 \cdot 8}{16} = \frac{48}{16} = 3
$$

So $x = 3$, not 2.

Therefore, the boxed answer $x = 2$ is incorrect.

But the student wrote “$x = 2$” in a green box — likely a mistake.

So correct answer is $x = 3$.

But why did they write $x = 2$?

Wait — maybe the scale factor was calculated differently.

They wrote:
> Scale Factor = $\frac{10}{16} = \frac{16}{4} = 4$ — nonsense

But if the large triangle has side 16, and small has side 4, then scale factor = 4.

But here, small has 6, large has 16? Then scale factor = $16/6 = 8/3$

Not 4.

So multiple errors in problem 4.

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✔ Final Summary

#### Problem 1 & 2:
All answers are correct.

#### Problem 3:
- Scale factor: $4$ (large to small)
- Proportion: $\frac{12}{3} = \frac{20}{x} \Rightarrow x = 5$
- ✔ Correct

#### Problem 4:
Student's work has errors:
- Proportion $\frac{6}{16} = \frac{x}{8}$ leads to $x = 3$
- But they boxed $x = 2$ — ✘ Incorrect
- Scale factor calculation is flawed

✔ Correct Answer for Problem 4: $x = 3$

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✔ Final Answers:

1. a. $\angle A$
b. $\overline{ON}$
c. $\angle N$
d. $\overline{CB}$

2. a. $\angle L$
b. $\overline{FL}$
c. $\angle T$
d. $\overline{AT}$

3. $x = 5$, Scale Factor = $4$

4. $x = 3$, Scale Factor = $\frac{16}{6} = \frac{8}{3}$ (if 6 ↔ 16), or better: if 6 ↔ 10, then scale factor = $10/6 = 5/3$, but need consistent labeling.

But since the student used $\frac{6}{16} = \frac{x}{8}$, then $x = 3$ is correct.

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✔ Conclusion:

- Problems 1–3 are solved correctly.
- Problem 4 has a mistake: $x = 3$, not $x = 2$.