Similar Triangles Notes and Worksheets - Lindsay Bowden - Free Printable

Problem Analysis and Solution

The image provided is a worksheet on similar triangles. Let's break it down step by step and solve the problems.

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#### Section 1: Similar Triangles Notes
The notes section defines similar triangles and provides key properties. Here are the blanks filled in:

1. Similar: same shape, different size
- Symbol: `~`

2. Similar Triangles: Triangles that have:
- 3 pairs of congruent corresponding angles
- 3 pairs of proportional corresponding sides

*Reminder*: Corresponding means in the same relative position.
*Proportional* means the sides will have the same ratio.

The example given shows two triangles, \( \triangle ABC \) and \( \triangle DEF \), with the following properties:
- Angles: \( \angle A = \angle D \), \( \angle B = \angle E \), \( \angle C = \angle F \)
- Sides: \( \frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{FD} \)

Thus, \( \triangle ABC \sim \triangle DEF \).

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#### Section 2: Examples

##### Example 1: \( \triangle JKL \sim \triangle MNO \)

We are given:
- \( \triangle JKL \sim \triangle MNO \)
- \( JK = 6 \), \( MN = 3 \)
- \( \angle J = 75^\circ \), \( \angle K = 60^\circ \)

We need to find all missing sides and angles.

Step 1: Determine the scale factor.
The scale factor between the two triangles is the ratio of the corresponding sides:
\[
\text{Scale factor} = \frac{JK}{MN} = \frac{6}{3} = 2
\]

Step 2: Find the missing sides.
Since \( \triangle JKL \sim \triangle MNO \), the sides are proportional:
\[
\frac{JK}{MN} = \frac{KL}{NO} = \frac{LJ}{OM} = 2
\]
Given \( MN = 3 \), we can find \( NO \) and \( OM \):
\[
KL = 2 \times NO \quad \text{and} \quad LJ = 2 \times OM
\]

However, the problem does not provide the lengths of \( KL \) or \( LJ \). Assuming we need to express the sides in terms of the given information:
\[
KL = 2 \times NO \quad \text{and} \quad LJ = 2 \times OM
\]

Step 3: Find the missing angles.
Since the triangles are similar, their corresponding angles are equal:
\[
\angle J = \angle M = 75^\circ, \quad \angle K = \angle N = 60^\circ
\]
The third angle in each triangle can be found using the fact that the sum of the angles in a triangle is \( 180^\circ \):
\[
\angle L = 180^\circ - \angle J - \angle K = 180^\circ - 75^\circ - 60^\circ = 45^\circ
\]
Thus:
\[
\angle O = \angle L = 45^\circ
\]

Final Answer for Example 1:
\[
\boxed{\angle M = 75^\circ, \angle N = 60^\circ, \angle O = 45^\circ, KL = 2 \times NO, LJ = 2 \times OM}
\]

##### Example 2: \( \triangle QRS \sim \triangle XYZ \)

We are given:
- \( \triangle QRS \sim \triangle XYZ \)

We need to identify the corresponding angles and sides.

Step 1: Identify corresponding angles.
Since the triangles are similar, their corresponding angles are equal:
\[
\angle Q \cong \angle X, \quad \angle R \cong \angle Y, \quad \angle S \cong \angle Z
\]

Step 2: Identify corresponding sides.
The corresponding sides are proportional:
\[
\frac{QR}{XY} = \frac{RS}{YZ} = \frac{SQ}{ZX}
\]

Final Answer for Example 2:
\[
\boxed{\angle Q \cong \angle X, \angle R \cong \angle Y, \angle S \cong \angle Z}
\]

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Final Answers:

1. For \( \triangle JKL \sim \triangle MNO \):
\[
\boxed{\angle M = 75^\circ, \angle N = 60^\circ, \angle O = 45^\circ, KL = 2 \times NO, LJ = 2 \times OM}
\]

2. For \( \triangle QRS \sim \triangle XYZ \):
\[
\boxed{\angle Q \cong \angle X, \angle R \cong \angle Y, \angle S \cong \angle Z}
\]