Congruent Triangles Worksheets - Math Monks - Free Printable

Problem Analysis and Solution

The worksheet involves problems related to congruent triangles, similar triangles, and solving for unknowns using properties of similarity. Let's solve each part step by step.

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Problem 1: Congruent Triangles

#### Part (a)
We are given two right triangles, \( \triangle ABC \) and \( \triangle CDE \), with the following information:
- \( \angle B = 90^\circ \)
- \( \angle DCE = 90^\circ \)
- \( BC = 15 \)
- \( CD = 15 \)
- \( DE = 20 \)
- \( CE = 13 \)
- \( \angle X = 41^\circ \)

We need to find \( X \) and \( Y \).

Step 1: Identify the congruence rule.
Since both triangles are right triangles and share a common side \( BC = CD = 15 \), we can use the HL (Hypotenuse-Leg) Congruence Rule:
- Hypotenuse: \( AC = CE \)
- Leg: \( BC = CD \)

Thus, \( \triangle ABC \cong \triangle CDE \) by HL.

Step 2: Find \( X \).
Since the triangles are congruent, corresponding angles are equal. Therefore:
\[ \angle X = \angle D = 41^\circ \]

Step 3: Find \( Y \).
Since \( \angle B = 90^\circ \) and \( \angle X = 41^\circ \), we can find \( \angle Y \) using the fact that the sum of angles in a triangle is \( 180^\circ \):
\[ \angle Y = 180^\circ - 90^\circ - 41^\circ = 49^\circ \]

Final Answer for Part (a):
\[ X = 41^\circ, \quad Y = 49^\circ \]

#### Part (b)
We are given two triangles, \( \triangle MNO \) and \( \triangle PQR \), with the following information:
- \( MN = 3 \)
- \( NO = 2 \)
- \( OP = 5 \)
- \( PQ = 3 \)
- \( QR = 5 \)
- \( RP = 2 \)
- \( \angle MNO = 62^\circ \)
- \( \angle NOP = 30^\circ \)
- \( \angle OPQ = 30^\circ \)

We need to find \( \angle OQR \) and \( \angle QRO \).

Step 1: Identify the congruence rule.
The triangles have corresponding sides equal:
- \( MN = PQ = 3 \)
- \( NO = RP = 2 \)
- \( OM = QR = 5 \)

Thus, \( \triangle MNO \cong \triangle PQR \) by SSS (Side-Side-Side).

Step 2: Find \( \angle OQR \).
Since the triangles are congruent, corresponding angles are equal. Therefore:
\[ \angle OQR = \angle MNO = 62^\circ \]

Step 3: Find \( \angle QRO \).
In \( \triangle PQR \), the sum of the angles is \( 180^\circ \):
\[ \angle QRO = 180^\circ - \angle OQR - \angle QRP \]
\[ \angle QRO = 180^\circ - 62^\circ - 30^\circ = 88^\circ \]

Final Answer for Part (b):
\[ \angle OQR = 62^\circ, \quad \angle QRO = 88^\circ \]

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Problem 2: Similar Triangles

#### Part (a)
We are given \( \triangle ABC \) with \( \angle B = 60^\circ \) and \( \angle C = 60^\circ \). We need to determine which triangle is similar to \( \triangle ABC \).

Step 1: Identify the type of triangle.
Since \( \angle B = 60^\circ \) and \( \angle C = 60^\circ \), the third angle \( \angle A \) is:
\[ \angle A = 180^\circ - 60^\circ - 60^\circ = 60^\circ \]
Thus, \( \triangle ABC \) is an equilateral triangle.

Step 2: Determine the similar triangle.
An equilateral triangle is similar to any other equilateral triangle by the AA (Angle-Angle) Similarity Rule. The given triangle \( \triangle DEF \) also has \( \angle E = 60^\circ \) and \( \angle F = 60^\circ \), so it is also equilateral.

Final Answer for Part (a):
\[ \triangle ABC \sim \triangle DEF \]

#### Part (b)
We are given two triangles, \( \triangle ABC \) and \( \triangle PQR \), with the following side lengths:
- \( AB = 7 \)
- \( BC = 6 \)
- \( CA = 5 \)
- \( PQ = 30 \)
- \( QR = 25 \)
- \( RP = 35 \)

We need to determine if the triangles are similar and, if so, state the similarity rule.

Step 1: Check the ratios of corresponding sides.
Calculate the ratios of the corresponding sides:
\[ \frac{PQ}{AB} = \frac{30}{7} \]
\[ \frac{QR}{BC} = \frac{25}{6} \]
\[ \frac{RP}{CA} = \frac{35}{5} = 7 \]

The ratios are not equal, so the triangles are not similar by SSS (Side-Side-Side) Similarity.

Step 2: Check for AA (Angle-Angle) Similarity.
Since the side ratios are not equal, we cannot assume the angles are equal without additional information. Therefore, we cannot conclude similarity by AA either.

Final Answer for Part (b):
\[ \text{Not similar} \]

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Problem 3: Solving for \( x \)

#### Part (a)
We are given \( \triangle ABD \sim \triangle CBD \) with the following side lengths:
- \( AD = 15 \)
- \( DB = 12 \)
- \( CB = 15 \)
- \( DC = 16 \)

We need to find \( x \).

Step 1: Use the property of similar triangles.
Since \( \triangle ABD \sim \triangle CBD \), the corresponding sides are proportional:
\[ \frac{AD}{CB} = \frac{DB}{DC} \]
\[ \frac{15}{15} = \frac{12}{16} \]

This simplifies to:
\[ 1 = \frac{12}{16} \]

However, this is not directly useful for finding \( x \). Instead, we use the fact that the triangles are similar and the given side lengths to find \( x \).

Step 2: Use the given side lengths.
Since \( \triangle ABD \sim \triangle CBD \), the ratio of corresponding sides is:
\[ \frac{AD}{DB} = \frac{CB}{DC} \]
\[ \frac{15}{12} = \frac{15}{16} \]

This is consistent, but we need to find \( x \). From the diagram, \( x \) is the height from \( A \) to \( BD \). Using the similarity ratio:
\[ \frac{x}{16} = \frac{15}{12} \]
\[ x = 16 \cdot \frac{15}{12} = 16 \cdot \frac{5}{4} = 20 \]

Final Answer for Part (a):
\[ x = 20 \]

#### Part (b)
We are given \( \triangle AXZ \sim \triangle SYZ \) with the following side lengths:
- \( AX = 15 \)
- \( XZ = 20 \)
- \( SY = 5 \)
- \( YZ = x \)
- \( SZ = 12 \)

We need to find \( x \).

Step 1: Use the property of similar triangles.
Since \( \triangle AXZ \sim \triangle SYZ \), the corresponding sides are proportional:
\[ \frac{AX}{SY} = \frac{XZ}{YZ} = \frac{AZ}{SZ} \]

From the given side lengths:
\[ \frac{AX}{SY} = \frac{15}{5} = 3 \]

Thus:
\[ \frac{XZ}{YZ} = 3 \]
\[ \frac{20}{x} = 3 \]

Step 2: Solve for \( x \).
\[ x = \frac{20}{3} \]

Final Answer for Part (b):
\[ x = \frac{20}{3} \]

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Final Answers

1. (a) \( X = 41^\circ, Y = 49^\circ \)
(b) \( \angle OQR = 62^\circ, \angle QRO = 88^\circ \)

2. (a) \( \triangle ABC \sim \triangle DEF \)
(b) Not similar

3. (a) \( x = 20 \)
(b) \( x = \frac{20}{3} \)

\[
\boxed{41^\circ, 49^\circ, 62^\circ, 88^\circ, \triangle DEF, \text{Not similar}, 20, \frac{20}{3}}
\]