Congruent Triangles Worksheets - Math Monks - Free Printable

Problem Analysis and Solution

The worksheet involves problems related to congruent triangles and similar triangles. 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:
- \( \angle B = 90^\circ \)
- \( \angle DCE = 90^\circ \)
- \( BC = 15 \)
- \( CD = 15 \)
- \( DE = 20 \)
- \( CE = 13 \)

We need to find the values of \( X \) and \( Y \).

##### Step 1: Identify the congruence rule
Both triangles are right triangles, and we can see that:
- \( BC = CD = 15 \) (one pair of corresponding sides are equal)
- \( \angle B = \angle DCE = 90^\circ \) (both are right angles)
- \( \angle ACB = \angle ECD \) (corresponding angles are equal because they are both complementary to \( \angle BCD \))

Thus, the triangles are congruent by the SAS (Side-Angle-Side) congruence rule.

##### Step 2: Find the unknown values
Since the triangles are congruent:
- \( AB = CE = 13 \) (corresponding sides)
- \( AC = CD = 15 \) (corresponding sides)

From the given information:
- \( \angle ACB = 41^\circ \)
- Since \( \triangle ABC \cong \triangle CDE \), \( \angle Y = \angle ACB = 41^\circ \).

Thus:
- \( X = 13 \)
- \( Y = 41^\circ \)

##### Final Answer for Part (a):
\[
\boxed{X = 13, Y = 41^\circ}
\]

#### Part (b)
We are given two triangles, \( \triangle MNO \) and \( \triangle QPR \), with the following information:
- \( MN = 3 \)
- \( NO = 2 \)
- \( MO = 5 \)
- \( QR = 3 \)
- \( PR = 2 \)
- \( PQ = 5 \)

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

##### Step 1: Identify the congruence rule
Both triangles have the same side lengths:
- \( MN = PQ = 3 \)
- \( NO = PR = 2 \)
- \( MO = QR = 5 \)

Thus, the triangles are congruent by the SSS (Side-Side-Side) congruence rule.

##### Step 2: Find the unknown angles
Since the triangles are congruent:
- Corresponding angles are equal. Therefore:
- \( \angle MNO = \angle QPR = 62^\circ \)
- \( \angle MON = \angle PQR = 30^\circ \)
- \( \angle NMO = \angle PRQ = 88^\circ \)

From the given information:
- \( \angle OQR = \angle MON = 30^\circ \)
- \( \angle QRO = \angle NMO = 88^\circ \)

##### Final Answer for Part (b):
\[
\boxed{\angle OQR = 30^\circ, \angle QRO = 88^\circ}
\]

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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: Analyze \( \triangle ABC \)
- Since \( \angle B = 60^\circ \) and \( \angle C = 60^\circ \), the third angle \( \angle A \) is:
\[
\angle A = 180^\circ - 60^\circ - 60^\circ = 60^\circ
\]
- Therefore, \( \triangle ABC \) is an equilateral triangle, meaning all its angles are \( 60^\circ \).

##### Step 2: Identify the similar triangle
The only triangle given in the options that has all angles equal to \( 60^\circ \) is \( \triangle DEF \).

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

##### Final Answer for Part (a):
\[
\boxed{\triangle DEF}
\]

#### Part (b)
We are given \( \triangle ABC \) with sides \( AB = 7 \), \( BC = 6 \), and \( AC = 5 \). We need to determine which triangle is similar to \( \triangle ABC \).

##### Step 1: Analyze the ratios of the sides
The sides of \( \triangle ABC \) are in the ratio:
\[
AB : BC : AC = 7 : 6 : 5
\]

##### Step 2: Compare with the given triangle
The other triangle has sides \( QR = 30 \), \( PR = 25 \), and \( PQ = 35 \). The ratios of the sides are:
\[
PQ : PR : QR = 35 : 25 : 30 = 7 : 5 : 6
\]

Rearranging the sides of \( \triangle PQR \) to match the order of \( \triangle ABC \):
\[
QR : PR : PQ = 30 : 25 : 35 = 6 : 5 : 7
\]

This matches the ratio \( AB : BC : AC = 7 : 6 : 5 \).

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

##### Final Answer for Part (b):
\[
\boxed{\triangle PQR}
\]

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Problem 3: Solve for \( x \) in the Given Figures

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

We need to find \( x \).

##### Step 1: Use the similarity ratio
Since \( \triangle ABD \sim \triangle CBD \), the corresponding sides are proportional:
\[
\frac{AD}{CB} = \frac{BD}{CD}
\]

Substitute the given values:
\[
\frac{15}{15} = \frac{12}{16}
\]

Simplify:
\[
1 = \frac{12}{16}
\]

This confirms the similarity ratio. Now, we need to find \( x \), which is the length of \( AB \).

##### Step 2: Use the similarity ratio for \( AB \) and \( CD \)
Since \( \triangle ABD \sim \triangle CBD \):
\[
\frac{AB}{CD} = \frac{AD}{CB}
\]

Substitute the known values:
\[
\frac{x}{16} = \frac{15}{15}
\]

Simplify:
\[
\frac{x}{16} = 1
\]

Solve for \( x \):
\[
x = 16
\]

##### Final Answer for Part (a):
\[
\boxed{x = 16}
\]

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

We need to find \( x \).

##### Step 1: Use the similarity ratio
Since \( \triangle AXZ \sim \triangle SYZ \), the corresponding sides are proportional:
\[
\frac{AX}{SY} = \frac{XZ}{YZ} = \frac{AZ}{SZ}
\]

From the diagram, \( XZ = XY + YZ = 20 + x \) and \( AZ = AX + XZ = 15 + (20 + x) = 35 + x \).

However, we can use the simpler ratio involving \( AX \) and \( SY \):
\[
\frac{AX}{SY} = \frac{XZ}{YZ}
\]

Substitute the known values:
\[
\frac{15}{5} = \frac{20 + x}{x}
\]

Simplify:
\[
3 = \frac{20 + x}{x}
\]

##### Step 2: Solve for \( x \)
Multiply both sides by \( x \):
\[
3x = 20 + x
\]

Subtract \( x \) from both sides:
\[
2x = 20
\]

Solve for \( x \):
\[
x = 10
\]

##### Final Answer for Part (b):
\[
\boxed{x = 10}
\]

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Final Answers:

1. (a) \( X = 13, Y = 41^\circ \)
(b) \( \angle OQR = 30^\circ, \angle QRO = 88^\circ \)
2. (a) \( \triangle DEF \)
(b) \( \triangle PQR \)
3. (a) \( x = 16 \)
(b) \( x = 10 \)

\[
\boxed{13, 41^\circ, 30^\circ, 88^\circ, \triangle DEF, \triangle PQR, 16, 10}
\]