Math Problem Statement

Solve problems 19, 20, 21, and 22 based on the provided triangle and given values. Prove certain relationships and find the values of x, y, and z.

Solution

Let's address each problem in turn based on the image provided.

Problem 19

We need to write a paragraph proof for Theorem 9.2.

Given:

  1. Triangle ADC\triangle ADC is a right triangle.
  2. DBDB is an altitude of ADC\triangle ADC.

Prove:

ABDB=DBCB\frac{AB}{DB} = \frac{DB}{CB}

Proof:

Since DBDB is an altitude in ADC\triangle ADC, it creates two smaller right triangles, ADB\triangle ADB and CDB\triangle CDB, each similar to the larger ADC\triangle ADC by AA similarity (they each have a right angle and share another angle with ADC\triangle ADC). By similarity, we know that corresponding side ratios are equal. Thus: ABDB=DBCB\frac{AB}{DB} = \frac{DB}{CB} This completes the proof.


Problem 20

This requires a two-column proof for Theorem 9.3.

Given:

  1. ADC\angle ADC is a right angle.
  2. DBDB is an altitude of ADC\triangle ADC.

Prove:

ABAD=ADAC,BCDC=DCAC\frac{AB}{AD} = \frac{AD}{AC}, \quad \frac{BC}{DC} = \frac{DC}{AC}

Two-Column Proof:

StatementsReasons
1. ADC\angle ADC is a right angle, DBDB is an altitude.Given
2. ADBADC\triangle ADB \sim \triangle ADC and CDBADC\triangle CDB \sim \triangle ADCAA Similarity
3. ABAD=ADAC\frac{AB}{AD} = \frac{AD}{AC}Corresponding side ratios of similar triangles
4. BCDC=DCAC\frac{BC}{DC} = \frac{DC}{AC}Corresponding side ratios of similar triangles

This completes the proof.


Problem 21

We are asked to find the values of xx, yy, and zz in the triangle.

Solution:

This involves using the Pythagorean theorem or similarity ratios, but we don’t have enough specific measurements in this part of the image to fully solve. Let me know if you have more details on side lengths or angles for a clearer solution.


Problem 22

We are to check the work of Aiden and Tia regarding finding the value of xx.

  1. Aiden's Calculation:

    • Set up the proportion 3.6x=x6\frac{3.6}{x} = \frac{x}{6}.
    • Cross-multiplying, we get x2=3.6×6x2=21.6x=21.64.6x^2 = 3.6 \times 6 \Rightarrow x^2 = 21.6 \Rightarrow x = \sqrt{21.6} \approx 4.6.
    • Conclusion: Aiden’s solution is correct.
  2. Tia's Calculation:

    • Set up the proportion 3.6x=x8\frac{3.6}{x} = \frac{x}{8}.
    • Cross-multiplying, we get x2=3.6×8x2=28.8x=28.85.4x^2 = 3.6 \times 8 \Rightarrow x^2 = 28.8 \Rightarrow x = \sqrt{28.8} \approx 5.4.
    • Conclusion: Tia’s solution is incorrect as she used the wrong proportion.

Would you like further details on any part of these solutions?

Related Questions

  1. How can similarity be used to prove more complex geometric relationships?
  2. Why does the altitude in a right triangle create similar triangles?
  3. Can we find unknown lengths without knowing angles in a triangle?
  4. How can we verify the accuracy of proportions in geometric figures?
  5. What methods can be used if we know only partial triangle side lengths?

Tip

When dealing with similar triangles, always look for corresponding side ratios. They provide the foundation for most similarity-based proofs and calculations.

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Math Problem Analysis

Mathematical Concepts

Geometry
Similarity of Triangles
Ratios and Proportions
Right Triangle Properties

Formulas

\( \frac{AB}{DB} = \frac{DB}{CB} \)
\( \frac{AB}{AD} = \frac{AD}{AC} \)
Pythagorean theorem \( a^2 + b^2 = c^2 \)
Cross multiplication for solving proportions

Theorems

Theorem 9.2
Theorem 9.3
AA Similarity Theorem

Suitable Grade Level

Grades 9-10