Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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What would a good explanation be for this problem?
![**Problem:**
Examine the work that someone else did to solve the problem below. Explain why you agree or disagree with the individual’s solving strategy and solution and include specific math vocabulary in your explanation. Saying “I disagree because I got a different answer” is not sufficient unless you also prove and explain that your answer is correct! You may write your explanation in the whiteboard space provided or type it out in the answer box. (2 points)
**Given:**
If \( \triangle JKL \sim \triangle NML \), find the value of \( x \).
**Diagram:**
- Two triangles \( \triangle JKL \) and \( \triangle NML \) are similar.
- Side \( KL \) corresponds to \( NM \), side \( JL \) corresponds to \( LM \), and side \( JK \) to \( NL \).
- \( KL = x + 7 \), \( JL = 24 \), \( NM = 9 \), \( LN = 6 \), \( JK = 21 \).
**Solution Explanation:**
1. The relation between the triangles due to their similarity is established as \( \frac{KL}{LN} = \frac{JL}{LM} \).
2. Set up the proportion: \(\frac{x+7}{9} = \frac{24}{6}\).
3. Simplify \( \frac{24}{6} \) to get 4.
4. The equation becomes \(\frac{x+7}{9} = 4\).
5. Cross-multiply to solve for \( x \):
\[
x + 7 = 36
\]
6. Subtract 7 from both sides:
\[
x = 29
\]
**Findings:**
The student correctly used the properties of similar triangles and the correct mathematical operations to solve for \( x \). The strategy involved setting up equivalent ratios and solving the resulting equation.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F64a909f3-bac2-427a-9419-941550f4323a%2F8290c720-23d3-452b-a53e-5e8ef44a716e%2Fpo75aa_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem:**
Examine the work that someone else did to solve the problem below. Explain why you agree or disagree with the individual’s solving strategy and solution and include specific math vocabulary in your explanation. Saying “I disagree because I got a different answer” is not sufficient unless you also prove and explain that your answer is correct! You may write your explanation in the whiteboard space provided or type it out in the answer box. (2 points)
**Given:**
If \( \triangle JKL \sim \triangle NML \), find the value of \( x \).
**Diagram:**
- Two triangles \( \triangle JKL \) and \( \triangle NML \) are similar.
- Side \( KL \) corresponds to \( NM \), side \( JL \) corresponds to \( LM \), and side \( JK \) to \( NL \).
- \( KL = x + 7 \), \( JL = 24 \), \( NM = 9 \), \( LN = 6 \), \( JK = 21 \).
**Solution Explanation:**
1. The relation between the triangles due to their similarity is established as \( \frac{KL}{LN} = \frac{JL}{LM} \).
2. Set up the proportion: \(\frac{x+7}{9} = \frac{24}{6}\).
3. Simplify \( \frac{24}{6} \) to get 4.
4. The equation becomes \(\frac{x+7}{9} = 4\).
5. Cross-multiply to solve for \( x \):
\[
x + 7 = 36
\]
6. Subtract 7 from both sides:
\[
x = 29
\]
**Findings:**
The student correctly used the properties of similar triangles and the correct mathematical operations to solve for \( x \). The strategy involved setting up equivalent ratios and solving the resulting equation.
![If triangles \( \triangle JKL \sim \triangle NML \), find the value of \( x \).
**Diagram Explanation:**
The diagram shows two triangles, \( \triangle JKL \) and \( \triangle NML \), with the following details:
- \( JL = 24 \)
- \( LM = 6 \)
- \( LN = 9 \)
- \( KL \) is labeled as \( x + 7 \)
**Solution:**
To find \( x \), use the similarity ratio of the triangles:
\[
\frac{KL}{LN} = \frac{JL}{LM}
\]
Substitute the given values into the equation:
\[
\frac{x+7}{9} = \frac{24}{6}
\]
Simplify the right side:
\[
\frac{24}{6} = 4
\]
The equation becomes:
\[
\frac{x+7}{9} = 4
\]
Cross-multiply to solve for \( x \):
\[
x+7 = 36
\]
Subtract 7 from both sides:
\[
x = 29
\]
Thus, the value of \( x \) is 29.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F64a909f3-bac2-427a-9419-941550f4323a%2F8290c720-23d3-452b-a53e-5e8ef44a716e%2Fx3t4t3l_processed.png&w=3840&q=75)
Transcribed Image Text:If triangles \( \triangle JKL \sim \triangle NML \), find the value of \( x \).
**Diagram Explanation:**
The diagram shows two triangles, \( \triangle JKL \) and \( \triangle NML \), with the following details:
- \( JL = 24 \)
- \( LM = 6 \)
- \( LN = 9 \)
- \( KL \) is labeled as \( x + 7 \)
**Solution:**
To find \( x \), use the similarity ratio of the triangles:
\[
\frac{KL}{LN} = \frac{JL}{LM}
\]
Substitute the given values into the equation:
\[
\frac{x+7}{9} = \frac{24}{6}
\]
Simplify the right side:
\[
\frac{24}{6} = 4
\]
The equation becomes:
\[
\frac{x+7}{9} = 4
\]
Cross-multiply to solve for \( x \):
\[
x+7 = 36
\]
Subtract 7 from both sides:
\[
x = 29
\]
Thus, the value of \( x \) is 29.
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