Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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Equations and inequalities describe the relationship between two mathematical expressions.
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A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.
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Find the value of the variable
![### Geometry Problem Set: Solving for Variables
#### Find the value of the variable
**Problem 34:**
This problem presents a triangle with two parallel segments. The triangle consists of three sides with lengths 12, 9, and 4. There is a fourth, non-parallel segment labeled \( x \).
- The side with length 12 is parallel to and directly above the segment \( x \).
- The side opposite to \( x \), with the length of 9, forms a part of the larger triangle.
- The smallest segment, with a length of 4, is parallel to the longest side of the triangle and divides it proportionally.
**Problem 35:**
This problem presents a right triangle with an altitude drawn from the right-angled vertex to the hypotenuse.
- The base of the triangle is divided into two segments by the altitude: one segment has a length of 4 units, and the other has a length of 9 units.
- The altitude, labeled \( x \), forms two smaller right triangles within the larger right triangle.
**Solution Explanation for Problem 34:**
Given that the sides are proportional due to the parallel segments, we can use the properties of similar triangles to find \( x \).
\[ \frac{x}{4} = \frac{9}{12} \]
Cross-multiplying to solve for \( x \):
\[ x = \frac{9 \cdot 4}{12} \]
\[ x = 3 \]
**Solution Explanation for Problem 35:**
In this right triangle with the altitude, the two smaller triangles formed are similar to the original triangle and to each other. We can use the geometric mean formula, which states that the altitude to the hypotenuse is the geometric mean of the segments it divides the hypotenuse into:
\[ x = \sqrt{4 \cdot 9} \]
\[ x = \sqrt{36} \]
\[ x = 6 \]
Thus, the value of the variable for both problems has been found using principles of geometry and properties of similar triangles.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff42d33e0-1c3e-419a-9ccd-c117b498d94b%2F28ce991e-ee57-4cff-ba35-8f6b205da139%2Fstsjse_processed.png&w=3840&q=75)
Transcribed Image Text:### Geometry Problem Set: Solving for Variables
#### Find the value of the variable
**Problem 34:**
This problem presents a triangle with two parallel segments. The triangle consists of three sides with lengths 12, 9, and 4. There is a fourth, non-parallel segment labeled \( x \).
- The side with length 12 is parallel to and directly above the segment \( x \).
- The side opposite to \( x \), with the length of 9, forms a part of the larger triangle.
- The smallest segment, with a length of 4, is parallel to the longest side of the triangle and divides it proportionally.
**Problem 35:**
This problem presents a right triangle with an altitude drawn from the right-angled vertex to the hypotenuse.
- The base of the triangle is divided into two segments by the altitude: one segment has a length of 4 units, and the other has a length of 9 units.
- The altitude, labeled \( x \), forms two smaller right triangles within the larger right triangle.
**Solution Explanation for Problem 34:**
Given that the sides are proportional due to the parallel segments, we can use the properties of similar triangles to find \( x \).
\[ \frac{x}{4} = \frac{9}{12} \]
Cross-multiplying to solve for \( x \):
\[ x = \frac{9 \cdot 4}{12} \]
\[ x = 3 \]
**Solution Explanation for Problem 35:**
In this right triangle with the altitude, the two smaller triangles formed are similar to the original triangle and to each other. We can use the geometric mean formula, which states that the altitude to the hypotenuse is the geometric mean of the segments it divides the hypotenuse into:
\[ x = \sqrt{4 \cdot 9} \]
\[ x = \sqrt{36} \]
\[ x = 6 \]
Thus, the value of the variable for both problems has been found using principles of geometry and properties of similar triangles.
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