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
Related questions
Question
![## Understanding Angles in a Parallelogram
### Problem Statement
In the given diagram, a parallelogram \(QRST\) is shown with its diagonals intersecting at point \(X\). The interior angles near the vertices \(Q\) and \(T\) are provided. The angle measures are as follows:
- \(\angle RQX = 81^\circ\)
- \(\angle QXR = 29^\circ\)
- \(\angle STR = 42^\circ\)
### Diagram Explanation
The parallelogram \(QRST\) has its diagonals \(RT\) and \(QS\) intersecting at point \(X\). The diagram also provides the measures of some angles formed at point \(X\) and between the sides of the parallelogram:
- \(\angle RQX\) is opposite to \(\angle QXR\) at point \(Q\).
- \(\angle STR\) is provided at vertex \(T\).
### Question
What is the measure of \(\angle RTQ\)?
### Calculation
To find the measure of \(\angle RTQ\):
1. Note that the sum of the angles in triangle \(QXR\) adds up to \(180^\circ\) (since the sum of angles in any triangle is \(180^\circ\)):
\[ \angle RQX + \angle QXR + \angle RXQ = 180^\circ \]
Plugging in the given angles:
\[ 81^\circ + 29^\circ + \angle RXQ = 180^\circ \]
\[ 110^\circ + \angle RXQ = 180^\circ \]
\[ \angle RXQ = 180^\circ - 110^\circ \]
\[ \angle RXQ = 70^\circ \]
2. Since \(X\) is the point where diagonals intersect and these diagonals bisect the angles in a parallelogram, \(\angle RTQ\) would have the same measure as \(\angle QXR\):
\[ \angle RTQ = \angle QXR = 29^\circ \]
### Answer
\[
m \angle RTQ = 29^\circ
\]
## Summary
The measure of \(\angle RTQ\) in the given parallelogram \(QRST\) is:
\[
m \angle RTQ = 29^\circ
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb294f3d7-c37e-4d19-872e-5bb2e2baac27%2F54496c8b-8db7-474e-84e6-f584b7dfb5ed%2F2tsqzrm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Understanding Angles in a Parallelogram
### Problem Statement
In the given diagram, a parallelogram \(QRST\) is shown with its diagonals intersecting at point \(X\). The interior angles near the vertices \(Q\) and \(T\) are provided. The angle measures are as follows:
- \(\angle RQX = 81^\circ\)
- \(\angle QXR = 29^\circ\)
- \(\angle STR = 42^\circ\)
### Diagram Explanation
The parallelogram \(QRST\) has its diagonals \(RT\) and \(QS\) intersecting at point \(X\). The diagram also provides the measures of some angles formed at point \(X\) and between the sides of the parallelogram:
- \(\angle RQX\) is opposite to \(\angle QXR\) at point \(Q\).
- \(\angle STR\) is provided at vertex \(T\).
### Question
What is the measure of \(\angle RTQ\)?
### Calculation
To find the measure of \(\angle RTQ\):
1. Note that the sum of the angles in triangle \(QXR\) adds up to \(180^\circ\) (since the sum of angles in any triangle is \(180^\circ\)):
\[ \angle RQX + \angle QXR + \angle RXQ = 180^\circ \]
Plugging in the given angles:
\[ 81^\circ + 29^\circ + \angle RXQ = 180^\circ \]
\[ 110^\circ + \angle RXQ = 180^\circ \]
\[ \angle RXQ = 180^\circ - 110^\circ \]
\[ \angle RXQ = 70^\circ \]
2. Since \(X\) is the point where diagonals intersect and these diagonals bisect the angles in a parallelogram, \(\angle RTQ\) would have the same measure as \(\angle QXR\):
\[ \angle RTQ = \angle QXR = 29^\circ \]
### Answer
\[
m \angle RTQ = 29^\circ
\]
## Summary
The measure of \(\angle RTQ\) in the given parallelogram \(QRST\) is:
\[
m \angle RTQ = 29^\circ
\]
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