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X is the midpoint of UV. Y is the midpoint of UW. If MZUXY = 60, find mZV. X V M. mZV=

X is the midpoint of UV. Y is the midpoint of UW. If MZUXY = 60, find mZV. X V M. mZV=

Elementary Geometry For College Students, 7e
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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**Problem Statement:** *X is the midpoint of \( \overline{UV} \). Y is the midpoint of \( \overline{UW} \). If \( m \angle UXY = 60^\circ \), find \( m \angle V \).* **Diagram Analysis:** - The diagram presents a triangle \( \triangle UVW \). - Points \( X \) and \( Y \) are on line segments \( \overline{UV} \) and \( \overline{UW} \) respectively. - \( X \) is the midpoint of \( \overline{UV} \). - \( Y \) is the midpoint of \( \overline{UW} \). - Line segments \( \overline{XY} \), \( \overline{XV} \), and \( \overline{YW} \) are drawn inside the triangle. - Angle \( \angle UXY = 60^\circ \). **Question:** Calculate \( m \angle V \). **Solution Box:** The answer should be provided by calculating the necessary angle in the triangle configuration, given the information about midpoints and angle \( \angle UXY \).
Transcribed Image Text:**Problem Statement:** *X is the midpoint of \( \overline{UV} \). Y is the midpoint of \( \overline{UW} \). If \( m \angle UXY = 60^\circ \), find \( m \angle V \).* **Diagram Analysis:** - The diagram presents a triangle \( \triangle UVW \). - Points \( X \) and \( Y \) are on line segments \( \overline{UV} \) and \( \overline{UW} \) respectively. - \( X \) is the midpoint of \( \overline{UV} \). - \( Y \) is the midpoint of \( \overline{UW} \). - Line segments \( \overline{XY} \), \( \overline{XV} \), and \( \overline{YW} \) are drawn inside the triangle. - Angle \( \angle UXY = 60^\circ \). **Question:** Calculate \( m \angle V \). **Solution Box:** The answer should be provided by calculating the necessary angle in the triangle configuration, given the information about midpoints and angle \( \angle UXY \).
### Transcription and Explanation for Educational Use #### Problem Statement: \( \overline{IJ} \) is a midsegment of \( \triangle FGH \). \\ Given: - \( IJ = 9 \) - \( FH = 12 \) - \( GH = 15 \) Find the perimeter of \( \triangle IJH \). #### Diagram Description: The diagram shows \( \triangle FGH \) with points \( F, G, \) and \( H \) being the vertices. Line segment \( \overline{IJ} \) is drawn parallel to side \( \overline{FH} \), indicating that it is a midsegment. Point \( I \) is on \( \overline{FH} \) and point \( J \) is on \( \overline{GH} \). #### Solution: To find the perimeter of \( \triangle IJH \): 1. Recognize that since \( \overline{IJ} \) is a midsegment, it is parallel to \( \overline{FG} \) and \( IJ = \frac{1}{2} FG \). 2. Given \( \overline{IJ} = 9 \), so \( FG = 2 \times 9 = 18 \). 3. Use the given lengths: - \( IJ = 9 \) - \( FH = 12 \) - \( GH = 15 \) The perimeter of \( \triangle IJH \) is simply the sum of all three segments: Perimeter \( = IJ + JH + HI \). However, given \( \overline{IJ} \) is a midsegment parallel and equal to half the side \( \overline{FG} \), you can calculate the remaining sides based on the triangle relationships given and midpoint properties if needed. #### Conclusion: The problem statement provides direct measures, and further midsegment properties guide the calculation of \( IJ \) being half of \( FG \). Since the problem setup names all necessary line segments for perimeter consideration, follow through with the computation based on known geometry principles for exact perimeter calculation.
Transcribed Image Text:### Transcription and Explanation for Educational Use #### Problem Statement: \( \overline{IJ} \) is a midsegment of \( \triangle FGH \). \\ Given: - \( IJ = 9 \) - \( FH = 12 \) - \( GH = 15 \) Find the perimeter of \( \triangle IJH \). #### Diagram Description: The diagram shows \( \triangle FGH \) with points \( F, G, \) and \( H \) being the vertices. Line segment \( \overline{IJ} \) is drawn parallel to side \( \overline{FH} \), indicating that it is a midsegment. Point \( I \) is on \( \overline{FH} \) and point \( J \) is on \( \overline{GH} \). #### Solution: To find the perimeter of \( \triangle IJH \): 1. Recognize that since \( \overline{IJ} \) is a midsegment, it is parallel to \( \overline{FG} \) and \( IJ = \frac{1}{2} FG \). 2. Given \( \overline{IJ} = 9 \), so \( FG = 2 \times 9 = 18 \). 3. Use the given lengths: - \( IJ = 9 \) - \( FH = 12 \) - \( GH = 15 \) The perimeter of \( \triangle IJH \) is simply the sum of all three segments: Perimeter \( = IJ + JH + HI \). However, given \( \overline{IJ} \) is a midsegment parallel and equal to half the side \( \overline{FG} \), you can calculate the remaining sides based on the triangle relationships given and midpoint properties if needed. #### Conclusion: The problem statement provides direct measures, and further midsegment properties guide the calculation of \( IJ \) being half of \( FG \). Since the problem setup names all necessary line segments for perimeter consideration, follow through with the computation based on known geometry principles for exact perimeter calculation.
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