Question
Geometry B Semester Exam
11 of 3011 of 30 Items
Question
Use the image to answer the question.
A circle has a center point marked as upper G. Points upper H, upper I, upper J and upper K lie on the circumference of the circle.
If HI≅IJ
and m∠KIJ=35
degrees, then what is m∠KGH
?
(1 point)
Responses
110 degrees
110 degrees
17.5 degrees
17.5 degrees
140 degrees
140 degrees
70 degrees
70 degrees
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11 of 3011 of 30 Items
Question
Use the image to answer the question.
A circle has a center point marked as upper G. Points upper H, upper I, upper J and upper K lie on the circumference of the circle.
If HI≅IJ
and m∠KIJ=35
degrees, then what is m∠KGH
?
(1 point)
Responses
110 degrees
110 degrees
17.5 degrees
17.5 degrees
140 degrees
140 degrees
70 degrees
70 degrees
Skip to navigation
Answers
GPT-4o mini
To find the measure of angle \( m\angle KGH \), we can use the properties of the circle and the information provided:
1. Since \( HI \cong IJ \), segments \( HI \) and \( IJ \) are equal in length. This indicates that triangle \( HIJ \) is isosceles, with \( H \) and \( J \) being the base angles.
2. Given that \( m\angle KIJ = 35^\circ \), and because \( HI \cong IJ \), we know that \( m\angle HJI = m\angle KIH \) (the other two angles in triangle \( HIJ \)).
3. In triangle \( HIJ \):
\[
m\angle HJI + m\angle KIJ + m\angle KIH = 180^\circ
\]
Since \( m\angle HJI = m\angle KIH \), we can denote them both as \( x \):
\[
x + 35^\circ + x = 180^\circ
\]
\[
2x + 35^\circ = 180^\circ
\]
\[
2x = 180^\circ - 35^\circ
\]
\[
2x = 145^\circ
\]
\[
x = 72.5^\circ
\]
Therefore, \( m\angle HJI = m\angle KIH = 72.5^\circ \).
4. Now, we move to determine \( m\angle KGH \). Since \( KGH \) is an angle formed at the center \( G \) by the points on the circumference \( H \) and \( J \):
- The measure of an angle at the center (such as \( \angle KGH \)) is twice the measure of the angle at the circumference subtended by the same arc. Thus, \( m\angle KGH = 2 \times m\angle KIJ \).
5. Therefore:
\[
m\angle KGH = 2 \times m\angle KIJ = 2 \times 35^\circ = 70^\circ
\]
Thus, the measure of \( m\angle KGH \) is \( 70^\circ \).
The correct response is:
**70 degrees**.
1. Since \( HI \cong IJ \), segments \( HI \) and \( IJ \) are equal in length. This indicates that triangle \( HIJ \) is isosceles, with \( H \) and \( J \) being the base angles.
2. Given that \( m\angle KIJ = 35^\circ \), and because \( HI \cong IJ \), we know that \( m\angle HJI = m\angle KIH \) (the other two angles in triangle \( HIJ \)).
3. In triangle \( HIJ \):
\[
m\angle HJI + m\angle KIJ + m\angle KIH = 180^\circ
\]
Since \( m\angle HJI = m\angle KIH \), we can denote them both as \( x \):
\[
x + 35^\circ + x = 180^\circ
\]
\[
2x + 35^\circ = 180^\circ
\]
\[
2x = 180^\circ - 35^\circ
\]
\[
2x = 145^\circ
\]
\[
x = 72.5^\circ
\]
Therefore, \( m\angle HJI = m\angle KIH = 72.5^\circ \).
4. Now, we move to determine \( m\angle KGH \). Since \( KGH \) is an angle formed at the center \( G \) by the points on the circumference \( H \) and \( J \):
- The measure of an angle at the center (such as \( \angle KGH \)) is twice the measure of the angle at the circumference subtended by the same arc. Thus, \( m\angle KGH = 2 \times m\angle KIJ \).
5. Therefore:
\[
m\angle KGH = 2 \times m\angle KIJ = 2 \times 35^\circ = 70^\circ
\]
Thus, the measure of \( m\angle KGH \) is \( 70^\circ \).
The correct response is:
**70 degrees**.