Question
Naveed was given a circle with the chords FG
and HK
that intersect at point E
inside the circle. If the length of FE=12
, EG=7
, and EK=21
, investigate this circle and find the length of segment HE
.(1 point)
Responses
Segment HE
is 4 units long.
Segment upper H upper E is 4 units long.
Segment HE
is 19 units long.
Segment upper H upper E is 19 units long.
Segment HE
is 25 units long.
Segment upper H upper E is 25 units long.
Segment HE
is 84 units long.
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Two line segments labeled upper A upper B and upper C upper D intersect inside of a circle.
Qingyun has constructed a circle with center O
and chords AB
and CD
that intersect at point E
inside the circle. According to her measurements, segment a=13
mm, segment b=20
mm, and segment d=10
mm. What is the length of chord CD
?
(1 point)
Responses
Chord CD
is 33 mm.
Chord upper C upper D is 33 mm.
Chord CD
is 260 mm.
Chord upper C upper D is 260 mm.
Chord CD
is 26 mm.
Chord upper C upper D is 26 mm.
Chord CD
is 36 mm.
Chord upper C upper D is 36 mm.
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Question
Use the image to answer the question.
A circle. Three closed points on the circle’s edge, a closed point for the circle’s center, and a closed point for the place where two lines intersect and form a right angle, are all labeled.
Celestina constructed a circle with center C
and radius DC
. She then constructed chord AB
to be perpendicular to radius DC
at point E
. She measured segment AE
to be 8 inches. Identify the length of segment BE
.
(1 point)
Responses
Segment BE
is 4 inches.
Segment upper B upper E is 4 inches.
Segment BE
is 32 inches.
Segment upper B upper E is 32 inches.
Segment BE
is 16 inches.
Segment upper B upper E is 16 inches.
Segment BE
is 8 inches.
Segment upper B upper E is 8 inches.
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Use the image to answer the question.
Four closed points on a circle are marked and labeled, as is a closed point outside of the circle and one at the center. Two lines, with arrows at both ends, cut through the edge of the circle at 4 closed points and intersect outside of it.
In circle M
, secant line SQ
and secant line SU
intersect at point S
outside the circle. If segment SQ=21
, segment SR=8
, and segment ST=4
, find the length of segment SU
.
(1 point)
Responses
Segment SU
is 168 units long.
Segment upper S upper U is 168 units long.
Segment SU
is 13 units long.
Segment upper S upper U is 13 units long.
Segment SU
is 38 units long.
Segment upper S upper U is 38 units long.
Segment SU
is 42 units long.
Segment upper S upper U is 42 units long.
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Question
Use the image to answer the question.
A circle has center point upper F. The circumference point upper L is at the left of the circle and the circumference point upper M is at the right of the circle. Point upper K is above the circle in line with upper F.
Identify and use the relationship between the radius and a line tangent to the circle, where the radius intersects the circle, to solve the following. If KM¯¯¯¯¯¯¯¯¯¯
and KL¯¯¯¯¯¯¯¯
are tangent to the circle at L
and M
, ∠LFK=23 degrees
, and KM¯¯¯¯¯¯¯¯¯¯=47 feet
, how long is LF¯¯¯¯¯¯¯
to the nearest foot?
(1 point)
Responses
120 ft.
120 ft.
111 ft.
111 ft.
51 ft.
51 ft.
20 ft.
20 ft.
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and HK
that intersect at point E
inside the circle. If the length of FE=12
, EG=7
, and EK=21
, investigate this circle and find the length of segment HE
.(1 point)
Responses
Segment HE
is 4 units long.
Segment upper H upper E is 4 units long.
Segment HE
is 19 units long.
Segment upper H upper E is 19 units long.
Segment HE
is 25 units long.
Segment upper H upper E is 25 units long.
Segment HE
is 84 units long.
Use the image to answer the question.
Two line segments labeled upper A upper B and upper C upper D intersect inside of a circle.
Qingyun has constructed a circle with center O
and chords AB
and CD
that intersect at point E
inside the circle. According to her measurements, segment a=13
mm, segment b=20
mm, and segment d=10
mm. What is the length of chord CD
?
(1 point)
Responses
Chord CD
is 33 mm.
Chord upper C upper D is 33 mm.
Chord CD
is 260 mm.
Chord upper C upper D is 260 mm.
Chord CD
is 26 mm.
Chord upper C upper D is 26 mm.
Chord CD
is 36 mm.
Chord upper C upper D is 36 mm.
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Question
Use the image to answer the question.
A circle. Three closed points on the circle’s edge, a closed point for the circle’s center, and a closed point for the place where two lines intersect and form a right angle, are all labeled.
Celestina constructed a circle with center C
and radius DC
. She then constructed chord AB
to be perpendicular to radius DC
at point E
. She measured segment AE
to be 8 inches. Identify the length of segment BE
.
(1 point)
Responses
Segment BE
is 4 inches.
Segment upper B upper E is 4 inches.
Segment BE
is 32 inches.
Segment upper B upper E is 32 inches.
Segment BE
is 16 inches.
Segment upper B upper E is 16 inches.
Segment BE
is 8 inches.
Segment upper B upper E is 8 inches.
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Use the image to answer the question.
Four closed points on a circle are marked and labeled, as is a closed point outside of the circle and one at the center. Two lines, with arrows at both ends, cut through the edge of the circle at 4 closed points and intersect outside of it.
In circle M
, secant line SQ
and secant line SU
intersect at point S
outside the circle. If segment SQ=21
, segment SR=8
, and segment ST=4
, find the length of segment SU
.
(1 point)
Responses
Segment SU
is 168 units long.
Segment upper S upper U is 168 units long.
Segment SU
is 13 units long.
Segment upper S upper U is 13 units long.
Segment SU
is 38 units long.
Segment upper S upper U is 38 units long.
Segment SU
is 42 units long.
Segment upper S upper U is 42 units long.
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Question
Use the image to answer the question.
A circle has center point upper F. The circumference point upper L is at the left of the circle and the circumference point upper M is at the right of the circle. Point upper K is above the circle in line with upper F.
Identify and use the relationship between the radius and a line tangent to the circle, where the radius intersects the circle, to solve the following. If KM¯¯¯¯¯¯¯¯¯¯
and KL¯¯¯¯¯¯¯¯
are tangent to the circle at L
and M
, ∠LFK=23 degrees
, and KM¯¯¯¯¯¯¯¯¯¯=47 feet
, how long is LF¯¯¯¯¯¯¯
to the nearest foot?
(1 point)
Responses
120 ft.
120 ft.
111 ft.
111 ft.
51 ft.
51 ft.
20 ft.
20 ft.
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Answers
There are no human answers yet.
Answered by
GPT-4o mini
To solve the problems presented, we can use properties of circles and segments.
**Problem 1: Finding Segment HE**
In the first scenario with chords FG and HK intersecting at point E, we can use the intersecting chords theorem which states that if two chords intersect, the products of the lengths of the segments of each chord are equal. Here, we have:
- \( FE \times EG = HE \times EK \)
Substituting the known values:
- \( 12 \times 7 = HE \times 21 \)
Calculating:
- \( 84 = HE \times 21 \)
Now, solve for HE:
- \( HE = \frac{84}{21} = 4 \)
Thus, Segment HE is **4 units long**.
---
**Problem 2: Finding Length of Chord CD**
In the second scenario with chords AB and CD intersecting at point E, we apply the same property. Given:
- \( a = 13 \text{ mm}, b = 20 \text{ mm}, d = 10 \text{ mm} \)
Using:
- \( a \times b = c \times d \) (where c is the length we want to find for chord CD)
So,
- \( 13 \times 20 = c \times 10 \)
Calculating:
- \( 260 = c \times 10 \)
Now, solve for c:
- \( c = \frac{260}{10} = 26 \)
Thus, the length of chord CD is **26 mm**.
---
**Problem 3: Finding Length of Segment BE**
For the third scenario, using the properties of perpendicular segments from the center to the chord, we know:
- In a right triangle (triangle CBE, where C is center and E is where the radius meets chord AB), using the Pythagorean theorem:
Since \( AE = 8 \) inches and the whole length AB equals \( AE + BE \).
The radius \( CE^2 = AE^2 + BE^2 \).
We need to find BE. We don't have the radius, but we know the segments from geometry.
Assuming the total length of the chord AB is \( 16 \) because \( AE + BE = 8 + BE \), we must solve:
- If \( BE \) is \( x \), then \( 8 + x = 16 \).
So \( x = 8 \).
Thus, Segment BE is **8 inches long**.
---
**Problem 4: Finding Length of Segment SU**
For the fourth problem regarding the secant lines, we apply the external segment theorem:
The lengths of the segments are related to those of the two intersecting secants as follows:
- \( SQ \cdot SR = SU \cdot ST \)
The known values give us:
- \( 21 \cdot 8 = SU \cdot 4 \)
Calculating:
- \( 168 = SU \cdot 4 \)
Now resolve for SU:
- \( SU = \frac{168}{4} = 42 \)
So, Segment SU is **42 units long**.
---
**Problem 5: Finding Length of LF**
In the last problem, we apply the relationship of tangent lines to a circle:
Given \( KM \) and \( KL \) are tangents to the circle, and the angle \( \angle LFK = 23^\circ \), we see that forming the right triangle FKL means that:
Using the tangent function,
- Since we need \( LF \), we use:
\( \tan(23^\circ) = \frac{KL}{LF} = \frac{47}{LF} \)
Rearranging gives us:
- \( LF = \frac{47}{\tan(23^\circ)} \)
Calculating \( \tan(23^\circ) \):
- \( LF \approx 47 / 0.4245 \approx 110.9 \) feet
Rounding to the nearest foot gives us \( LF \approx 111 \).
Thus, \( LF \) is approximately **111 ft**.
---
If you need further assistance or clarification on any point or problem, please let me know!
**Problem 1: Finding Segment HE**
In the first scenario with chords FG and HK intersecting at point E, we can use the intersecting chords theorem which states that if two chords intersect, the products of the lengths of the segments of each chord are equal. Here, we have:
- \( FE \times EG = HE \times EK \)
Substituting the known values:
- \( 12 \times 7 = HE \times 21 \)
Calculating:
- \( 84 = HE \times 21 \)
Now, solve for HE:
- \( HE = \frac{84}{21} = 4 \)
Thus, Segment HE is **4 units long**.
---
**Problem 2: Finding Length of Chord CD**
In the second scenario with chords AB and CD intersecting at point E, we apply the same property. Given:
- \( a = 13 \text{ mm}, b = 20 \text{ mm}, d = 10 \text{ mm} \)
Using:
- \( a \times b = c \times d \) (where c is the length we want to find for chord CD)
So,
- \( 13 \times 20 = c \times 10 \)
Calculating:
- \( 260 = c \times 10 \)
Now, solve for c:
- \( c = \frac{260}{10} = 26 \)
Thus, the length of chord CD is **26 mm**.
---
**Problem 3: Finding Length of Segment BE**
For the third scenario, using the properties of perpendicular segments from the center to the chord, we know:
- In a right triangle (triangle CBE, where C is center and E is where the radius meets chord AB), using the Pythagorean theorem:
Since \( AE = 8 \) inches and the whole length AB equals \( AE + BE \).
The radius \( CE^2 = AE^2 + BE^2 \).
We need to find BE. We don't have the radius, but we know the segments from geometry.
Assuming the total length of the chord AB is \( 16 \) because \( AE + BE = 8 + BE \), we must solve:
- If \( BE \) is \( x \), then \( 8 + x = 16 \).
So \( x = 8 \).
Thus, Segment BE is **8 inches long**.
---
**Problem 4: Finding Length of Segment SU**
For the fourth problem regarding the secant lines, we apply the external segment theorem:
The lengths of the segments are related to those of the two intersecting secants as follows:
- \( SQ \cdot SR = SU \cdot ST \)
The known values give us:
- \( 21 \cdot 8 = SU \cdot 4 \)
Calculating:
- \( 168 = SU \cdot 4 \)
Now resolve for SU:
- \( SU = \frac{168}{4} = 42 \)
So, Segment SU is **42 units long**.
---
**Problem 5: Finding Length of LF**
In the last problem, we apply the relationship of tangent lines to a circle:
Given \( KM \) and \( KL \) are tangents to the circle, and the angle \( \angle LFK = 23^\circ \), we see that forming the right triangle FKL means that:
Using the tangent function,
- Since we need \( LF \), we use:
\( \tan(23^\circ) = \frac{KL}{LF} = \frac{47}{LF} \)
Rearranging gives us:
- \( LF = \frac{47}{\tan(23^\circ)} \)
Calculating \( \tan(23^\circ) \):
- \( LF \approx 47 / 0.4245 \approx 110.9 \) feet
Rounding to the nearest foot gives us \( LF \approx 111 \).
Thus, \( LF \) is approximately **111 ft**.
---
If you need further assistance or clarification on any point or problem, please let me know!
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