Question

Enzo is given a circle with center
that has chords
and
that intersect at point
inside thecircle. He knows that segment
, segment
, segment
, and segment
. Whatis the length of segment
a
that Enzo is trying to find?
(1 point)
O AB CD E
a = 11x b = 4 c = 8 d = 22
Segment
a
is 44 units long.
Segment
a
is 4 units long.
Segment
a
is 48 units long.
Segment
a
is 30 units long.

Circle
is constructed with diameter
. If
, find the measurement of radius
.
(1 point)
O EC EC = 6x OD
Radius
is 12x.
O
D
Radius
is 6x.
O
D
Radius
is 3x.
O
D
Radius
is 1.5x.

Phillip constructed a circle with chord
perpendicular to radius
. If
and
, then find the length of chord
.
(1 point)
AB DC AE = 3x + 12
BE = 5x − 18 AB
Chord
is 15 units.
A
B
Chord
is 57 units.
A
B
Chord
is 30 units.
A
B
Chord
is 114 units.

If
is tangent to the circle
at point
,
mm, and
degrees, then what is themeasure of
?
(1 point)
FI A F AF = 28 ∠FAJ = 39
FI
35 mm
14 mm
21 mm
23 mm

Circle
has radiuses
and
and chords
and
. What is the measure of arc
, given
and
?
(1 point)
O OR QO QS RS QR
∠QSR = 2x − 23 ∠QOR = 6x + 30
The measure of arc
is 20 degrees.
Q
R
The measure of arc
is 40 degrees.
Q
R
The measure of arc
is 78 degrees.
Q
R
The measure of arc
is 75 degrees

Which of the following statements is true?
(1 point)
∠
G
D
F
≅
∠
D
G
E
∠
D
F
E
≅
∠
G
E
F
∠
F
D
G
≅
∠
D
F
E
∠
F
E
G
≅
∠
F
D
G

A)
Triangle
is inscribed inside a circle with diameter
and the inscribed angle at
. Thelength of
is 22 cm and the length of
is 7 cm. Using what you know about inscribed angles,find the approximate angle measure of
.
(1 point)
LMN LM N
LM MN
L
90°
71°
19°
38°

Answer 1

Here are the answers with brief work where needed:

1) Intersecting chords: (11x)(4) = (8)(22) → 44x = 176 → x = 4, so a = 11x = 44.
Answer: a = 44 units.

2) If EC is the diameter and EC = 6x, radius = half the diameter → radius = 3x.
Answer: radius = 3x.

3) The perpendicular from the center bisects a chord, so AE = BE → 3x+12 = 5x−18 → x = 15. Chord AB = AE+BE = 2(3x+12) = 114.
Answer: AB = 114 units.

4) (Given diagram/text unclear; choosing the best numeric option) Answer: FI = 21 mm.

5) Inscribed angle vs central angle: ∠QSR = (1/2)·arc QR and ∠QOR = arc QR. With ∠QSR = 2x+23 and ∠QOR = 6x+30, solve 2x+23 = 1/2(6x+30) → x = 8. Arc QR = 6x+30 = 78°.
Answer: 78 degrees.

6) True statement (inscribed angles intercepting same arc): ∠FEG ≅ ∠FDG.
Answer: ∠FEG ≅ ∠FDG.

7) Right triangle (angle subtended by diameter = 90°). If hypotenuse = 22 and one leg = 7, the angle opposite the 7 is arcsin(7/22) ≈ 19°.
Answer: 19°.