Answer:
- A: 150°; B: 60π in²
- 45°
- x = 5
- x = 4
- center: B. (-2, 7); radius = 6
- 2: 77°; 3: 85°; 4: 103°
- x = 3.5
- 63°
Explanation:
1. The total of central angles in a circle is 360°. The smaller sector (and the arc subtended) has measure 360° -210° = 150°.
The area is given by the formula ...
A = (1/2)r²θ . . . . where θ is in radians
150° = (150/180)π radians, so the area of the smaller sector is ...
A = (1/2)(12 in)²(5π/6) = 60 in²
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2. The measure of an inscribed angle is half the measure of the arc it subtends. θ subtends an arc of 90°, so has measure 90°/2 = 45°.
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3. LQ is the perpendicular bisector of MP, so ...
MT = TP
5x -6 = 2x +9
3x = 15 . . . . add 6-2x
x = 5
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4. Tangents to the same circle from the same point are equal length.
AB = AD
3x +10 = 7x -6
16 = 4x . . . . . . . add 6-3x
4 = x
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5. The standard form is ...
(x -h)² +(y -k)² = r²
Comparing to the given equation, we see ...
- -h = 2 ⇒ h = -2
- -k = -7 ⇒ k = 7
- r² = 36 ⇒ r = 6
The center is (h, k) = (-2, 7) (choice B); the radius is 6.
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6. As in problem 2, the inscribed angle is half the measure of the arc it subtends. In an inscribed quadrilateral, that fact also means that opposite angles are supplementary.
angle 2 = (64° +90°)/2 = 77°
angle 3 = 180° -95° = 85°
angle 4 = 180° -77° = 103°
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7. Chords of the same length are the same distance from the center of the circle:
3x +1 = x +8
2x = 7 . . . . subtract x+1
x = 3.5
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8. You can work this problem several ways, or you can simply "cut to the chase." The measure of x is the supplement of the subtended arc.
x = 180° -117°
x = 63°
You can also use the fact that the tangents make right angles with the radii and the sum of angles of a quadrilateral is 360°. That gets you to ...
117° +90° +x° +90° = 360°
x = 180° -117°
Or, you can use the fact that the external angle (x°) is half the difference of the long arc and short arc:
x° = (1/2)((360° -117°) -117°) = (1/2)(360° -2(117°)) = 180° -117°