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Tangent Lines

Tangent Lines. Lesson 12-1. Check Skills You’ll Need. (For help, go to the Skills Handbook, page 754 and Lesson 8-1.). Find each product. 1. ( p + 3) 2 2. ( w + 10) 2 3. ( m – 2) 2 Find the value of x . Leave your answer in simplest radical form. 4. 5. 6. Check Skills You’ll Need.

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Tangent Lines

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  1. Tangent Lines Lesson 12-1 Check Skills You’ll Need (For help, go to the Skills Handbook, page 754 and Lesson 8-1.) Find each product. 1. (p + 3)22. (w + 10)23. (m – 2)2 Find the value of x. Leave your answer in simplest radical form. 4.5.6. Check Skills You’ll Need 12-1

  2. Tangent Lines Lesson 12-1 Notes 12-1

  3. Tangent Lines Lesson 12-1 Notes 12-1

  4. Tangent Lines Lesson 12-1 Notes A common tangent is a line that is tangent to two circles. 12-1

  5. Tangent Lines Lesson 12-1 Notes A common tangent is a line that is tangent to two circles. 12-1

  6. Tangent Lines Lesson 12-1 Notes 12-1

  7. Tangent Lines Lesson 12-1 Notes 12-1

  8. Tangent Lines Lesson 12-1 Notes When a circle is inscribed in a triangle, as in the diagram, the triangle is circumscribed about the circle. Each side of the triangle is tangent to the circle. 12-1

  9. Tangent Lines Lesson 12-1 Notes 12-1

  10. . . Because BA is tangent to C, A must be a right angle. Use the Triangle Angle-Sum Theorem to find x. m A + m B + m C = 180 Triangle Angle-Sum Theorem Tangent Lines Lesson 12-1 Additional Examples Finding Angle Measures BA is tangent to C at point A. Find the value of x. 90 + 22 + x = 180 Substitute. 112 + x = 180 Simplify. x = 68 Solve. Quick Check 12-1

  11. . A belt fits tightly around two circular pulleys, as shown below. Find the distance between the centers of the pulleys. Round your answer to the nearest tenth. Draw OP. Then draw OD parallel to ZW to form rectangle ODWZ, as shown below. Because OZ is a radius of O, OZ = 3 cm. Tangent Lines Lesson 12-1 Additional Examples Real-World Connection Because opposite sides of a rectangle have the same measure, DW = 3 cm and OD = 15 cm. 12-1

  12. . (continued) Because ODP is the supplement of a right angle, ODP is also a right angle, and OPD is a right triangle. Because the radius of P is 7 cm, PD = 7 – 3 = 4 cm. OP 15.524175 Use a calculator to find the square root. Tangent Lines Lesson 12-1 Additional Examples Quick Check OD2 + PD2 = OP2Pythagorean Theorem 152 + 42 = OP2Substitute. 241 = OP2Simplify. The distance between the centers of the pulleys is about 15.5 cm. 12-1

  13. . . . . . . Draw the situation described in the problem. For PA to be tangent to O at A, A must be a right angle, OAP must be a right triangle, and PO2 = PA2 + OA2. PO2PA2 + OA2Is OAP a right triangle? 122 132 + 52Substitute. Because PO2PA2 + OA2, PA is not tangent to O at A. Tangent Lines Lesson 12-1 Additional Examples Finding a Tangent Quick Check O has radius 5. Point P is outside O such that PO = 12, and point A is on O such that PA = 13. Is PA tangent to O at A? Explain. 144 194 Simplify. 12-1

  14. . . Because QS and QT are tangent to O, QS QT, so QS = QT. Tangent Lines Lesson 12-1 Additional Examples Using Theorem 12-3 QS and QT are tangent to O at points S and T, respectively. Give a convincing argument why the diagonals of quadrilateral QSOT are perpendicular. Theorem 12-3 states that two segments tangent to a circle from a point outside the circle are congruent. OS = OT because all radii of a circle are congruent. Two pairs of adjacent sides are congruent. Quadrilateral QSOT is a kite if no opposite sides are congruent or a rhombus if all sides are congruent. By theorems in Lessons 6-4 and 6-5, both the diagonals of a rhombus and the diagonals of a kite are perpendicular. Quick Check 12-1

  15. . C is inscribed in quadrilateral XYZW. Find the perimeter of XYZW. XU = XR = 11 ft YS = YR = 8 ft ZS = ZT = 6 ft WU = WT = 7 ft By Theorem 11-3, two segments tangent to a circle from a point outside the circle are congruent. Tangent Lines Lesson 12-1 Additional Examples Circles Inscribed in Polygons p = XY + YZ + ZW + WXDefinition of perimeter p = XR + RY + YS + SZ + ZT + TW + WU + UXSegment Addition Postulate = 11 + 8 + 8 + 6 + 6 + 7 + 7 + 11 Substitute. = 64 Simplify. The perimeter is 64 ft. Quick Check 12-1

  16. . . . Trapezoid; the tangent line forms right angles at vertices H and J, so HA || JB. Because HA = JB, AHJB is not a parallelogram but a trapezoid. / Tangent Lines Lesson 12-1 Lesson Quiz PA and PB are tangent to C. Use the figure for Exercises 1–3. 1. Find the value of x. 2. Find the perimeter of quadrilateral PACB. 3. Find CP. HJ is tangent to A and to B. Use the figure for Exercises 4 and 5. 4. Find AB to the nearest tenth. 5. What type of special quadrilateral is AHJB? Explain how you know. 87.2 82 cm 29 cm 20.4 cm 12-1

  17. Tangent Lines Lesson 12-1 Check Skills You’ll Need Solutions 1. (p + 3)2 = (p + 3)(p + 3) = p2 + 3p + 3p + 32 = p2 + 6p + 9 2. (w + 10)2 = (w + 10)(w + 10) = w2 + 10w + 10w + 102 = w2 + 20w + 100 3. (m – 2)2 = (m – 2)(m – 2) = m2 – 2m – 2m + (–2)2 = m2 – 4m + 4 4. Use the Pythagorean Theorem: a2 + b2 = c2 (8)2 + (16)2 = x2 64 + 256 = x2x2 = 320 x = 320 = 26 • 5 = 23 5 = 8 5 5. Use the Pythagorean Theorem: a2 + b2 = c2x2 + (13)2 = (17)2 x2 + 169 = 289 x2 = 289 – 169 = 120 x = 120 = 22 • 2 • 3 • 5 = 2 2 • 3 • 5 = 2 30 6. Use the Pythagorean Theorem: a2 + b2 = c2x2 + (16)2 = (20)2 x2 + 256 = 400 x2 = 400 – 256 = 144 x = 144 = 122 = 12 12-1

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