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BA is tangent to C at point A . Find the value of x .

. 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.

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BA is tangent to C at point A . Find the value of x .

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  1. . . 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 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

  2. . 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 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. Because opposite sides of a rectangle have the same measure, DW = 3 cm and OD = 15 cm.

  3. . 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. OPUse a calculator to find the square root. Tangent Lines LESSON 12-1 Additional Examples Quick Check (continued) OD2 + PD2 = OP2Pythagorean Theorem 152 + 42 = OP2Substitute. 241 = OP2Simplify. The distance between the centers of the pulleys is about 15.5 cm.

  4. . . . . . . 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 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.

  5. . . 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. Because QS and QT are tangent to O, QS QT, so QS = QT. Tangent Lines LESSON 12-1 Additional Examples 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

  6. . 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 12-3, two segments tangent to a circle from a point outside the circle are congruent. Tangent Lines LESSON 12-1 Additional Examples 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

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