Presentation on theme: "Section 6.4 Inscribed Polygons"— Presentation transcript:
1Section 6.4 Inscribed Polygons USING INSCRIBED ANGLESUSING PROPERTIES OF INSCRIBED POLYGONS
2USING INSCRIBED ANGLES An inscribed angle is an angle whose vertex is on a circle and whose sides contain chords of the circle. The arc that lies in the interior of an inscribed angle and has endpoints on the angle is called the intercepted arc of the angle.inscribedangleinterceptedarc
3USING INSCRIBED ANGLES THEOREMTHEOREM : Measure of an Inscribed AngleIf an angle is inscribed in a circle, then its measure ishalf the measure of its intercepted arc.CABDm ADB = m AB12mAB = 2m ADB
4Finding Measures of Arcs and Inscribed Angles Find the measure of the blue arc or angle.RSTQCSOLUTIONmQTS = 2m QRS = 2(90°) = 180°
5USING INSCRIBED ANGLES THEOREMADCBIf two inscribed angles of a circleintercept the same arc, then theangles are congruent.C D
6USING PROPERTIES OF INSCRIBED POLYGONS If all of the vertices of a polygon lie on a circle, the polygonis inscribed in the circle and the circle is circumscribedabout the polygon. The polygon is an inscribed polygonand the circle is a circumscribed circle.
7USING PROPERTIES OF INSCRIBED POLYGONS THEOREMS ABOUT INSCRIBED POLYGONSTHEOREM 10.10If a right triangle is inscribed in a circle, then the hypotenuse is a diameter ofthe circle. Conversely, if one side ofan inscribed triangle is a diameter of the circle, then the triangle is a right triangle and the angle opposite the diameter is the right angle.CABB is a right angle if and only ifAC is a diameter of the circle.
8. USING PROPERTIES OF INSCRIBED PLOYGONS THEOREMS ABOUT INSCRIBED POLYGONSTHEOREM 10.11CEDFGA quadrilateral can be inscribed in a circle if and only if its opposite angles are supplementary.D, E, F, and G lie on some circle, C, if and only ifm D + m F = 180° and m E + m G = 180°..
9. In the diagram, ABCD is inscribed in P. Using an Inscribed QuadrilateralPBCDA2y °3y °5x °3x °In the diagram, ABCD is inscribed in P.Find the measure of each angle..
10Using an Inscribed Quadrilateral SOLUTIONPBCDA2y °3y °5x °3x °ABCD is inscribed in a circle, so opposite angles are supplementary.3x + 3y = 1805x + 2y = 180
11y = 60 – x. Substitute this expression into the second equation. Using an Inscribed Quadrilateral3x + 3y = 1805x + 2y = 180PBCDA2y °3y °5x °3x °To solve this system of linear equations,you can solve the first equation for y to gety = 60 – x. Substitute this expression into the second equation.5x + 2y = 180Write second equation.5x + 2(60 – x) = 180Substitute 60 – x for y.5x – 2x = 180Distributive property3x = 60Subtract 120 from each side.x = 20Divide each side by 3.y = 60 – 20 = 40Substitute and solve for y.
12Using an Inscribed Quadrilateral PBCDA2y °3y °5x °3x °x = 20 and y = 40,so m A = 80°, m B = 60°,m C = 100°, and m D = 120°.
13HomeworkPage 207 – 20816 – 22 allReview for Quiz