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TODAY IN GEOMETRY⦠Review: Learning Target : 10.4 You will use inscribed angles of circles Independent practice CH.10 QUIZ - FRIDAY
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REVIEW: Graph β³π΄π΅πΆ. π΄ 3, 4 π΅ β1, 5 πΆ 1, β2
Translate the Pre-Image and record the image points. (π₯, π¦)β(π₯β2, π¦+3) π΄β² π΅β² πΆβ² πππππ‘ π΄: 3, 4 β 3β2, 4+3 (3, 4)β(1, 7) π¨β²(π, π) πππππ‘ π΅: β1, 5 β β1β2, 5+3 (β1, 5)β(β3, 8) π©β²(βπ, π) πππππ‘ πΆ: 1, β2 β 1β2, β2+3 1, β2 β β1, 1 πͺβ²(βπ, π) π΄ π΅ πΆ
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REVIEW: Graph β³π΄π΅πΆ. π΄ 3, 4 π΅ β1, 5 πΆ 1, β2
Reflect the pre-image across the y-axis. RULE FOR RELECTION ACROSS Y-AXIS: (π₯, π¦)β(βπ₯, π¦) π΄β² π΅β² πΆβ² π΄ π΅ πΆ πππππ‘ π΄: 3, 4 β β3, 4 πππππ‘ π΅: β1, 5 β 1, 5 πππππ‘ πΆ: 1, β2 β β1, β2
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REVIEW: Graph β³π΄π΅πΆ. π΄ 3, 4 π΅ β1, 5 πΆ 1, β2
Reflect the pre-image across the line π₯=1. EACH POINT IN A REFLECTION IS EQUAL DISTANT FROM THE LINE OF REFLECTION π΄β² π΅β² πΆβ² πππππ‘ π΄: 7, 1 β β5, 1 πππππ‘ π΅: 3, 2 β β1, 2 πππππ‘ πΆ: 5, β5 β β3, β5 π΄ π΅ πΆ π₯=1
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SOME IMPORTANT DEFINITIONS-not on your notes:
INSCRIBED ANGLE: An angle whose vertex is on circle. INTERCEPTED ARC: The arc made from an inscribed angle.
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INSCRIBED ARC FORMULA: The measure of an inscribed angle is one half the measure of its intercepted arc. Intercepted arc Inscribed arc π₯ 2π₯
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PRACTICE:
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PRACTICE:
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Two inscribed angles that intercept the same arc are congruent.
π΅ Be able to see that these figures are the same! Shared Intercepted arc π΄ π΄ π΅ πΆ π· π· πΆ
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PRACTICE: Β° =ππΒ° 2 38Β° =ππΒ° ππΒ°
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The circle that contains the vertices is a circumscribed circle.
A polygon is an inscribed polygon if all of its vertices lie on the circle. The circle that contains the vertices is a circumscribed circle. Circumscribed circle Inscribed triangle Inscribed polygon
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A quadrilateral can be inscribed in a circle if and only if its opposite angles are supplementary.
πΉ πβ π·+πβ πΉ=180Β° πΊ πΈ πβ πΈ+πβ πΊ=180Β° π·
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Opposite angles of an inscribed polygon are supplementary:
PRACTICE: Opposite angles of an inscribed polygon are supplementary: a. π¦+75Β°=180Β° β β 75 π=πππΒ° π₯+80Β°=180Β° β β 80 π=πππΒ° b. 2π+2π=180Β° 4π=180Β° π=ππ 4π+2π=180Β° 6π=180Β° π=ππ
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Opposite angles of an inscribed polygon are supplementary:
PRACTICE: Opposite angles of an inscribed polygon are supplementary: 6. 8π₯+10π₯=180Β° 18π₯=180Β° π=ππ 5. π¦+68Β°=180Β° β β 68 π=πππΒ° π₯+82Β°=180Β° β β 82 π=ππΒ° π+2πβ6=180Β° 3cβ6=180Β° 3π=186 π=ππ
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HOMEWORK #4: Pg. 676: 3-16 Pg. 679: 40-46 If finished, work on other assignments: HW #1: Pg. 655: 3-20, 24-26, 30 HW #2: Pg. 661: 3-14, 17, 23 HW #3: Pg. 667: 3-15
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