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TODAY IN GEOMETRY… Review: Arc Length

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Presentation on theme: "TODAY IN GEOMETRY… Review: Arc Length"β€” Presentation transcript:

1 TODAY IN GEOMETRY… Review: Arc Length Learning Target : 11.3 You will use ratios to find areas of similar figures Independent practice CH.10 TEST – THURSDAY/FRIDAY!

2 REVIEW: Find π‘šπΆπ΅ Arc is twice the inscribed arc. 2. Angles around a circle add to 360Β° 6π‘₯ π‘₯+2 5π‘₯βˆ’11 =πŸ‘πŸ”πŸŽ 6π‘₯ π‘₯+10π‘₯βˆ’22=360 23π‘₯+15=360 βˆ’ 15 βˆ’ 15 23π‘₯=345 𝒙=πŸπŸ“ 3. π‘šπΆπ΅=6π‘₯+20 2(5π‘₯βˆ’11) = =𝟏𝟏𝟎°

3 AREAS OF SIMILAR POLYGONS:
π‘ƒπ‘œπ‘™π‘¦π‘”π‘œπ‘› 𝐼~π‘ƒπ‘œπ‘™π‘¦π‘”π‘œπ‘› 𝐼𝐼 𝑏 π‘Ž Polygon I Polygon II POLYGON I π‘Ž RATIO OF SIDES: = POLYGON II 𝑏 POLYGON I π‘Ž RATIO OF PERIMETERS: = POLYGON II 𝑏 POLYGON I π‘Ž 2 RATIO OF AREAS: = POLYGON II 𝑏 2

4 EXAMPLE: In the diagram, βˆ†π΄π΅πΆ~βˆ†π·πΈπΉ. Find the indicated ratio.
Ratio (red to blue) of the perimeters of the areas 8 = 2 3 a. Ratio of Perimeter = 12 2 2 = 4 9 b. Ratio of Areas = 3 2

5 PRACTICE: In the diagram, πΉπ‘–π‘”π‘’π‘Ÿπ‘’ 1~πΉπ‘–π‘”π‘’π‘Ÿπ‘’ 2. Find the indicated ratio.
Ratio of the perimeters Ratio of the areas 18 10 Figure 1 Figure 2 18 = 9 5 a. Ratio of Perimeter = 10 9 2 = 81 25 b. Ratio of Areas = 5 2

6 EXAMPLE: Rectangles I and II are similar
EXAMPLE: Rectangles I and II are similar. The perimeter of Rectangle I is 66 inches. Rectangle II has a perimeter of 110 inches and an area of 700 𝑖𝑛 2 . Find the area of Rectangle I. 66 = 3 5 Ratio of Perimeter = π‘…π‘’π‘π‘‘π‘Žπ‘›π‘”π‘™π‘’ 𝐼 π‘…π‘’π‘π‘‘π‘Žπ‘›π‘”π‘™π‘’ 𝐼𝐼 = Ratio of Areas = π‘…π‘’π‘π‘‘π‘Žπ‘›π‘”π‘™π‘’ 𝐼 π‘…π‘’π‘π‘‘π‘Žπ‘›π‘”π‘™π‘’ 𝐼𝐼 = Set up proportions of areas = Cross multiply π‘₯=9(700) 25π‘₯=6300 Divide 𝒙=πŸπŸ“πŸ π’Šπ’ 𝟐 110 3 2 = 9 25 5 2 9 π‘₯ 25 700

7 EXAMPLE 2: Find the ratio of the perimeter (red to blue) and area
EXAMPLE 2: Find the ratio of the perimeter (red to blue) and area. Then find the unknown area. 12𝑓𝑑 𝐴=64 𝑓𝑑 2 16𝑓𝑑 16 = 4 3 Ratio of Perimeter = π‘Ÿπ‘’π‘‘ 𝑏𝑙𝑒𝑒 = Ratio of Areas = π‘Ÿπ‘’π‘‘ 𝑏𝑙𝑒𝑒 = Set up proportions of areas = Cross multiply 9(64)=16π‘₯ 576=16π‘₯ Divide 𝒙=πŸ‘πŸ” 𝒇𝒕 𝟐 12 4 2 = 16 9 3 2 16 64 9 π‘₯

8 PRACTICE: Find the ratio of the perimeter (red to blue) and area
PRACTICE: Find the ratio of the perimeter (red to blue) and area. Then find the unknown area. 10𝑓𝑑 14𝑓𝑑 10 = 5 7 Ratio of Perimeter = π‘Ÿπ‘’π‘‘ 𝑏𝑙𝑒𝑒 = Ratio of Areas = π‘Ÿπ‘’π‘‘ 𝑏𝑙𝑒𝑒 = Set up proportions of areas = Cross multiply π‘₯=25(441) 49π‘₯=11025 Divide 𝒙=πŸπŸπŸ“ 𝒇𝒕 𝟐 𝐴=441 𝑓𝑑 2 14 5 2 = 25 49 7 2 25 π‘₯ 49 441

9 PRACTICE: Find the ratio of the perimeter (red to blue) and area
PRACTICE: Find the ratio of the perimeter (red to blue) and area. Then find the unknown area. 𝐴=108 𝑓𝑑 2 12𝑓𝑑 8𝑓𝑑 12 = 3 2 Ratio of Perimeter = π‘Ÿπ‘’π‘‘ 𝑏𝑙𝑒𝑒 = Ratio of Areas = π‘Ÿπ‘’π‘‘ 𝑏𝑙𝑒𝑒 = Set up proportions of areas = Cross multiply 4(108)=9π‘₯ 432=9π‘₯ Divide 𝒙=πŸ’πŸ– 𝒇𝒕 𝟐 8 3 2 = 9 4 2 2 9 108 4 π‘₯

10 HOMEWORK #10: Pg. 740: 3-14 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 HW #4: Pg. 676: 3-16 Pg. 679: 40-46 HW #5: Pg. 683: 3-13 HW #6: Pg. 692: 3, 4, 6, 9-14 HW #7: Pg. 702: 3-25 HW #8: Pg. 749: 3-7, 11-13, 15-23 HW #9: Pg. 758: 11-17, 26-31


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