Download presentation
Presentation is loading. Please wait.
1
TODAY IN GEOMETRYβ¦ Review: Arc Length Learning Target : 11.3 You will use ratios to find areas of similar figures Independent practice CH.10 TEST β THURSDAY/FRIDAY!
2
REVIEW: Find ππΆπ΅ Arc is twice the inscribed arc. 2. Angles around a circle add to 360Β° 6π₯ π₯+2 5π₯β11 =πππ 6π₯ π₯+10π₯β22=360 23π₯+15=360 β 15 β 15 23π₯=345 π=ππ 3. ππΆπ΅=6π₯+20 2(5π₯β11) = =πππΒ°
3
AREAS OF SIMILAR POLYGONS:
ππππ¦πππ πΌ~ππππ¦πππ πΌπΌ π π Polygon I Polygon II POLYGON I π RATIO OF SIDES: = POLYGON II π POLYGON I π RATIO OF PERIMETERS: = POLYGON II π POLYGON I π 2 RATIO OF AREAS: = POLYGON II π 2
4
EXAMPLE: In the diagram, βπ΄π΅πΆ~βπ·πΈπΉ. 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, πΉπππ’ππ 1~πΉπππ’ππ 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 ππ 2 . Find the area of Rectangle I. 66 = 3 5 Ratio of Perimeter = π
πππ‘πππππ πΌ π
πππ‘πππππ πΌπΌ = Ratio of Areas = π
πππ‘πππππ πΌ π
πππ‘πππππ πΌπΌ = Set up proportions of areas = Cross multiply π₯=9(700) 25π₯=6300 Divide π=πππ ππ π 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. 12ππ‘ π΄=64 ππ‘ 2 16ππ‘ 16 = 4 3 Ratio of Perimeter = πππ πππ’π = Ratio of Areas = πππ πππ’π = Set up proportions of areas = Cross multiply 9(64)=16π₯ 576=16π₯ Divide π=ππ ππ π 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. 10ππ‘ 14ππ‘ 10 = 5 7 Ratio of Perimeter = πππ πππ’π = Ratio of Areas = πππ πππ’π = Set up proportions of areas = Cross multiply π₯=25(441) 49π₯=11025 Divide π=πππ ππ π π΄=441 ππ‘ 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. π΄=108 ππ‘ 2 12ππ‘ 8ππ‘ 12 = 3 2 Ratio of Perimeter = πππ πππ’π = Ratio of Areas = πππ πππ’π = Set up proportions of areas = Cross multiply 4(108)=9π₯ 432=9π₯ Divide π=ππ ππ π 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
Similar presentations
© 2024 SlidePlayer.com Inc.
All rights reserved.