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Chapter 11.

Chapter 11. Camilo Henao Dylan Starr. Postulate 17 & 18. Postulate 17: The area of a square is the square of the length of a side (pg.423) A=s 2 Postulate 18 (Area congruence Postulate): If two figures are congruent, then they have the same area. 3. Example Side= 3 Area=3 Area=9 .

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Chapter 11.

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  1. Chapter 11. Camilo Henao Dylan Starr

  2. Postulate 17 & 18 • Postulate 17: The area of a square is the square of the length of a side (pg.423) • A=s2 • Postulate 18(Area congruence Postulate): If two figures are congruent, then they have the same area 3 Example Side= 3 Area=3 Area=9 3 2 By SAS similarity these triangles are similar and postulate 18 they must have the same area

  3. Postulate 19 (Area addition postulate) • The are of a region is the sum of the areas of its non-overlapping parts (pg.424) Area 1 Area 2 Area 3 Area 1 + Area 2 + Area 3= area of the overall figure

  4. Practice Problem 5 8 8 Sector 1 4 4 Area of sector 1: 5 8=40 Area of sector 2: 6 13= 78 Area of sector 3: 5 5= 25 Area = 143 6 Sector 2 6 4 4 Sector 3 5 5 5

  5. 11-1 • Theorem 11-1: the area of a rectangle equals the product of its base and height. (pg.424) • (A=b h) 15 Example A=bh A=(15)(4) A=60 4

  6. 11-2 • Theorem 11-2: The area of a parallelogram equals the product of a base and the height to that base.(pg.429) • (A=bh) Example 10 A=bh A=(10)(5) A=50 5

  7. Practice problem 4 In this parallelogram the right triangle forms a 45-45-90 triangle By the ITT Th. The height of the parallelogram is 4 A= bh A=(10)(4) A=40 45 10

  8. 11-3 • Theorem 11-3: the area of a triangle equals half the product of a base and its height to the base.(pg.429) • (A= ½ bh) Example 4 A= ½ bh A=(1/2)(10)(4) A=(5)(4) A=20 5 5 I___________________I 10

  9. Practice problem In this triangle to find the height you have to use the 30-60-90 triangle that has been formed by the height The height will end up being 4 by dividing side 8 by 2 So the are would be A: ½ bh A: 1/2(8)(4) A:16 8 30 8

  10. 11-4 • Theorem 11-4: The area of a rhombus equals half the product of its diagonals.(pg.430) • A= ½ x (d1 x d2) Example A= ½ x (10x4) A=20 2 5 5 2

  11. 11-5 • Theorem 11-5:the area of a trapezoid equals half the product of the height and the sum of the bases.(pg.435) • A= ½ h (b1 + b2) 5 8 Example A: ½ 8(17) A:68 12

  12. Practice problem 6 To find the height and only given the angle 72 and the side 5 you must use the sin of 72 = x/5 (Sin 72)x5=X Then with the height you can continue with the normal Area formula A: ½ h (b1+b2) A: ½ (12) 72 5 8

  13. Formulas for circles • Circumference of a circle C=2∏r or C= ∏d • C=2 ∏4 or C=∏8 • C=8∏ • Area of a circle A= ∏r2 • A=∏42 • A= 16∏ 4

  14. Arc lengths and Areas of sectors • A sector of a region bounded by two radii and an arc of the circle • Formulas: • Length of AB: • x/360 2∏r • Area of sector AOB: • x/360 ∏r2 A x o B

  15. Practice problems • Find the area of the shaded area • find the area of the triangle and the area of the sector then subtract the Area of the triangle from the area of the sector Area of the triangle: ½ bh A: ½ (4)(4) A:8 Area of the sector: 90/360 ∏ 42 A:4∏ Area of shaded area : 4∏ -8 A 90 Shaded area 4 B

  16. 11-6 • Theorem 11-6: The area of a regular polygon is equal to half the product of the apothem and the perimeter.(pg.441) • A= ½ ap • A= ½ 4 (36) • A=72 6 NOT APOTHEM! Apothem=4

  17. Practice problem What is the area of the polygon ? Given: Apothem =6√3 By the 30-60-90 postulate One of the sides of the polygon is 2 = 12 A= ½ ap A= ½ (6√3) (72) A=216√3 30 6√3

  18. 11-7 • If the scale factor of two similar figures is a:b, then: (pg.457) • (1) the ratio of the perimeters is a:b • (2) the ratio of the areas is a2: b2 • Example: the ratio of the perimeter of two polygons is 2:3 what is the ratio of their areas • 22: 32 • 4 : 9

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