1 / 18

Geometric Postulates and Theorems: Understanding Area Calculations

Explore key postulates and theorems related to calculating areas of figures such as squares, triangles, rectangles, parallelograms, circles, and polygons. Understand how these formulas can be applied to solve practice problems efficiently.

jalene
Download Presentation

Geometric Postulates and Theorems: Understanding Area Calculations

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related