3. What is area? Area is a count of how many one by one unit squares cover a figure. A unit square is a square that is one unit long by one unit wide. It can be 1'x1', 1 m x 1 m, 1 yd x 1 yd, 1" x 1",...

4. Rectangle: Square: Parallelogram: Triangle: Circle: BH 1/2BH  r2 r2

5. What is a 3- Dimensional Shape?

6. What is the base?

7. A measure of the number of square units needed to cover the outside of a figure.

8. Ask yourself 4 Questions What type of prism it is? What is the Base Shape? What is the Area of the Base? What is the Perimeter of the Base? The Surface Area is… (Area of the Base) 2 + Perimeter(Height of Prism) 2(A) + P(H)

10. What type of prism it is? -Rectangular prism What is the Base Shape? -Base shape: Rectangle, base ‘B’ and height‘H ' What is the Area of the Base? -Area of base: B × h What is the Perimeter of the Base? - P=2B+2h The Surface Area is… 2Bh+ (2B+2h)H Base Height 2(Area) + P(Height of the Prism) Height Base

11. SA=2(Area of the Base)+ (Perimeter of the Base)Height 2BH+ 2(B+H)H The Base is a Area=B*H= 12cm * 4cm = 48c m² The Perimeter of the Base: 2(12cm)+2(4cm) = 32cm SA=2(48c m² ) + 32cm(6cm) = 96c m² +192cm ² = 288c m² 12cm 4cm SA=2(Area) + P(H) : Rectangle 12cm 4cm

12. Net

13. What shapes do you see? Two Squares Four Rectangles

14. Area of Side= BH =6cm*4cm = 24 cm 2 Area of Rectangle= BH Bottom= 12cm*4cm = 48 cm 2 Front face= 12cm*6cm = 72 cm 2 Front Bottom Side 12cm 6cm 4cm 6cm 4cm

15. But FIRST How many squares do we have? 2(24 cm 2 )= 48 cm 2 How many front face rectangles do we have? 2(72 cm 2 )= 144 cm 2 How many bottom rectangles do we have? 2(48 cm 2 )= 96 cm 2 To find the surface area we need to add them all together!

16. 2(144 cm 2 ) =288 cm 2 We have: 2 bottoms +2 fronts+ 2 sides 2(48cm 2 ) + 2(72cm 2 ) + 2(24cm 2 ) 2(48cm 2 +72cm 2 + 24cm 2 ) How did we get: 48 = Bb * Hb 72=Bf*Hf 24=Bs*Hs 2(Bb*Hb) + 2(Bf*Hf) + 2(Bs+Hs) 2( Bb*Hb + Bf*Hf + Bs* Hs)

17. Surface Area Formula SA=2( Bb*Hb + Bf*Hf + Bs* Hs)

18. Find the Surface Area of the Rectangular Prism 2(Area of the Base) + (Perimeter of base) Height =2(7cm*4cm) + 2(7cm+4cm) 10cm =2(28c m² )+ (22cm) 10cm = 56c m² +220c m² =276c m² =2(10cm*4cm) + 2(10cm+4cm) 7cm =2(40c m² )+ (28cm) 7cm =80c m² +196c m² =276c m² 7cm 4cm 2(Area of the Base) + (Perimeter of base) Height 4cm 10cm

19. Find the Surface Area of the Rectangular Prism You can either : Add up all the areas OR 2(Area of the Base) + (Perimeter of base) Height Add up all the areas: Two squares- 2(7cm*4cm)= 56cm^2 Two front Rectangles- 2( 10cm*7cm)= 140cm^2 Two side Rectangles- 2( 1ocm*4cm)= 80cm^2 SA =56cm^2+ 140cm^2 + 80cm^2 = 276cm^2 2(Area of the Base) + (Perimeter of base) Height =2(4cm*7cm) + 2(7cm+4cm) 10cm =2(28cm^2)+ (22cm) 10cm =56cm^2+220cm^2 =276cm^2 =2(10cm*4cm) + 2(10cm+4cm) 7cm =2(40cm^2)+ (28cm) 7cm =80cm^2+196cm^2 =276cm^2

20. Ask your self those 4 questions What type of prism it is? What is the Base Shape? What is the Area of the Base? What is the Perimeter of the Base? The Surface Area is… -Triangular based prism -Triangle: base 'b', height 'h', and sides S 1, S 2 and S 3 -Area of base: ½B×H -Perimeter of base: S 1 + S 2 + S 3 BH + (S 1 + S 2 + S 3 )H 2(Area ) + P(H) Height Base S1 S2 S3

21. SA=2(Area of the Base)+ (Perimeter of the Base)Height BH + (S 1 + S 2 + S 3 )H The Base is a: ½(BH)=½( 6mm*5mm) =15m m² The Perimeter of the Base: 6mm+7mm+7mm =20mm SA=2(15m m² )+(20mm)8mm =30m m² +160m m² =190m m² 6mm 5mm 7mm SA=2(Area) + P(H) Triangle

22. What shapes do you see? Triangle Rectangle

23. Net

24. Area of Rectangle= BH Area of Triangle= 1/2BH 6mm 8mm 7mm 8mm 5mm =½(6mm*5mm) = 15mm 2 = 6mm*8mm = 48mm 2 = 8mm*7mm = 56mm 2 Area of Rectangle= BH Bottom Side

25. We have: 2 triangles+ bottom rectangle+ 2 side rectangle How did we get: 30mm 2 = 2(BH/2)= BH 48mm 2 =Bb*H 2(Bt*H) 2*(6*5)/2 + (8*6) + 2(8*7) 2(15) + 48+ 2(56) 30+48+112=190mm 2 112mm 2 =

26. SA=(B*H) +(Bb*H)+2(Bs*H)

27. Find the Surface Area of the Triangular Prism 2(Area of the Bases) + (Perimeter Base) Height =2(1/2*4cm*3cm) + (3cm+4cm+5cm)11cm =12c m² + 13cm(11cm) =144 c m² 4cm 3cm 5cm

28. Find the Surface Area of the Triangular Prism Area of the Triangle= 1/2BH= ½(4cm*3cm)= 6cm^2 Area of Bottom Rectangle= BbH= 11cm*4cm=44cm^2 Area of the Side Rectangle= BsH= 11cm* 5cm= 55cm^2 Area of the Side Rectangle= BsH= 11cm*3cm=33cm^2 SA=12cm^2+44cm^2+55cm^2+33cm^2=144cm^2 2(Area of the Bases) + (Perimeter Base) Height =2(1/2*4cm*3cm) + (3cm+4cm+5cm)11cm =12cm^2 + 13cm(11cm) =144 cm^2 Add all the areas:

29. What shapes do you see? Circle Rectangle

30. Ask yourself those 4 questions What type of prism it is? What is the Base Shape? What is the Area of the Base? What is the Perimeter of the Base? The Surface Area is… -Circular based prism aka Cylinder -Base shape: Circle, radius 'R‘ -Area of base:  r² -Perimeter of base: 2  r 2  r² + 2  r H 2(Area) + P(H) radius

31. SA=2(Area of the Base)+ (Perimeter of the Base)Height The Base is a:  rea=  r² =  5in)² =25  in² The Perimeter of the Base= Circumference of the Base: 2  r =2  (5in) =10  in SA=2( 25  in² )+(  in  in =50  in² +  in²  in² 2  r² + 2  r H 5in SA=2(Area) + P(H) Circle

32. Rectangle

33. net

34. 7 in 5in 2r2r Area of a Circle=  r² =  (5in)² =25  in² Area of a Rectangle= BH= 2  rH =2  (5in)*(7in) =10  in  in = 70  in²

35. We have: 2 circle + 1 rectangle 2(25  in²)+ 70  in² SA=5 0  in² + 70  in²= 120in² How did we get: 5 0  in² =  r²) 70  in²=2  rH

36. SA=  r²) + 2  rH

37. Find the Surface Area of the Cylinder SA= 2(Area) + (Perimeter of base) Height =2(  (4cm)² ) + ( 2  (4cm)) 12cm =32  cm²+ 96  cm² =128  cm² 4cm

38. Find the Surface Area of the Cylinder Add up all the areas: Two circles= 2(  r² )= 2(  (4cm)² =32  cm² One Rectangle= BH=2  rH= 2  cm  cm) =96  cm² SA=32  cm² + 96  cm² SA= 128  cm² SA= 2(Area of the Base) + (Perimeter of base) Height =2(  (4cm)² ) + ( 2  (4cm)) 12cm =32  cm²+ 96  cm² =128  cm²

39. 2BH+ 2(B+H)H BH + (S 1 + S 2 + S 3 )H 2  r² + 2  r H In general the SA =Area of the Bases + (Perimeter Base) Height 2( Bb*Hb + Bf*Hf + Bs* Hs) (B*H) +(Bb*H)+2(Bs*H)  r²) + 2  rH

41. Find the Surface Area

47. Identify the Slant Height

48. Write a plan for finding the surface area of the pyramid

49. SA= Area of the Base + ½ Perimeter of the Base( Slant Height)

50. Why do we use Perimeter SA= Area of the Base + ½ Perimeter of Base(S) (8inx4)10in (32in)(10in) Area triangle= ½ BH = (8in)(10in) 8(10+10+10+10)= 8(40)= 320 Since I have four triangles: 8(10)+8(10)+8(10)+8(10) 10(8+8+8+8)= 10(32)= 320 320in ²

51. Why do we use Perimeter SA= Area of the Base + ½ Perimeter of Base(S) The perimeter of the Base times the Slant Height is the same as: Finding the area of each triangle How many triangles do I have? What is the height of each triangle? So if I add up all the sides of the square I get 32. And 32(10) is the same as 8(10)+8(10)+8(10)+8(10) If I factor out a 10_> 10(8+8+8+8)= 10(32) or if I factor out an 8(10+10+10+10)= 8(40)=320

52. What is the (S)lant height? SA= Area of the Base + ½ Perimeter of the Base( Slant Height) The Base is a square: Area= BH A= 8in(8in) A= 64 in ² The Perimeter of a square: =(8in)4 =32in SA= 64 in ² + ½ (32in)(10in) 10 in SA= 64 in ² + (16in)(10in) SA= 64 in ² + 160in ² SA= 224in ²

53. If the Slant Height is not given then we need to find it How do we find it? Pythagorean Thm : Leg^2+ leg^2= Hyptenuse^2

54. SA= Area of the Base + ½ Perimeter of Base(S) What is the (S)lant height? Use the Pythagorean Thm Leg ² + Leg ² = Hypotenuse ² (12in) ² + (16in) ² = Slant Height ² 144in ² +256in ² = Slant Height ² 400in ² = Slant Height ² Slant Height= 20in 16in 24in

55. SA= Area of the Base + ½ Perimeter of Base(S) Slant Height= 20 in Area of the Base=24in(24in) =576in ² Perimeter of the Base=24in(4 ) =96in SA= 576in ² + ½(96in)(20in) SA= 576in ² + (48*20) 24in 16in SA= 576in ² + 960 in ² SA= 1536in ²

57. A Cone is a Rotated Triangle A cone is made by rotating a triangle! The triangle has to be a right triangle, and it gets rotated around one of its two short sides. The side it rotates around is the axis of the cone.

59. The Base shape is a Circle Area=  r2 r2  rL SA=  r 2 +  r (Slant Height) radius Slant height

60. SA=  r 2 +  r (Slant Height) Find the Surface Area:

62. SA= 4πr 2