1. Friday, 03 July 2015 of a Prism and Cylinder Surface area

2. The prism has 3 rectangular and 2 triangular faces. This is one way of drawing the net. This is a triangular prism. To find the surface area we need to be able to visualise the faces.

3. This is one way of drawing the net. This is a triangular prism. To find the surface area we need to be able to visualise the faces. The surface area is the sum of the areas of 3 rectangles and 2 triangles.

4. The cylinder has 2 flat surfaces and 1 curved surface. top How many surfaces does a solid cylinder have? base The curved surface unwraps to a rectangle. curved surface Unwrap the cylinder... to get this e.g.

5. Tip: No square, it’s a length ! To find the surface area of a cylinder we need some circle facts. Reminders: r Area =   r 2 Tip: Area, so a square! Circumference =   d When we are calculating the values, unless we are asked for an estimate, we use the  button on the calculator. We must work to at least one more decimal place or significant figure than we need in the final answer. ( Reminder: d = 2r )

6. top base 3 cm 8 cm r = 3 curved surface 8 cm r = 3 Area of the base =   r 2 = 28·27 cm 2 Length of rectangle =  d =  (6) = 18·85 cm 2 Total area = 150·80 + 28·27 + 28·27 = 207·3 cm 2 ( 1 d.p. ) Tip: Leave this answer on your calculator so you don’t have to type it in again at the next stage. e.g.Find the surface area, giving the answer correct to 1 decimal place. =  (3) 2 = 150·80 cm 2 dd Area of rectangle = 18·85  8

7. SUMMARY  The surface area of a solid cylinder is made up of: 2 circles 1 curved surface that unwraps to a rectangle. The area of each circle is  r 2. The length of the rectangle is  d.  The surface area of a triangular prism is made up of: 3 rectangles 2 triangles

8. EXERCISE 1.(a) Draw and label, with names and lengths, the triangle and 3 rectangles that make the different faces of the prism shown below. ( The drawings need not be to scale. ) (b)How many faces has the prism? (c)Find the surface area of the prism. 1m1m 30 cm 40 cm

9. Solution: EXERCISE 30 cm 40 cm By Pythagoras c 2 = 30 2 + 40 2 = 2500 = 50 cm c = √ 2500 cm c 50 cm 1m1m 30 cm 40 cm end face

10. 1m1m 30 cm 40 cm Solution: Area of 1 end = ½  30  40= 600 cm 2 Area of sloping face = 100  50= 5000 cm 2 Area of base = 100  30 = 3000 cm 2 Surface area = 600 + 600 + 5000 + 3000 + 4000 = 13200 cm 2 EXERCISE 1m1m 50 cm 40 cm 30 cm Area of vertical face = 100  40 = 4000 cm 2 The prism has 5 faces 30 cm 40 cm end face sloping face vertical face base 50 cm

11. EXERCISE 2(i)Draw and label the 3 surfaces of the cylinders shown in the diagram. ( They need not be to scale. ) (ii)Find the surface area of each cylinder giving your answers correct to 1 decimal place. 50 cm 10 cm 9m9m 3m3m (a) (b)

12. EXERCISE 50 cm 10 cm Solutions: (a) Area of base or top =  r 2 =  (10) 2 = 314·16 Length of rectangle =  d =  (20  = 62·83 Area of rectangle = 62·83  50 = 3141·59 Total area = r = 10 cm base and top curved surface dd 50 cm r = 10 cm

13. EXERCISE 50 cm 10 cm Solutions: (a) Area of base or top =  r 2 =  (10) 2 = 314·16 Length of rectangle =  d =  (20  = 62·83 Area of rectangle = 62·83  50 = 3141·59 2  314·16 + 3141·59 = 3769·9 cm 2 ( 1 d.p. ) Total area = r = 10 cm base and top curved surface dd 50 cm r = 10 cm

14. EXERCISE 9m9m 3m3m Solutions: (b) base and top curved surface dd 9m9m r = 3 m Area of base or top =  r 2 =  (3) 2 = 28·27 Length of rectangle =  d =  (6  = 18·85 Area of rectangle = 18·85  9 = 169·65 Total area = r = 3 m

15. EXERCISE 9m9m 3m3m Solutions: (b) curved surface dd 9m9m Area of base or top =  r 2 =  (3) 2 = 28·27 Length of rectangle =  d =  (6  = 18·85 Area of rectangle = 18·85  9 = 169·65 2  28·27 + 169·65 = 226·2 m 2 ( 1 d.p. ) Total area = base and top r = 3 m