LESSON 2: TRIANGULAR PRISMS, CYLINDERS, AND SPHERES Unit 9: Figures and Solids
base parallel two The sum of the area of the lateral faces (al sides except for the bases) The sum of all the area (lateral and bases)
B The area of one of the bases h Height of the prism (think about the length of the lateral sides) The perimeter of the base
Triangular Prisms Triangles Volume formula: V = Bh height of prism 1 b height of 2 Surface Area formula: SA = hp + 2B 1 b height of 2 height of prism perimeter of
Shade the bases of the prism. height of = 4cm base of = 3cm hyp. of = 5cm height of prism = 12cm What is the height of the prism? h = 12cm What is the area of the base? A = 1 2 bh = 1 2 3 4 = 6cm2 What is the perimeter of the base? a 2 + b 2 = c 2 4 2 + 3 2 = c 2 25 = c 2 5 = c p = 4 + 3 + 5 = 12 cm Volume: Bh area of base height of the prism V = Bh = 6 12 = 72cm 3 Area of = 6m 2 Surface Area: hp + 2B height of prism = 12cm height of prism perimeter of base + 2(area of base) SA = hp + 2B = 12 12 + 2 6 = 144 + 12 = 156cm 2 height of prism perimeter of Area of = 6cm 2
height of = 12ft base of = 10ft hyp. of = 13ft height ofprism = 22ft 12ft Shade the bases of the prism. What is the height of the prism? h = 22ft What is the area of the base? A = 1 2 bh = 1 2 10 Volume: Bh area of base height of the prism V = Bh = 60 22 = 1320ft 3 Area of = 60ft 2 Surface Area: hp + 2B 12 = 60ft2 height of prism What is the perimeter of the base? a 2 + b 2 = c 2 12 2 + 5 2 = c 2 169 = c 2 13ft = c p = 13 + 13 + 10 = 36ft height of prism perimeter of base + 2(area of base) SA = hp + 2B = 22 36 + 2 60 = 792 + 120 = 912ft 2 height of prism perimeter of Area of = 60ft 2
This is the triangular prism on top of a rectangular prism. We have already found the V & SA of the triangular prism from above. What is the volume of the greenhouse? height of = 12ft base of = 10ft hyp. of = 13ft lenght of prism = 22 ft V = V + V V = 1320 + 2200 = 3520 ft 3 What is the surface area of the greenhouse? p = 13 + 13 + 10 + 10 + 10 = 56ft V = 1320ft 3 V = lwh = 10 22 10 = 2200ft 3 A = 60ft 2 A = bh = 10 10 = 100ft 2 B = A + A = 60 + 100 = 160ft 2 SA = hp + 2B = 22 56 + 2 160 = 1552ft 3 height of prism perimeter of Area of = 160ft 2
Right Cylinders Circles Volume formula: V = πr 2 h Lateral Area formula: LA = 2πrh Area of circle Height of cylinder Circumference of circle Height of cylinder Surface Area formula: 2πr 2 + 2πrh Area of circle Lateral Area
Volume: r = 4cm h = 11cm A = πr 2 = 4 2 π = 16π cm 2 C = 2πr = 2 4 π = 8π cm V = πr 2 h = 16π 11 = 176π cm 3 552.92 cm 3 Area of circle Height of cylinder Lateral Area: LA = 2πrh = 8π 11 = 88π cm 2 276.46 cm 2 Circumference of circle Height of cylinder Surface Area: SA = 2πrh + 2πr 2 = 88π + 2(176π) Lateral Area Area of circle
Example B: Method 1: Find the volume of the whole, then dividing by 2 A cylindrical water contain is 3.2 m high and has a diameter of 5.8 m. Approximately how many cubic meters of water will the container be holding when it is half full? d = 5.8 m r = 2.9 m h = 3.2 m A = πr 2 = 2.9 2 π = 8.41π m 2 V = πr 2 h = 8.41π 3.2 = 26.912π m 3 Area of circle Height of cylinder V 2 = 26.912π 2 = 13.456π m 2 42.27 m 2
Example B: Method 2: Half the original height (h ) then find the volume. A cylindrical water contain is 3.2 m high and has a diameter of 5.8 m. Approximately how many cubic meters of water will the container be holding when it is half full? d = 5.8 m r = 2.9 m h = 1.6 m A = πr 2 = 2.9 2 π = 8.41π m 2 V = πr 2 h = 8.41π 1.6 = 13.456π m 3 42.27 m 2 Area of circle Height of cylinder half full
Example C: A cylinder has a diameter that is 12 in and a height that is three times its radius. What is its surface area? d = 12 in r = 6 in h = 3r = 3 6 = 18 in A = πr 2 = 6 2 π = 36π in 2 C = 2πr = 2 6π = 12π in SA = 2πrh + 2πr 2 = 12π 18 + 2 36π = 216π + 72π Area of circle Circumference of circle Height of cylinder = 288π in 2 904.78 in 2
Spheres Volume: V = 4 3 πr3 (Area of circle)(radius of the circle) Surface Area: SA = 4πr 2 Area of circle
Volume: V = 4 3 πr3 = 4 3 512π = 2048π 3 in 3 (Area of circle)(radius of the circle) r = 8 in r 2 = 64 in 2 r 3 = 512 in 3 A = πr 2 = 64π in 2 = 2048 3 π in3 2144.66 in 3 Surface Area: SA = 4πr 2 = 4 64π = 256π in 2 804.25 in 2 Area of circle
Volume: d = 30 cm r = 15 cm r 2 = 225 cm 2 r 3 = 3375 cm 3 A = πr 2 = 225π cm 2 V = 4 3 πr3 = 4 3 (Area of circle)(radius of the circle) 3375π = 4500π cm3 4137.17 cm 3 Surface Area: SA = 4πr 2 = 4 225π = 900π in 2 2827.43 in 2 Area of circle
Example C: A spherical paintball measures 12.5 cm in diameter. Approximately how much paint is in it? d = 12.5 cm r = 6.25 cm r 2 = 39.0625 cm 2 r 3 = 244.140625 cm 3 A = πr 2 = 39.0625π cm 2 V = 4 3 πr3 = 4 3 6.25 3 π = 976.5625 π cm 3 3 1022.62 cm 3 (Area of circle)(radius of the circle)
Example D: Find the surface area of a globe with a 14 m radius? r = 14 m r 2 = 196 m 2 A = πr 2 = 196π m 2 SA = 4πr 2 = 4 196π = 784π m 2 2463.01 m 2 Area of circle
Example E: Method 1: Find the volume of the whole and then divide by 2. A spherical watermelon has diameter of 14 in. If the watermelon is shared equality between two people, how many cubic inches of watermelon would each person get? d = 14 in r = 7 in r 2 = 49 in 2 r 3 = 343 in 3 V = 4 3 πr3 = 4 3 7 3 π = 4(343) 3 π in 3 = 1372 3 π in3 (Area of circle)(radius of the circle) 1 2 V = 1 2 1372π 3 = 686π 3 in 3 718.38 in 3
Example E: Method 2: Build in ½ into the equation. A spherical watermelon has diameter of 14 in. If the watermelon is shared equality between two people, how many cubic inches of watermelon would each person get? d = 14 in r = 7 in r 2 = 49 in 2 r 3 = 343 in 3 Formula for volume of a sphere 1 2 V = 1 2 4 3 πr3 = 4(343) 2 3 π in3 = 686 3 π in3 718.38 in 3 (Area of circle)(radius of the circle)
Example F: How much paper is needed to make a label for this can of beans which has a radius of 1.5 in and a height of 6 in? r = 1.5 in h = 6 in C = 2πr = 2 1.5 π = 3π in LA = 2πrh = 3π 6 = 18π in 2 56.55 in 2 Circumference of circle
**There is half a sphere on top of a cylinder. Example G: Method 1: Find the volume of the sphere and then divide by 2. Find the volume of the grain silo pictured. V = 4 3 πr3 = 4 3 (64)π = 256 3 π ft3 (Area of circle)(radius of the circle) r = 4 ft r 2 = 16 ft 2 r 3 = 64 ft 3 h = 10 ft A = πr 2 = 4 2 π = 16π ft 2 1 2 V = 1 2 256 128 π = π ft3 3 3 V C = πr 2 h = 16π 10 = 160π ft 3 V W = 1 2 V + V C = 128 3 626.70 ft 3 π + 160π = 608 3 π ft3
**There is half a sphere on top of a cylinder. Example G: Method 2: Build in ½ into the equation. Find the volume of the grain silo pictured. 1 2 V = 1 2 4 3 πr3 = 2 3 Formula for volume of a sphere (Area of circle)(radius of the circle) (64)π = 128 3 π ft3 r = 4 ft r 2 = 16 ft 2 r 3 = 64 ft 3 h = 10 ft A = πr 2 = 4 2 π = 16π ft 2 V C = πr 2 h = 16π 10 = 160π ft 3 V W = 1 2 V + V C = 128 π + 160π 3 = 608 3 π ft3 626.70 ft 3