1
2 Classifying Solids
PROBLEM 1 PROBLEM 2
Surface Area of Cylinders Volume of a Right Cylinder
3
PRISM PYRAMID
CYLINDER CONE SPHERE
4
Standards 8, 10, 11
h
base
base
h
Lateral Area: 2 r
L = h
2 rh
r
r
r
Total Surface Area = Lateral Area + 2(Base Area) T= 2 rh + 2 r2
r2
r2
h= height r= radius
5 VOLUME OF CYLINDERS
h
r r2
B=
V = Bh V = r2h
6 Find the lateral area, the surface area and volume of a right cylinder with a radius of 20 in and a height of 10 in.
10 in
20 in
Lateral Area: 2 r
L = h
L = 2 ( )( )20 in 10 in
Total Surface Area = Lateral Area + 2(Base Area)
T= 2 rh + 2 r2
T = 2 ( )( ) + 2 ( )20 in 10 in 20 in 2
T= 400 in + 2(400 in )2 2
L=400 in2
T = 400 + 800in2 in2
T = 1200 in2
Volume:
V = r2 h
V = ( 20 in )2( )10 in V= (400 in )(10 in) 2
7 Find the lateral area and the surface area of a cylinder with a circumference of 14 cm. and a height of 5cm.
C=2 r 2 2
r= C 2
r= 2 r=7 cm
Finding the radius:
14
5 cm 7 cm
Lateral Area: 2 r
L = h
L = 2 ( )( )7 cm 5 cm L= 70 cm 2
Total Surface Area = Lateral Area + 2(Base Area)
T= 2 rh + 2 r2
T = 2 ( )( ) + 2 ( )7 cm 5 cm 7 cm 2
T= 70 cm + 2(49 cm )2 2
T = 70 + 98cm2 cm2
8 Find the Volume for the cylinder below:
2 5
First we find the height: 4
h
h
4 5
5 = 4 + h2 2 2 25 = 16 + h2
-16 -16
h = 92 h = 92
h = 3
Volume:
V = r2 h
V = ( 2 )2( )3 V= ( 4 )(3)
9 The surface area of a right cylinder is 400 cm. If the height is 12 cm., find the radius of the base.
Total Surface Area:
T= 2 rh + 2 r2
We substitute values:
6.28 6.28
75.4 75.4 -400
+
-Using the Quadratic Formula:
10 6
3 SIMILARITY IN SOLIDS
4
8
Are this two cylinders similar?
These cylinders are NOT SIMILAR =
4 6
11
VOLUME 1 VOLUME 2
IF THEN
The ratio of the radii of two similar cylinders is 2:5. If the volume of the smaller cylinder is 40 units, what is the volume of the larger cylinder.3
V2 =
They are similar
12
13 Classifying Solids
PROBLEM 1 PROBLEM 2 Surface Area of Cones Volume of a Right Cone
PROBLEM 3 PROBLEM 4 PROBLEM 5
14
area of sector area of circle
perimeter of cone’s base
perimeter of circle
=
2
C= l
2 l
=
area of sector
2 r C=
Area of Circle
2
l
2 r C=
perimeter of cone’s base
r
2
l l
=
area of sector 2
l
2
l
TOTAL SURFACE AREA:
T = area of sector + area of cone’s base
Standards 8, 10, 11
h= height r = radius
15 h
r
VOLUME OF A RIGHT CIRCULAR CONE
2
r B=
V = Bh1 3
V = 1 r2
3 h
16 Find the lateral area, the surface area and volume of a right cone with a height of 26 cm and a radius of 12 cm. Round your answers to the nearest tenth.
h
we need to find the slant height, using the Pythagorean Theorem:
l
l 2= 26 + 122 2
l = 676 + 1442
l = 8202
l 28.6 cm
Calculating the base area:
2
Calculating surface area:
T = L + B
L= ( )( )12 cm 28.6 cm L = 1077.7 cm2
T = 1077.7 cm + 452.2 cm2 2
Calculating the volume:
17 Find the lateral area and the surface area and volume of a right cone whose slant height is 9 ft and whose circumference at the base is 4 ft. Round your
answers to the nearest tenth.
h r
l
=9ft
We need to find the radius: C=2 r
we need to find the height, using the Pythagorean Theorem:
C=4πft
Calculating the base area:
2
Calculating surface area: T = L + B
T = 56.5 ft + 12.6 ft2 2
T = 69.1 ft2
Calculating the volume:
18 Find the lateral area, the surface area, and the volume of a right cone whose height is 18 m and whose slant height is 22 m. Round your answers to the nearest unit.
h = 18 m r
l =22 m
we need to find the radius, using the Pythagorean Theorem:
22 = r + 182 2 2
Calculating the base area:
2
Calculating surface area: T = L + B
T = 898 m + 531 m2 2
T = 1429 m2
Calculating the volume:
19 Find the lateral surface of a cone whose volume is 900 mm and whose radius is 15 mm. Round your answers to the closest tenth.
3
Now we draw the cone:
h
r 15=
we need to find the slant height, using the Pythagorean Theorem:
20 The ratio of the radii of two similar cones is 3:8. If the volume of the larger cone is 2090 units, what is the approximate volume of the smaller cone? 3
VOLUME 1 > VOLUME 2
2
r V = 1
3 h
Volume: IF VOLUME 1 VOLUME 2
V = 1 r2h
They are similar
V 110 units3
What can you conclude about the ratio of the volumes and the ratio of the
