Chapter 12
12.1 Surface Areas of Prisms
12.2 Surface Areas of Pyramids
Surface Areas of Circular Solids
12.1 Surface Areas of Prisms
A prism is a 3 dimensional figure with 2 congruent parallel polygons which are called bases and sides that connect the bases that are parallelograms.
Right prisms have sides that are rectangles.
Oblique Prisms - sides that are non rectangular parallelograms
Faces - flat surfaces of the prism
Bases = parallel congruent sides
Lateral Faces - parallelograms that connect the faces (rectangles on a right prism)
Edges - segments where the faces intersect (number of sides on the base times 3)
Vertices - points where the 3 faces intersect (number of sides on the base times 2)
Lateral surface area is the area of the parallelograms on the sides (rectangles on a right prism)
Total surface area = lateral surface area + the areas of the two bases
Formulas for Right Prisms
L.A.= ph (perimeter of the base times the height of the prism)
T. A. = L.A. + 2 B (B= area of the base)
Examples: Find the no. of faces, bases, lateral faces, edges, vertices, lateral area and total area for each of the following prisms
1.
Faces = 6 Bases = 2 Lateral faces= 4
Edges = 12 Vertices = 8
L.A. = (8+6+8+6)(5) = 28 x 5 = 140 sq units
T. A. = 140 + 2(8 x6) = 140 + 2(48) = 140 + 96 = 236 sq units
2.
Faces = 5 Bases = 2 Lateral faces = 3 edges = 9 vertices =6
LA = 32 x 15 = 480 sq units
TA = 480 + 2 Triangle base areas
To find the area of the isosceles triangles, drop a perpendicular to the base of the triangle. This will bisect the base . Use the Pythagorean theorem to find the height of the triangle. Then use the formula for finding the area of a triangle A = 1/2 bh
TA = 480 + 2 ( 1/2 x 12 x 8) = 480 + 2(48) = 480 + 96 = 576
3.
Faces = 8, Bases = 2, Lateral faces = 6, Edges = 18, Vertices = 12
LA = 36 x 10 = 360
TA = 360 + area of the two hexagons
To find the area of the regular hexagons A = 1/2 a P
4.
Faces = 5 Bases = 2 Lateral Faces = 3 Vertices = 6 Edges = 9
Lateral Surface Area - find the hypotenuse of the right triangle using the Pythagorean Theorem or notice that it is a triple. 6-8-10
Lateral Area = 24(12) = 288 sq units
Total Area = 288 + 2 right triangle areas (1/2 leg x leg)= 288 +2(1/2 x 6 x 8) = 288 + 2(24)= 288 +48 = 336 sq units
5. Right Trapezoidal Prism
Faces = 6 Bases = 2 Lateral Faces = 4 Edges = 12 Vertices = 8
Lateral Surface Area = 74(20) = 1480 sq units
Total Surface Area = Lateral Area + 2 Trapezoid Areas
To find the trapezoid areas drop perpendiculars from the top vertices. The middle section becomes a rectangle with a base of 10. This leaves 24 units of the 34 on the base to divide evenly between the two end triangles. Use the Pythagorean theorem on the triangle to solve for the height which results in 9. Since the area of a trapezoid = 1/2 h (b1+ b2) then the area of the trap = 1/2 (9)(34+10) = 1/2(9)(44) = 9 ( 22) = 198
Total Surface Area = 1480 + 2(198) = 1480 + 396 = 1876 sq units
12.2 Surface Areas of Pyramids
Pyramid - A three dimensional figure with a polygonal base and sides that are triangle converging on a point.
Regular Right Pyramids - These are the type of pyramids we will be studying. Each has a base that is a regular polygon. The sides of the pyramid are congruent isosceles triangles and the vertex of the pyramid is directly over the center of the pyramids base.
Below are pictures of 3 different pyramids.
The number of faces on each is determined by the number of sides on the base
Triangular pyramid - 4 faces (one more than the number of sides)
Square pyramid - 5 faces
Hexagonal pyramid - 7
The number of vertices (points where three faces intersect) is the same as the number of faces.
Vertices: Triangular pyramid = 4 Square pyramid =5 Hexagonal pyramid = 7
The number of edges (segments where two faces intersect) = twice the number of sides on the base.
Triangular pyramid has 6 edges ( 3 x 2)
Square pyramid has 8 edges
Hexagonal pyramid =12
The slant height ( l ) of a pyramid is the altitude of each triangle. Since in a regular pyramid the triangles are all congruent, the slant heights are the same.
Most problems on the surface area of a pyramid have you solve for the length of the slant height by using the pythagorean theorem. Below are two examples of solving for the slant height.
Ex. 1
Find the length of the slant height given that the base edge = 24 and the lateral edge = 15
Ans. The perpendicular splits the base edge into two congruent segments of 12 each. Using the pythagorean theorem 152=122+ l2 , l = 9
Ex. 2
Find the length of the slant height given the base edge = 12 and the height of the pyramid =8
Ans. The right triangle formed by the height of the pyramid, the slant height and the segment from the center of the square to the side is what will be used in the Pythagorean theorem. The length of the segment from the center to the side is 6. The equation will be l2 =82+ 62 , l = 10
Lateral Surface Area - area of the triangles on the pyramid.
To find the lateral surface area of a pyramid, find the areas of each of the triangles on the sides. Since we are studying only the regular right pyramids, a short cut formula may be use.
Lateral Area = 1/2 pl where p = perimeter of the base, and l is the slant height.
Total Area = Lateral Area + B B = area of the base
Ex. 1 Find the L.A. and T.A. of the pyramid below.
Ans. First find the slant height by dividing the base edge into two congruent pieces of 6 each. Use the pythagorean theorem 102 =62+ l2 l=8 LA = 1/2(36)(8) = 144 TA = 144 + Area of the equilateral triangle which is the base.
Ex. 2 Find the L.A. and T.A. of the regular square pyramid
Ans. First find the slant height by dividing the base edge into two congruent pieces of 6 each. Use the pythagorean theorem 132 =122+ l2 l=5 LA = 1/2(96)(5) = 240 TA = 240 + Area of the square which is the base. Total Area = 240 + s2 = 240 + 242 = 240 +576=816
Surface Areas of Circular Solids
The three circular solids we will be studying are the cylinder, cone and sphere.
Surface Area of Cylinder
Example 1. Given the
cylinder with dimensions as shown, find the L.A. and T.A.
Surface Area of a Cone =
