Find lateral areas of regular pyramids. Find surface areas of regular pyramids.

All pyramids have the following characteristics. AAll of the faces, except the base, intersect at one point called the vertex. TThe base is always a polygon. TThe faces that intersect at the vertex are called lateral faces and form triangles. The edges of the lateral faces that have the vertex as an endpoint are called lateral edges. TThe altitude is the segment from the vertex perpendicular to the base.

Pyramids If the base of a pyramid is a regular polygon and the segment with endpoints that are the center of the base and the vertex is perpendicular to the base is called a regular pyramid. An altitude is the segment with endpoints that are at the center of the base and the vertex. All of the lateral faces are congruent isosceles triangles. The height of each lateral face is called the slant height l of the pyramid.

Parts of a Pyramid Vertex Square Pyramid Lateral Edge BaseAltitude Lateral Face Regular Square Pyramid Base (a regular polygon) Slant Height Altitude

Key Concepts If a regular pyramid has a lateral area of L square units, a slant height of l units, and its base has a perimeter of P units, then L = 1/2P l. If a regular pyramid has a surface area of T square units, a slant height of l units, and its base has a perimeter of P units and an area of B square units, then T = 1/2P l + B.

Example 1: Use Lateral Area to Solve Problem A farmer is building a hexagonal barn. The base of the roof has sides of 15 feet, and the slant height of the roof is 30 feet. The farmer needs to know exactly how much wood to use for the roof. Find the amount of wood used for the roof.

Example 1: Continued L = ½P l Lateral area of a regular pyramid = ½(90)(30) P = 90, l = 30 = 1350 Multiply So, 1350 square feet of wood are used to cover the roof of the barn.

Example 2: Surface Area of a Square Pyramid Find the surface area of the square pyramid. To find the surface area, first find the slant height of the pyramid. The slant height is the hypotenuse of a right triangle with legs that are the altitude and a segment with a length that is one-half the side measure of the base. 24 m 1 8 m l

Example 2: Continued c 2 = a 2 + b 2 Pythagorean Theorem l 2 = a = 9, b = 24, c = l l = √ 657 Simplify Now find the surface area of a regular pyramid. The perimeter of the base is 4(18) or 72 meters, and the area of the base is 182 or 324 square meters. T = ½ P l + B Surface area of a regular pyramid T = ½ ( 72 )√ P = 72, l = √ 657, B = 324 T ≈ Use a calculator

Example 3: Surface Area of Pentagonal Pyramid Find the surface area of the regular pyramid. The altitude, slant height, and apothem form a right triangle. Use Pythagorean Theorem to find the apothem. Let a represent the length of the apothem.

Example 3: Continued c 2 = a 2 +b 2 Pythagorean Theorem (17) 2 = a 2 =15 2 b =15, c = 17 8 = aSimplify Now find the length of the sides of the base. The central angle of the pentagon measures 360º/5 or 72º. Let x represent the measure of the angle formed by a radius and the apothem. Then, x = 72/2 or 36. Tan 36 = ( ½ s)/8tan x= 8(tan 36) = ½ sMultiply each side by 8 16(tan 36) = sMultiply each side by ≈ sUse a calculator opposite adjacent

Example 3: Continued #2 Next, find the perimeter and area of the base. P = 5s ≈ 5(11.6) or 58 B = ½Pa ≈ ½(58)(8) or 232 Finally, find the surface area. T = ½P l + B Surface area of a regular pyramid ≈ ½(58)(17)+232 P ≈ 58, l =232,B ≈232 ≈ Simplify The surface area is approximately square inches

Pre-AP Geometry Pg. 663 # 7-16, 18-23, 27