<<

10.3 of Regular

Geometry Mr. Peebles Spring 2013 Daily Learning Target (DLT) Monday January 14, 2013 • “I can remember, apply, and understand to find the and of a regular .” Finding the area of an equilateral • The area of any triangle with length b and height h is given by A = ½bh. The following formula for equilateral ; however, uses ONLY the side length. * Look at the next slide * Theorem 11.3 Area of an • The area of an equilateral triangle is one fourth the of the length of the side times 3 s s

A = ¼ s2 s

A = ¼ s2 Ex. 1: Finding the area of an Equilateral Triangle • Find the area of an equilateral triangle with 8 inch sides.

A = ¼ 3 s2 Area of an equilateral Triangle

A = ¼ 82 Substitute values.

A = ¼ • 64 Simplify.

A = • 16 Multiply ¼ times 64. A = 16 Simplify. Using a calculator, the area is about 27.7 square inches. Ex. 2: Finding the area of an Equilateral Triangle • Find the area of an equilateral triangle with 12 inch sides.

A = ¼ 3 s2 Area of an equilateral Triangle

A = ¼ 122 Substitute values.

A = ¼ • 144 Simplify.

A = • 36 Multiply ¼ times 144. A = 36 Simplify. Using a calculator, the area is about 62.4 square inches. More . . .

• The is the F A height of a triangle between the center H and two consecutive a E vertices of the G B polygon. • You can find the area of any regular n-gon by dividing the D C polygon into ABCDEF with center G, GA, congruent triangles. and apothem GH More . . .

A = Area of 1 triangle • # of triangles OR F A

= ½ • apothem • # of sides • side length OR H a E = ½ • apothem • perimeter of a polygon G B

This approach can be used to find the area of any .

D C Hexagon ABCDEF with center G, radius GA, and apothem GH Theorem 11.4 Area of a Regular Polygon • The area of a regular n-gon with side lengths (s) is half the product of the apothem (a) and the perimeter (P), so The number of congruent triangles formed will be A = ½ aP, or A = ½ a • ns. the same as the number of sides of the polygon.

NOTE: In a regular polygon, the length of each side is the same. If this length is (s), and there are (n) sides, then the perimeter P of the polygon is n • s, or P = ns Example 3

Find the area of a regular with an apothem of 4 Feet and side length of 3 Feet.

= ½ • apothem • # of sides • side length OR

= ½ • apothem • perimeter of a polygon

Apothem = 4 Feet Numbers of Sides = 5 Side Length = 3

Example 3

Find the area of a regular pentagon with an apothem of 4 Feet and side length of 3 Feet.

= ½ • apothem • # of sides • side length OR

= ½ • apothem • perimeter of a polygon

Apothem = 4 Feet Numbers of Sides = 5 Side Length = 3 So ½ • 4 • 3 • 5 = 30 Square Feet

Example 4

Find the area of a regular STOP SIGN with an apothem of 9 Feet and side length of 12 Feet.

= ½ • apothem • # of sides • side length OR

= ½ • apothem • perimeter of a polygon

Apothem = 9 Feet Numbers of Sides = 8 Side Length = 12

Example 4

Find the area of a regular octagon STOP SIGN with an apothem of 9 Feet and side length of 12 Feet.

= ½ • apothem • # of sides • side length OR

= ½ • apothem • perimeter of a polygon

Apothem = 9 Feet Numbers of Sides = 8 Side Length = 12 So ½ • 9 • 8 • 12 = 432 Square Feet

Ex. 5: Finding the area of a regular • Pendulums. The enclosure on the floor underneath the Foucault Pendulum at the Houston Museum of Natural Sciences in Houston, Texas, is a regular dodecagon with side length of about 4.3 feet and a radius of about 8.3 feet. What is the floor area of the enclosure? Solution:

• A dodecagon has 12 sides. So, the perimeter of the enclosure is S

P = 12(4.3) = 51.6 feet 8.3 ft.

A B Solution: S

• In ∆SBT, BT = ½ (BA) = ½ (4.3) = 2.15 8.3 feet feet. Use the Pythagorean 2.15 ft. Theorem to find the A B apothem ST. T 4.3 feet a = 8.32  2.152 a  8 feet So, the floor area of the enclosure is: A = ½ aP  ½ (8)(51.6) = 206.4 ft. 2 Assignment: 10-3

Pages 548-551 (4-9, 11, 24) Due Tomorrow Tuesday January 15, 2013 Closure

On the whiteboards, write down in your own words what an Apothem is.