Areas of Regular Polygons and Composite Figures Section 11.4 Areas of Regular Polygons and Composite Figures

In the figure, a regular pentagon is inscribed in circle P, and circle P is circumscribed about the pentagon. The center of a regular polygon and the radius of a regular polygon are also the center and the radius of its circumscribed circle. A segment drawn from the center of a regular polygon perpendicular to a side of the polygon is called an apothem. Its length is the height of an isosceles triangle that has two radii as legs. A central angle of a regular polygon has its vertex at the center of the polygon and its sides pass through consecutive vertices of the polygon. The measure of each central angle of a regular n-gon is

Example 1: In the figure, pentagon PQRST is inscribed in circle X Example 1: In the figure, pentagon PQRST is inscribed in circle X. Identify the center, a radius, an apothem, and a central angle of the polygon. Then find the measure of a central angle. center: point X radius: XR or XQ apothem: XN central angle: RXQ A pentagon is a regular polygon with 5 sides. Thus, the measure of each central angle of pentagon PQRST is or 72º.

Example 2: The top of the table shown is a regular hexagon with a side length of 3 feet and an apothem of 1.7 feet. What is the area of the tabletop to the nearest tenth? Since the polygon has 6 sides, the polygon can be divided into 6 congruent isosceles triangles, each with a base of 3 ft and a height of 1.7 ft. Find the area of one triangle. Area of a triangle b = 3 and h = 1.7 = 2.55 ft2 Simplify. Multiply the area of one triangle by the total number of triangles. Since there are 6 triangles, the area of the table is 2.55 ● 6 or 15.3 ft2.

From Example 2, we can develop a formula for the area of a regular n-gon with side length s and apothem a.

Example 3: a) Find the area of the regular hexagon. Round to the nearest tenth. Find the measure of a central angle. A regular hexagon has 6 congruent central angles, so mQPR = 60º. Find the apothem. Apothem PS is the height of isosceles ΔQPR. It bisects QPR, so mSPR = 30º. It also bisects QR, so SR = 2.5 meters. ΔPSR is a 30°-60°-90° triangle with a shorter leg that measures 2.5 meters, so

Use the apothem and side length to find the area. Area of a regular polygon ≈ 65.0 m2 Use a calculator.

Example 3: b) Find the area of the regular pentagon. Round to the nearest tenth. A regular pentagon has 5 congruent central angles, so mACB = 72º Apothem CD is the height of isosceles ΔBCA. It bisects BCA, so mBCD = 36º. Use trigonometric ratios to find the side length and apothem of the polygon. AB = 2DB or 2(9 sin 36°). So, the pentagon’s perimeter is 5 ● 2(9 sin 36°). The length of the apothem CD is 9 cos 36°.

Use the apothem and side length to find the area. Area of a regular polygon = 192.6 cm2 Use a calculator.

A composite figure is a figure that can be separated into regions that are basic figures, such as triangles, rectangles, trapezoids, and circles. To find the area of a composite figure, find the area of each basic figure and then use the Area Addition Postulate.

Example 4:   a) The dimensions of an irregularly shaped pool are shown. What is the area of the surface of the pool? The figure can be separated into a rectangle with dimensions 16 feet by 32 feet, a triangle with a base of 32 feet and a height of 15 feet, and two semicircles with radii of 8 feet. Area of composite figure = area of rectangle + area of triangle + area of 2 semicircles = 512 + 240 + 64π  953.1 square feet

Example 4: b) Find the area of the figure in square feet. Round to the nearest tenth if necessary. The figure can be separated into a rectangle with dimensions 18 feet by 8 feet, two triangles with a base of 5 feet and a height of 8 feet, and a semicircles with radius of 9 feet. Area of composite figure = area of rectangle + area of triangles + area of semicircle = 144 + 40 + 40.5π  311.2 square feet

Example 5: a) Find the area of the shaded figure To find the area of the figure, subtract the area of the smaller rectangle from the area of the larger rectangle. The length of the larger rectangle is 25 + 100 + 25 or 150 feet. The width of the larger rectangle is 25 + 20 + 25 or 70 feet. area of shaded figure = area of larger rectangle – area of smaller rectangle = b1h1 – b2h2 Area formulas = 150(70) – 100(20) Substitution = 10,500 – 2000 Simplify = 8500 square feet

Example 5: b) Cara wants to wallpaper one wall of her family room. She has a fireplace in the center of the wall. Find the area of the wall around the fireplace. To find the area of the figure, subtract the area of the smaller rectangle from the area of the larger rectangle. The length of the larger rectangle is 10 feet. The width of the larger rectangle is 18 feet. area of wall minus the fireplace = area of larger rectangle – area of smaller rectangle = b1h1 – b2h2 Area formulas = 10(18) – 6(4) Substitution = 180 – 24 Simplify = 156 square feet