# Areas of Rectangles and Parallelograms

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1 CONDENSED LESSON 8.1 Areas of Rectangles and Parallelograms In this lesson you will Review the formula for the area of a rectangle Use the area formula for rectangles to find areas of other shapes Discover the formula for the area of a parallelogram The area of a plane figure is the numer of square units that can e arranged to fill the figure completely. You proaly already know several area formulas. The investigations in this chapter will help you understand and rememer the formulas. Pages 422 and 423 of your ook discuss the formula for Area 15 square units the area of a rectangle. Read this text carefully. Make sure that you understand the meaning of ase and height and that the area formula makes sense to you, then complete the Rectangle Area Conjecture in your ook. Example A in your ook shows how the area formula for rectangles can help you find areas of other shapes. Here is another example. Area 11 square units EXAMPLE A Find the area of this square. Solution Surround the given slanted square with a 7-y-7 square with horizontal and vertical sides. Then sutract the area of the four right triangles formed from the area of the surrounding square. Each of the four triangles is half of a 2-y-5 rectangle, so each has area , or 5 square units. Therefore, the area of the original square is (7 7) (4 5) 29 square units. Discovering Geometry Condensed Lessons CHAPTER 8 103

2 Lesson 8.1 Areas of Rectangles and Parallelograms Just as with a rectangle, any side of a parallelogram can e called the ase. An altitude of a parallelogram is any segment from one side of the parallelogram, perpendicular to that side, to a line through the opposite side. The height of a parallelogram is the length of the altitude. Study the diagrams of altitudes on page 424 of your ook. Altitude Base Investigation: Area Formula for Parallelograms Follow Steps 1 and 2 of the investigation in your ook. In Step 2, each new shape you form has the same area as the original parallelogram ecause you have simply rearranged pieces, without adding or removing any cardoard. Form a rectangle with the two pieces. Notice that the ase and height of the rectangle are the same as h the ase and height of the original parallelogram. Because the area of the rectangle and the parallelogram are the same, the area of the parallelogram is h. This can e summarized as a conjecture. s s Parallelogram Area Conjecture The area of a parallelogram is given y the formula A h, where A is the area, is the length of the ase, and h is the height of the parallelogram. C-75 If the dimensions of a figure are measured in inches, feet, or yards, the area is measured in in 2 (square inches), ft 2 (square feet), or yd 2 (square yards). If the dimensions are measured in centimeters or meters, the area is measured in cm 2 (square centimeters) or m 2 (square meters). Read Example B in your ook, and then read the example elow. EXAMPLE B A parallelogram has height 5.6 ft and area 70 ft 2. Find the length of the ase. Solution A h Write the formula. 70 (5.6) Sustitute the known values Solve for the ase length Divide. The length of the ase is 12.5 ft. 104 CHAPTER 8 Discovering Geometry Condensed Lessons

3 CONDENSED LESSON 8.2 Areas of Triangles, Trapezoids, and Kites In this lesson you will Discover area formulas for triangles, trapezoids, and kites You can use the area formulas you already know to derive new area formulas. In the first investigation you will focus on triangles. Investigation 1: Area Formula for Triangles Follow Step 1 in your ook to create and lael a pair of congruent triangles. You know the area formula for rectangles and parallelograms. Arrange the two congruent triangles to form one of these figures. Write an expression for the area of the entire figure. Then write an expression for the area of one of the triangles. Summarize your findings y completing the conjecture elow. Triangle Area Conjecture The area of a triangle is given y the formula, where A is the area, is the length of the ase, and h is the height of the triangle. C-76 Next, you ll consider the area of a trapezoid. Investigation 2: Area Formula for Trapezoids Follow Steps 1 and 2 in your ook to make and lael two congruent trapezoids. You can arrange the trapezoids to form a parallelogram. 2 1 h s s h 1 2 What is the ase length of the parallelogram? What is the height? Use your answers to the questions to write an expression for the area of the parallelogram. Then use the expression for the area of the parallelogram to write an expression for the area of one trapezoid. Summarize your findings y completing this conjecture. Trapezoid Area Conjecture The area of a trapezoid is given y the formula, where A is the area, 1 and 2 are the lengths of the two ases, and h is the height of the trapezoid. C-77 Discovering Geometry Condensed Lessons CHAPTER 8 105

4 Lesson 8.2 Areas of Triangles, Trapezoids, and Kites Finally, you will consider the area of a kite. Investigation 3: Area Formula for Kites Draw a kite. Draw its diagonals. Let d 1 e the length of the diagonal connecting the vertex angles, and let d 2 e the length of the other diagonal. d 2 d 1 Recall that the diagonal connecting the vertex angles of a kite divides it into two congruent triangles. Consider the diagonal laeled d 1 to e the ase of one of the triangles. Then, ecause the diagonal connecting the vertex angles of a kite is the perpendicular isector of the other diagonal, the height of the triangle is 1 2 d 2. _ 1 d _ d 2 d 1 d 1 Write an expression for the area of one of the triangles. Then use the expression for the area of the triangle to write an expression for the area of the kite. Summarize your findings y completing this conjecture. Kite Area Conjecture The area of a kite is given y the formula, where A is the area and d 1 and d 2 are the lengths of the diagonals. C CHAPTER 8 Discovering Geometry Condensed Lessons

5 CONDENSED LESSON 8.3 Area Prolems In this lesson you will Use a variety of strategies to approximate the areas of irregularly shaped figures Use the area formulas from the previous two lessons to find the areas of more complex figures You have discovered formulas for areas of rectangles, parallelograms, triangles, trapezoids, and kites. In this lesson you will use these formulas, along with other methods, to find the approximate areas of irregularly shaped figures. Investigation: Solving Prolems with Area Formulas On the next page, you ll find eight geometric figures. For each figure, find a way to calculate the approximate area. Then record the area and write a sentence or two explaining how you found it. It may help to trace a figure onto another sheet of paper. Below are some hints for how you might find the area of each figure. Read these hints only if you get stuck. There are lots of ways to find each area. The methods you use may e very different from those descried here. Figure A Figure B the area. Divide the figure into rectangles. This figure is a kite. Use what you learned in Lesson 8.2 to find Figure C This figure is a parallelogram. Use what you learned in Lesson 8.1 to find the area. Figure D Divide the figure into triangles. Figure E This figure is a trapezoid. Use what you learned in Lesson 8.2 to find the area. Figure F Find the area of the two squares. Cut out the other two pieces and rearrange them to form a recognizale shape. Figure G Divide this dodecagon into 12 identical, isosceles triangles with vertex angles at the center of the polygon. Figure H Trace the figure onto a sheet of graph paper. Estimate the numer of squares that fit inside the figure. Or draw as large a rectangle as will fit in the shape. Cut off the remaining pieces and arrange them to form recognizale shapes. Discovering Geometry Condensed Lessons CHAPTER 8 107

6 Lesson 8.3 Area Prolems E G B A H C D F 108 CHAPTER 8 Discovering Geometry Condensed Lessons

7 CONDENSED LESSON 8.4 Areas of Regular Polygons In this lesson you will Discover the area formula for regular polygons You can divide a regular polygon into congruent isosceles triangles y drawing segments from the center of the polygon to each vertex. The center of the polygon is actually the center of the circumscried circle, so each of these congruent segments is called a radius of the regular polygon. In the investigation you will divide regular polygons into triangles. Then you will write a formula for the area of any regular polygon. Investigation: Area Formula for Regular Polygons The apothem of a regular polygon is a perpendicular segment from the center of the polygon s circumscried circle to a side of the polygon. The apothem is also the length of the segment. Follow the steps in your ook to find the formula for the area of a regular n-sided polygon with sides of length s and apothem a. Your findings can e summarized in this conjecture. Regular Polygon Area Conjecture The area of a regular polygon is given y the formula A 1 2 asn or A = 1 ap, 2 where A is the area, P is the perimeter, a is the apothem, s is the length of each side, and n is the numer of sides. C-79 The examples elow show you how to apply your new formulas. EXAMPLE A The area of a regular nonagon is aout cm 2 and the apothem is aout 9.6 cm. Find the approximate length of each side. Solution Because you are trying to find the side length, s, it is proaly easier to use the formula A 1 2 asn. You could also use A 1 ap, 2 solve for P, and then divide the result y 9 (the numer of sides). A 1 2 asn Write the formula (9.6)(s)(9) Sustitute the known values s Multiply s Solve for s. 7 s Divide. Each side is aout 7 cm long. Discovering Geometry Condensed Lessons CHAPTER 8 109

8 Lesson 8.4 Areas of Regular Polygons EXAMPLE B Find the shaded area of the regular pentagon PENTA. The apothem measures aout 2.0 cm. Segment PE measures aout 2.9 cm. E N P T A Solution First, find the area of the entire pentagon. A 1 2 asn A 1 2 (2.0)(2.9)(5) A 14.5 Write the formula. Sustitute the known values. Multiply. The area of the pentagon is aout 14.5 cm 2. The shaded portion makes up 3 5 of the pentagon. (If you divide the pentagon into five isosceles triangles, three will e shaded.) So, the shaded area is aout cm 2, or 8.7 cm CHAPTER 8 Discovering Geometry Condensed Lessons

10 Lesson 8.5 Areas of Circles EXAMPLE A The circumference of a circle is 22 ft. What is the area of the circle? Solution Use the circumference formula to find the radius. Then use the area formula to find the area. C 2 r 22 2 r Write the formula for circumference. Sustitute the known values. 11 r Solve for r. A r 2 A (11) 2 A 121 Write the formula for area. Sustitute the known values. Simplify. The area is 121 ft 2, or aout ft 2. EXAMPLE B At Maria s Pizzeria, a pepperoni pizza with diameter 10 inches costs \$8, and a pepperoni pizza with diameter 12 inches costs \$10. Which size is a etter uy? 10 in. 12 in. \$8 \$10 Solution Find the area of each pizza, and then find the price per square inch. 10-inch pizza 12-inch pizza A r 2 A r 2 (5) 2 (6) The area is 25 in 2. To find the The area is 36 in 2. To find the cost per square inch, divide the cost per square inch, divide the price y the area. price y the area The 10-inch pizza costs aout The 12-inch pizza costs aout 10 per square inch. 9 per square inch. The 12-inch pizza costs less per square inch, so the 12-inch pizza is a etter uy. 112 CHAPTER 8 Discovering Geometry Condensed Lessons

11 CONDENSED LESSON 8.6 Any Way You Slice It In this lesson you will Learn how to find the area of a sector, a segment, and an annulus of a circle In Lesson 8.5, you discovered the formula for calculating the area of a circle. In this lesson you ll learn how to find the areas of three types of sections of a circle. A sector of a circle is the region etween two radii and an arc of the circle. A segment of a circle is the region etween a chord and an arc of the circle. An annulus is the region etween two concentric circles. Examples of the three types of sections are pictured elow. Sector of a circle Segment of a circle Annulus The picture equations elow show you how to calculate the area of each type of section. r a h r r a a 360 r a 360 r 2 A sector r a h a r 2 1_ h A segment R r R R 2 r 2 A annulus Read the examples in your ook carefully. Then read the examples elow. EXAMPLE A R 9 cm and r 3 cm. Find the area of the annulus. R r Discovering Geometry Condensed Lessons CHAPTER 8 113

12 Lesson 8.6 Any Way You Slice It Solution A R 2 r 2 The area formula for an annulus. (9) 2 (3) 2 Sustitute the values for R and r Evaluate the exponents. Sutract. The area of the annulus is 72 cm 2, or aout 226 cm 2. EXAMPLE B The shaded area is 21 cm 2. The radius of the large circle is 12 cm, and the radius of the small circle is 9 cm. Find x, the measure of the central angle. x Solution First, find the area of the whole annulus. A R 2 r 2 The area formula for an annulus. (12) 2 π(9) 2 Sustitute the values for R and r. 63 Simplify. The shaded area, 21 cm 2 x, is 36 0 of the area of the annulus. Use this information to write and solve an equation. x x 120 x The measure of the central angle is CHAPTER 8 Discovering Geometry Condensed Lessons

13 CONDENSED LESSON 8.7 Surface Area In this lesson you will Learn how to find the surface areas of prisms, pyramids, cylinders, and cones You can use what you know aout finding the areas of plane figures to find the surface areas of prisms, pyramids, cylinders, and cones. The surface area of each of these solids is the sum of the areas of all the faces or surfaces that enclose the solid. The faces include the solid s ases and its lateral faces. In a prism, the ases are two congruent polygons and the lateral faces are rectangles or other parallelograms. In a pyramid, the ase can e any polygon and the lateral faces are triangles. Lateral face Bases Base Lateral face Read the Steps for Finding Surface Area on page 462 of your ook. Example A shows how to find the surface area of a rectangular prism. Read the example carefully. Then read Example B, which shows how to find the surface area of a cylinder. Notice that, to find the area of the cylinder s lateral surface, you need to imagine cutting the surface and laying it flat to get a rectangle. Because the rectangle wraps exactly around the circular ase, the length of the rectangle s ase is the circumference of the circular ase. The surface area of a pyramid is the area of the ase, plus the areas of the triangular faces. The height of each triangular face is called the slant height. Use l for the slant height and h for the height of the pyramid. Height Slant height h l Discovering Geometry Condensed Lessons CHAPTER 8 115

14 Lesson 8.7 Surface Area Investigation 1: Surface Area of a Regular Pyramid The lateral faces of a regular pyramid are identical isosceles triangles, and the ase is a regular polygon. l a Each lateral face is a triangle with ase length and height l. What is the area of each face? If the ase is an n-gon, then there are n lateral faces. What is the total lateral surface area of the pyramid? What is the area of the ase in terms of a,, and n? Use your expressions to write a formula for the surface area of a regular n-gon pyramid in terms of the numer of sides n, ase length, slant height l, and apothem a. Using the fact that the perimeter of the ase is n, write another formula for the surface area of a regular n-gon pyramid in terms of slant height l, apothem a, and perimeter of the ase, P. In the next investigation you will find the surface area of a cone with radius r and slant height l. Investigation 2: Surface Area of a Cone As the numer of faces of a pyramid increases, it egins to look like a cone. You can think of the lateral surface as many thin triangles or as a sector of a circle. You can rearrange the triangles to form a rectangle. l r l l r 2 r r Use the diagrams to help you write a formula for the lateral surface area in terms of r and l. Using the expression for the lateral surface area and an expression for the area of the ase, write a formula for the surface area of the cone. 116 CHAPTER 8 Discovering Geometry Condensed Lessons

15 Lesson 8.7 Surface Area Example C in your ook shows you how to apply the formula for the surface area of a cone. Read Example C carefully. Then read the example elow. EXAMPLE Find the surface area of this solid. D 10, d 6, h 14. d h D Solution The surface area is the lateral surface area of the outside cylinder, plus the lateral surface area of the inside cylinder, plus the area of the two ases, which are annuluses. Lateral surface area of outside cylinder 2 D 2 h (14) 140 cm 2 Lateral surface area of inside cylinder 2 d 2 h (14) 84 cm2 Area of one ase D 2 2 d 2 2 So, cm 2 Total surface area (16 ) 256 cm cm 2. Discovering Geometry Condensed Lessons CHAPTER 8 117

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