A Polygon Is Convex If No Line That Contains a Side of the Polygon Passes ,Through the Interior of the Polygon

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A polygon is convex if no line that contains a side of the polygon passes ,through the interior of the polygon. A polygon that is not convex is called concave. A polygon is equilateral if all of its sides are congruent. A polygon is equiangular if all of its interior angles are congruent. A polygon is regular if it is both equilateral and equiangular.

Decide whether the polygon is convex or ,concave.

None of the extended sides pass through the interior. So, the polygon is convex.

At least one extended side passes through the interior. So, the polygon is concave.

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Decide whether the polygon is regular. Explain your answer.

a. The polygon is equilateral because all of the sides are congruent. The polygon is not equiangular because not all of the angles are congruent. So, the polygon i~s not regular. b. The polygon is equilateral because all of the sides are congruent. The polygon is equiangular because all of the angles are congruent. So, the polygon is regular.

The polygon is regular. Find the value of x. B__C

Because the polygon is regular, all of its sides are congruent. +1 AB = DE The sides of the polygon are congruent. E 25 = 6x+ 1 Substitute 25 for AB and 6x + 1 for DE. 24 = 6x Subtract 1 from each side. 4--X Divide each side by 6.

10.

For use wi~h pa~es

Theorem 8.1 Polygon Interior Angles Theorem The sum of the measures of the interior angles of a convex polygon with n sides is (n - 2) ¯ 180°. Theorem 8.2 Polygon Exterior Angles Theorem ¯ The sum of the measures of the exterior angles of a convex polygon, one angle at each vertex, is 360°. ~

Find the sum of the measures of the interior angles of the polygon.

The polygon has six sides (hexagon). Use the Polygon Interior Angles Theorem and substitute 6 forn. (n - 2). 180° = (6 - 2) ¯ 180° Substitute 6 for n. = 4 ¯ 180° Simplify. = 720° Multiply...... Find the sum of the measures of the interior angles of the polygon. 1. quadrilateral 2, pentagon 3o octagon

Find the measure of Z A in the diagram. A B

~135° 83°~, D C

Far use with pages ~,16-423

The polygon has five sides, so the sum of the measures of the interior angles is (n - 2). 180° = (5 - 2). 180° = 3. 180° = 540°: Add the measures of the interior angles and set the sum equal to 540°. 90° + 135° + 83° + 96° + mZA = 540° The sum is 540°. 404° + mZA = 540° Simplify. mZA = 136° Subtract 404° from each side. Exercises for Example 2 F~d the ~easure ~f Z A. 4. Bj 5. E F C

A D B

c D D c E

Find the value of x.

Using the Polygon Exterior Angles Theorem, set the sum of the exterior angles equal to 360°. 130°+x°+100°+30°+x°=360° Polygon Exterior Angles Theorem 260 + 2x = 360 Add like terms. 2x = 100 Subtract 260 from each side. x = 50 Divide each side by 2. Answer: The value of x is 50. , Exercises fer Example 3 Find the va~ue of x.

<70°\ 147°"

For use with pages 424-429

Find the area of squares and rectangles.

The amount of surface covered by a figure is its area. Area of a Square: Area = (side)2 Area of a Rectangle: Area = (base)(height)

Find the Area of a Square and the Area of a RectanOle

a. Find the area of the square. b. Find the area of the rectangle.

3 cm 22 cm

a. Use the formula for the area of a square and substitute 6 for s, A = s2 Formula for the area of a square = 62 Substitute 6 for s. = 36 Simplify. Answer: The area of the square is 36 square inches. b. Use the formula for the area of a rectangle and substitute 22 for b and 3 for h. A = bh Formula for the area of a rectangle , = 22 ¯ 3 Substitute 22 for b and 3 for h. = 66 Simplify. Answer: The area of the rectangle is 66 square centimeters. Exercises for Example I , ...... Find the area of the square or rectangle. 1. rectangle 2. square 3. rectangle

6 cm 4in. lOft 15 crn

8ft

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COi~Tli~! U ~= D For use with pages a,2o,-42g

The rectangle has an area of 132 square feet. Find its base. llft A = bh Formula for the area of a rectangle 132 = b ¯ 11 Substitute 132 for A and 11 for h. 12 = b Divide each side by 11. Answer: The base of the rectangle is 12 feet...... ¯ ...... A gives the area of the rectangJe. Find the missing side Jength. 6. h h 20 cm 9ft A = 12 in.2 A = 81 ft2 A = 140 cm2

Find the area of the polygon made up of rectangles, t~ 10 in.

4in. Add the areas of the two rectangles. Notice that the 1 in. base and the height of rectangle A are given in the 4in. diagram. The base of rectangle B is 10 inches minus 4 inches. The height of rectangle B is 4 inches minus 1 inch. Area = Area of A + Area of B =4.4+(10-4).(4- 1) =4"4+6.3 = 16 + 18 = 34 in.2 Exercises for Example 3

2 cm 5cm

~ 15 in. ~1 I 11 cm I

For use with pages ~0-o,38

The height ef a .triangle is the perpendicular segment from a vertex to the line containing the opposite side, called the base of the triangle. Area of a ~iangle: ~ea = ~(base)(height) Theorem 803 Areas of S~mflar P~lyg~ns If two polygons simil~with a scale factor of ~, then the ratio of ~2 their ~eas is -- b~"

Find the Area of a Rioht Triangle

Find the area of the right triangle.

Use the formula for the area of a triangle and 15 yd substitute 15 for b and 8 for h. A = ~bh Formula for the area of a triangle ’ = ~(15)(8) Substitute 15 for b and 8 for h. = 60 Simplify. Answer." The right triangle has an area of 60 square yards.

Find the area of the right triangle. 1. 2. 3. 5 in. 8cm

12 in. 18 cm 4ft

Find the area of the triangle.

A = ½bh Formula for the area of a triangle 8 cm = ½(8)(4) Substitute 8 for b and 4 for h. = 16 Simplify. Answer: The triangle has an area of 16 square centimeters. Copyright © McD0ugal Littell Inc. Chapter 8 Resource Book All rights reserved. NAME DATE

For use with pages ~0-~38

Exercises fer Example 2

¸4. 14ft =

20 cm 9in.

Find the base of the triangle, given that its area is 42 square feet.

A = ½bh Formula for the area of a triangle 1 42 = ~b. 6 Substitute 42 for A and 6 for h. 84=b.6 Multiply each side by 2. 14 = b Divide each side by 6. Answer." The triangle has a base of 14 feet. Exercises fer Exampie 3 , A gives the area of the triangle. Find the missing measure. 7. A= 15in.2 8. A= 126cm2 9. A=6ft2 6ft

10 in.

Areas of Similar Tdang~es

/~ABC ~/~DEF. Find the scale factor of /~DEF to/~ABC. Then find the ratio of their areas. 5 C D ~;~TB~ 10 F The scale factor of ADEF to/~ABC is 4 _ 2 Then by Theorem 8.3, the ratio of the 2 1" 22 -J-’4 You can verify this by finding the~ ~eas. ~eas of ~DEF to ~ABC is ~ = Exercise for E~amp~e 4 ~ ~. ~ABC ~ GD£F. FJnd th~ sca]~ ~acto~ o~ B E ~DEF to ~ABC. Then find the ~ado o~ the~ ~eas. F 10 C

For use with pages 439-4a,5

Find the area of parallelograms.

Either pair of parallel sides of a parallelogram are called the bases la parallelogram. The shortest distance between the bases of a parallelogram is called the height ot" a parallelogram. Area of a Parallelogram: Area = (base)(height) i 1 Area of a Rhombus: Area = =-(productz- of diagonals)

Find the Area

Find the area of the parallelogram.

Use the formula for the area of a parallelogram and 8 yd substitute 8 for b and 6 for h. A = bh Formula for the area of a parallelogram = (8)(6) Substitute 8 for b and 6 for h. = 48 Multiply. Answer: The parallelogram has an area of 48 square yards...... Find the area of the para~e~ogramo :2.

12 cm

17 cm 5m .

4. 12 yd =

10 yd

14ft

For use with pages

Find the base of the parallelogram given that its area is 105 square inches.

Use the formula for the area of a parallelogram and substitute 105 for A and 7 for h. A = bh Formula for the’ area of a parallelogram 105 = b ¯ 7 Substitute 105 for A and 7 for h. 15 = b Divide each side by 7. Answer: The parallelogram has a base of 15 inches. Exorcises for Example 2 A ~ves ~he area ~f ~he parallelogram, F~nd ~he m~ss~ng measure. 5. A=63m2 & A= 144ft2 7. A=55cm2

16ft

Find the area of the rhombus.

Use the formula for the area of a rhombus. Add the segment lengths to find the values of d~ and d2. 1 A = Formula for the area of a rhombus

= ½(6 + 6)(3 + 3) Substitute (6 + 6) for d~ and (3 + 3) for d2. ½(12)(6) Simplify within parentheses. =. 36 Multiply. Answer." The area of the rhombus is, 36 square feet.

Find the area ~f the rhombus.

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F~nd the area o~ trapez~s,

The shortest distance between the bases of a trapezoid is the height of the trapezoid. 1 Area of a ~lYapezoid: Area = -~(height)(sum of bases)

Find the area of the trapezoid. 15m

A = -~h(bI + b2) Formula for the area of a trapezoid 13m = ½(9)(15 + 13) Substitute 9 for h, 15 for ba, and 13 for b2. = ½(9)(28) Simplify within parentheses. = 126 mz Multiply.

Find the area of the trapezoid° 1. 2. 15 cm 3. 17 yd

6 in. 8yd 10 in. 11 yd 12 crn

Use the Area of a Trapezoid , Given that the area of the trapezoid b1 is 17,0 square inches, find b1.

16in.

A = ~h(bI + b2) Formula for the area of a trapezoid 170 = ½(lO)(bI + 16) Substitute 170 for A, 10 for h, and 16 for b2. 170 = 5(bI + 16) Simplify -~ (10). 34 = b1 ÷ 16 Divide each side by 5. 18 = bl Subtract 16 from each side. Answer: The value of b~ is 18 inches.

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Exercises fer Example 2

4. A=:lSm2 5. A=S6ft2 6. A=34:2yd~ ,,

7 m bI b1 12 yd 31 yd 5m 6ft

Use the Pj~hag~man Theorem

Find the height using the Pythagorean Theorem. 22 Then find the area of the trapezoid.

t----- 12 -- 38 Find the height of the trapezoid by using the Pythagorean Theorem on the right triangle. 15 a2 + b~ = c2 Pythagorean Theorem h2 + 122 = 152 Substitute 12 for b, 15 for c, and h for a. 12 h2 + 144 = 225 Simplify. h2 = 81 Subtract 144 from each side. h = 9 Take the positive square root of each side. So, the height of the trapezoid is 9 units. Now use the formula for the area of the trapezoid.

_._ A ~h(b1 d- b2)= ½(9)(22 + 38)= ½(9)(60)-- 270 Answer: The area of the trapezoid is 270 square units.

Find the height of the trapezoid using the Pythagorean Theerer~. Then find the area of the t~apez~d. 8. 18 25 15 30 22 34

For use with pages ~,51-a,59

A circle is the set of all points in a plane that are the same distance from a given point, called the center of the circle. The distance from the , center to a point on the circle is the radius. The distance across the circle, through the center, is the diameter. The circumference of a circle is the distance around the circle. An angle whose vertex is the center of a circle is a central angle of the circle. A region of a circle determined by two radii and a part of the circle is called a sector of the circle. Circumference ~of a Circle: Circumference = ~r (diameter) = 27r (radius) Area of a Circle: Area = ~r(radius)2

Find the circumference of the circle.

~’~L~’~~ Use the formula for the circumference of a circle and substitute 7 for r. C = 2~rr Formula for the circumference of a circle = 2 r(7) Substitute 7 for r. = 14~r Simplify. ~ 14(3.14) Use 3.14 as an approximation for ~r. = 43.96 Multiply. Answer: The circumference of the circle is about 44 meters. Exercises for E~amp~e 1 Find the circumference of the ¢~r~e. ~eund your answer to the nearest where number.

[] For use with pages 451-~J59

Find the area of a circle with a radius of 6 feet.

A ’ ,rrr2 Formula for the area of a circle = ,rr (6)2 Substitute 6 for r. = ~r. (36) Simplify. --~ (3.14) ¯ (36) Use 3.14 as an approximation for "rr. ~ 113 ft2 Multiply.

Find the area of the ~zirc~e. Round y~ur answer t~ the nearest

Find the Ame of, Sector Find the area of the shaded sector.

First find the area of the circle. A = 7rr2 = "rr(10)2 ~ 314 yd2 Now find the area of the sector. Let x equal the area of the sector.

Area of sector ___ Measure of central angle Area of entire circle Measure of entire circle x 55° 314 360° Substitute. 360x = 17,270 Cross product property x ~ 48 yd2 Divide each side by 360. Exercises for Exempie 3 . ’