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1. Find the area and perimeter of (a) a square of side 5.83 in (b) a square of side 4.82 m. (c) a rectangle measuring 384 cm x 734 cm.

1. Find the area and perimeter of (a) a square of side 5.83 in (b) a square of side 4.82 m. (c) a rectangle measuring 384 cm x 734 cm.

Elementary Geometry For College Students, 7e
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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b (1) Trapezoid FIGURE 6-55 Quadrilaterals. Area= FIGURE 6-56 One solar panel. 453M Interior Angles of Any Polygon So Adding the six given angles gives us Subtracting yields collecting area per panel 4190 41.44149 in.² There are six panels, so Section 3 + Quadrilaterals You may have found that for a quadrilateral the sum of the interior angles is 360°. for a pentagon it is 540°, and for a polygon of n sides it is 190 total collecting area = 6(4149) 24,890 in.2 Converting now to square feet, there are 144 square inches in a square foot so 24,890 in.² - 24,890 in.² x +++ Example 14: Find angle in Fig. 6-58. Solution: The polygon shown has seven sides, so n = 7. By Eq. 72. sum of angles (7-2)(180°) = 900° Exercise 3. Quadrilaterals 1. Find the area and perimeter of (a) a square of side 5.83 in. (b) a square of side 4.82 m. Sum of the Interior Angles ■ Exploration: Try this. (a) Draw any quadrilateral. Then select a point P inside the quadrilateral and connect it to each vertex, Fig. 6-57. The sum of the angles in each triangle is 180°, so the four triangles contain a total of 4 x 180°, or 720°. From this subtract the four angles around P to get the sum of the interior angles of the quadrilateral. What do you get? Sum of angles=(n-2)180° (b) Try this again with a pentagon, a polygon of 5 sides. (c) Try this again with a polygon of n sides. Can you generalize your result? ■ 278° +62° + 123° +99° +226° +43° = 831° Solution: The sum of the angles of the pentagon are, with n = 5, sum of angles (n - 2)180° 8900831° = 69° ++ Example 15: An Application. The miter angle is the angle at which a saw or miter box must be set to cut a piece of stock to form a joint. It is equal to half the angle between the pieces to be joined. Find the miter angle for the pentagonal window in Fig. 6-59. (52)180= 540° Each interior angle is then 540/5= 108. The miter angle is half of that, or 0-54° (c) a rectangle measuring 384 cm x 734 cm (d) a rectangle measuring 55.4 in. by 73.5 in. 72 *** 10² 144 in 2173 ² *** 2. Find the area of (a) a parallelogram whose base is 4.52 ft and whose altitude is 2.95 ft. (b) a parallelogram whose base is 16.3 m and whose altitude is 22.6 m. (c) a rhombus whose base is 14.2 cm and whose altitude is 11.6 cm (d) a rhombus whose base is 382 in. and whose altitude is 268 in. (e) a trapezoid whose bases are 3.83 m and 2.44 m and whose altitude is 1.86 m. (f) a trapezoid whose bases are 33.6 ft and 24.7 ft and whose altitude is 15.3 ft. Applications 3. What will be the cost, to the nearest dollar, of flagging a sidewalk 312 ft long and 6.5 ft wide, at $13.50 per square yard? How many 9-in.-square tiles will cover a floor 48 ft by 12 ft? Chapter 6 Geometry 5. What will it cost to carpet a floor, 6.25 m by 7.18 m, at $7.75 per square meter? 6. How many rolls of paper, each 8.00 yd long and 18.0 in. wide, will paper the sides of a room 16.0 ft by 14.0 ft and 10.0 ft high, deducting 124 ft for doors and windows? FIGURE 6-57 99 FIGURE 6-58 *** FIGURE 6-59 189 43 7. What is the cost of plastering the walls and ceiling of a room 40 ft long, 36 ft wide, and 22 ft high, at $8.50 per square yard, allowing 1375 ft² for doors, windows, and baseboard? 8. What will it cost to cement the floor of a cellar 25.3 ft long and 18.4 ft wide, at $3.50 per square foot? 9. Find the cost of lining a topless rectangular tank 68 in. long, 54 in. wide, and 48 in. deep with zinc, weighing 5.2 lb per square foot, at $1.55 per pound installed. 1o. A parter of 1000 1 een two paramer sueens, Fig. 6-60. How many acres will
Transcribed Image Text:b (1) Trapezoid FIGURE 6-55 Quadrilaterals. Area= FIGURE 6-56 One solar panel. 453M Interior Angles of Any Polygon So Adding the six given angles gives us Subtracting yields collecting area per panel 4190 41.44149 in.² There are six panels, so Section 3 + Quadrilaterals You may have found that for a quadrilateral the sum of the interior angles is 360°. for a pentagon it is 540°, and for a polygon of n sides it is 190 total collecting area = 6(4149) 24,890 in.2 Converting now to square feet, there are 144 square inches in a square foot so 24,890 in.² - 24,890 in.² x +++ Example 14: Find angle in Fig. 6-58. Solution: The polygon shown has seven sides, so n = 7. By Eq. 72. sum of angles (7-2)(180°) = 900° Exercise 3. Quadrilaterals 1. Find the area and perimeter of (a) a square of side 5.83 in. (b) a square of side 4.82 m. Sum of the Interior Angles ■ Exploration: Try this. (a) Draw any quadrilateral. Then select a point P inside the quadrilateral and connect it to each vertex, Fig. 6-57. The sum of the angles in each triangle is 180°, so the four triangles contain a total of 4 x 180°, or 720°. From this subtract the four angles around P to get the sum of the interior angles of the quadrilateral. What do you get? Sum of angles=(n-2)180° (b) Try this again with a pentagon, a polygon of 5 sides. (c) Try this again with a polygon of n sides. Can you generalize your result? ■ 278° +62° + 123° +99° +226° +43° = 831° Solution: The sum of the angles of the pentagon are, with n = 5, sum of angles (n - 2)180° 8900831° = 69° ++ Example 15: An Application. The miter angle is the angle at which a saw or miter box must be set to cut a piece of stock to form a joint. It is equal to half the angle between the pieces to be joined. Find the miter angle for the pentagonal window in Fig. 6-59. (52)180= 540° Each interior angle is then 540/5= 108. The miter angle is half of that, or 0-54° (c) a rectangle measuring 384 cm x 734 cm (d) a rectangle measuring 55.4 in. by 73.5 in. 72 *** 10² 144 in 2173 ² *** 2. Find the area of (a) a parallelogram whose base is 4.52 ft and whose altitude is 2.95 ft. (b) a parallelogram whose base is 16.3 m and whose altitude is 22.6 m. (c) a rhombus whose base is 14.2 cm and whose altitude is 11.6 cm (d) a rhombus whose base is 382 in. and whose altitude is 268 in. (e) a trapezoid whose bases are 3.83 m and 2.44 m and whose altitude is 1.86 m. (f) a trapezoid whose bases are 33.6 ft and 24.7 ft and whose altitude is 15.3 ft. Applications 3. What will be the cost, to the nearest dollar, of flagging a sidewalk 312 ft long and 6.5 ft wide, at $13.50 per square yard? How many 9-in.-square tiles will cover a floor 48 ft by 12 ft? Chapter 6 Geometry 5. What will it cost to carpet a floor, 6.25 m by 7.18 m, at $7.75 per square meter? 6. How many rolls of paper, each 8.00 yd long and 18.0 in. wide, will paper the sides of a room 16.0 ft by 14.0 ft and 10.0 ft high, deducting 124 ft for doors and windows? FIGURE 6-57 99 FIGURE 6-58 *** FIGURE 6-59 189 43 7. What is the cost of plastering the walls and ceiling of a room 40 ft long, 36 ft wide, and 22 ft high, at $8.50 per square yard, allowing 1375 ft² for doors, windows, and baseboard? 8. What will it cost to cement the floor of a cellar 25.3 ft long and 18.4 ft wide, at $3.50 per square foot? 9. Find the cost of lining a topless rectangular tank 68 in. long, 54 in. wide, and 48 in. deep with zinc, weighing 5.2 lb per square foot, at $1.55 per pound installed. 1o. A parter of 1000 1 een two paramer sueens, Fig. 6-60. How many acres will
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