#2 (f) TrapezoidFIGURE 6-55 Quadrilaterals.Area-FIGURE 6-56 One solar panel.Section 3 Quadrilaterals453MInterior Anglesof Any PolygonSoAdding the six given angles gives usSubtracting yieldscollecting area per panel 4190414-4149 in.²There are six panels, soYou 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 is190total collecting area= 6(4149) 24,890 in.2Converting now to square feet, there are 144 square inches in a square foot so24,890 in.²2 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°) - 900Exercise 3. Quadrilaterals1. 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 AnglesExploration: Try this.(a) Draw any quadrilateral. Then select a point P inside the quadrilateral andconnect it to each vertex, Fig. 6-57. The sum of the angles in each triangleis 180°, so the four triangles contain a total of 4 x 180°, or 720". From thissubtract the four angles around P to get the sum of the interior angles of thequadrilateral. What do you get?Sum of angles (-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°8900831° 69°** Example 15: An Application. The miter angle is the angle at which a saw or miterbox must be set to cut a piece of stock to form a joint. It is equal to half the anglebetween the pieces to be joined. Find the miter angle for the pentagonal window inFig. 6-59.5.Solution: The sum of the angles of the pentagon are, withsum of angles(n-2)180°(52)180-540°Each interior angle is then 540/5= 108. The miter angle is half of that, or0-54°(c) a rectangle measuring 384 cm x 734 cm.(d) a rectangle measuring 55.4 in. by 73.5 in.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.72******(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.Applications3. What will be the cost, to the nearest dollar, of flagging a sidewalk 312 ft longand 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 Geometry144 in 2 173 ²5. 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 thesides of a room 16.0 ft by 14.0 ft and 10.0 ft high, deducting 124 ft for doorsand windows?FIGURE 6-57FIGURE 6-58***FIGURE 6-591894397. What is the cost of plastering the walls and ceiling of a room 40 ft long, 36 ftwide, 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.verweer ter pourme, Fig. 6-60. How many acres will10. is parcer of mind nies

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#2 (f) TrapezoidFIGURE 6-55 Quadrilaterals.Area-FIGURE 6-56 One solar panel.Section 3 Quadrilaterals453MInterior Anglesof Any PolygonSoAdding the six given angles gives usSubtracting yieldscollecting area per panel 4190414-4149 in.²There are six panels, soYou 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 is190total collecting area= 6(4149) 24,890 in.2Converting now to square feet, there are 144 square inches in a square foot so24,890 in.²2 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°) - 900Exercise 3. Quadrilaterals1. 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 AnglesExploration: Try this.(a) Draw any quadrilateral. Then select a point P inside the quadrilateral andconnect it to each vertex, Fig. 6-57. The sum of the angles in each triangleis 180°, so the four triangles contain a total of 4 x 180°, or 720&#34;. From thissubtract the four angles around P to get the sum of the interior angles of thequadrilateral. What do you get?Sum of angles (-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°8900831° 69°** Example 15: An Application. The miter angle is the angle at which a saw or miterbox must be set to cut a piece of stock to form a joint. It is equal to half the anglebetween the pieces to be joined. Find the miter angle for the pentagonal window inFig. 6-59.5.Solution: The sum of the angles of the pentagon are, withsum of angles(n-2)180°(52)180-540°Each interior angle is then 540/5= 108. The miter angle is half of that, or0-54°(c) a rectangle measuring 384 cm x 734 cm.(d) a rectangle measuring 55.4 in. by 73.5 in.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.72******(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.Applications3. What will be the cost, to the nearest dollar, of flagging a sidewalk 312 ft longand 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 Geometry144 in 2 173 ²5. 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 thesides of a room 16.0 ft by 14.0 ft and 10.0 ft high, deducting 124 ft for doorsand windows?FIGURE 6-57FIGURE 6-58***FIGURE 6-591894397. What is the cost of plastering the walls and ceiling of a room 40 ft long, 36 ftwide, 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.verweer ter pourme, Fig. 6-60. How many acres will10. is parcer of mind nies

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