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Section 6–2 ◆ 155 Triangles 2.16 2.01 B Parallel x 5.43 A 90° 35° Line FIGURE 6–15 FIGURE 6–14 ight of s Angle of elevation Horizontal Observer Ra y (a) Angle of elevation D A x C Horizontal 5 2.5 B m Observer 2.25 m 46.3° Line Angle of depression of si ght 4.06 m (b) Angle of depression FIGURE 6–16 6–2 FIGURE 6–17 FIGURE 6–18 Triangles Polygons A polygon is a plane figure formed by three or more line segments, called the sides of the polygon, joined at their endpoints, as in Fig. 6–19. The points where the sides meet are called vertices. We say that the sides of a polygon are equal if their measures (lengths) are equal. If all of the sides and angles of a polygon are equal, it is called a regular polygon, as in Fig. 6–20. The perimeter of a polygon is simply the sum of its sides. Vertex A Side Interior angle FIGURE 6–19 B A polygon Equilateral triangle Square Pentagon Hexagon FIGURE 6–20 Some regular polygons. Sum of Interior Angles A polygon of n sides has n interior angles, such as those shown in Fig. 6–21. Their sum is equal to the following: Interior Angles of a Polygon Sum of angles (n 2)180 112 Modern definitions of plane figures exclude the interior as part of the figure. Thus in Fig. 6–20, point A is not on the square, while point B is on the square. The interior is referred to as a region. 156 Chapter 6 123˚ ◆ Geometry ◆◆◆ 99˚ Example 4: Find angle in Fig. 6–21. Solution: The polygon shown in Fig. 6–21 has seven sides, so n 7. By Eq. 112, Sum of angles (7 2)180 900 62˚ 278˚ Adding the six given angles gives us 226˚ 278 62 123 99 226 43 831 43˚ So 900 831 69 FIGURE 6–21 Scalene triangle Acute triangle Isosceles triangle ◆◆◆ Triangles A triangle is a polygon having three sides. The angles between the sides are the interior angles of the triangle, usually referred to as simply the angles of the triangle. As shown in Fig. 6–22, a scalene triangle has no equal sides; an isosceles triangle has two equal sides; and an equilateral triangle has three equal sides. An acute triangle has three acute angles; an obtuse triangle has one obtuse angle and two acute angles; and a right triangle has one right angle and two acute angles. Obtuse triangle Altitude and Base The altitude of a triangle is the perpendicular distance from a vertex to the opposite side called the base, or an extension of that side (Fig. 6–23). Equilateral triangle Right triangle FIGURE 6–22. Types of triangles. The measuring of areas, volumes, and lengths is sometimes referred to as mensuration. Area of a Triangle Altitude h Altitude h Base b Area of a Triangle We use the following familiar formula: Base b Area equals one-half the product of the base and the altitude to that base bh A 2 137 ◆◆◆ Example 5: Find the area of the triangle in Fig. 6–23 if the base is 52.0 and the altitude is 48.0. Solution: By Eq. 137, FIGURE 6–23 52.0(48.0) area 1250 2 ◆◆◆ (rounded) If the altitude is not known but we have instead the lengths of the three sides, we may use Hero’s formula. If a, b, and c are the lengths of the sides, the following formula applies: This formula is named for Hero (or Heron) of Alexandria, a Greek mathematician and physicist of the 1st century A.D. Hero’s Formula Area of triangle s(s a)(s b)(s c) where s is half the perimeter, or abc s 2 138 Section 6–2 ◆◆◆ ◆ 157 Triangles Example 6: Find the area of a triangle having sides of lengths 3.25, 2.16, and 5.09. Solution: We first find s, which is half the perimeter. 3.25 2.16 5.09 s 5.25 2 Thus the area is area 5.25(5.25 3.25)(5.25 2.16)(5.25 5.09) 2.28 ◆◆◆ Sum of the Angles An extremely useful relationship exists between the interior angles of any triangle. The sum of the three interior angles A, B, and C of any triangle is 180 degrees. Sum of the Interior Angles ◆◆◆ 139 A B C 180 Example 7: Find angle A in a triangle if the other two interior angles are 38 and 121. Solution: By Eq. 139, A 180 121 38 21 ◆◆◆ Exterior Angles An exterior angle is the angle between the side of a triangle and an extension of the adjacent side, such as angle in Fig. 6–24. The following theorem applies to exterior angles: The exterior angle of a triangle is the sum of the two angles in the triangle that are not the angle along the extension of the adjacent side. Exterior Angle of a Triangle B =A+B A C FIGURE 6–24 An exterior angle. 142 AB Congruent Triangles Two triangles (or any other polygons, for that matter) are said to be congruent if the angles and sides of one are equal to the angles and sides of the other, as in Fig. 6–25. 2 4 5 5 2 4 FIGURE 6–25 Congruent triangles. Similar Triangles Two triangles are said to be similar if they have the same shape, even if one triangle is larger than the other. This means that the angles of one of the triangles must equal the angles of the other triangle, as in Fig. 6–26. Sides that lie between the same pair of equal angles are called corresponding sides, such as sides a and d. Sides b and e, as well as sides c and f, are also corresponding sides. Thus we have the following two theorems: c a b f d e b c a = = e f d FIGURE 6–26 Similar triangles. 158 Chapter 6 We will see in a later chapter that these relationships also hold for similar figures other than triangles, and for similar solids as well. A Geometry Similar Triangles B 5.14 m ◆ 143 Corresponding sides of similar triangles are in proportion. 144 Example 8: Two beams, AB and CD, in the framework of Fig. 6–27 are parallel. Find distance AE. ◆◆◆ Solution: By statement 104, we know that angle AEB equals angle DEC. Also, by statement 105, angle ABE equals angle ECD. Thus triangle ABE is similar to triangle CDE. Since AE and ED are corresponding sides, E 5.8 AE 5.87 7m 7.25 m C If two angles of a triangle equal two angles of another triangle, the triangles are similar. 5.14 7.25 5.14 AE 5.87 p q 4.16 m 7.25 D ◆◆◆ FIGURE 6–27 Right Triangles In a right triangle (Fig. 6–28), the side opposite the right angle is called the hypotenuse, and the other two sides are called legs. The legs and the hypotenuse are related by the well-known Pythagorean theorem. Pythagorean Theorem B c A a + b2 145 a2 b2 c2 C b a2 The square of the hypotenuse of a right triangle is equal to the sum of the squares of the two legs. = c2 ◆◆◆ FIGURE 6–28. The Pythagorean theorem, named for the Greek mathematician Pythagoras (ca. 580–500 B.C.). Example 9: A right triangle has legs of length 6 units and 11 units. Find the length of the hypotenuse. Solution: Letting c the length of the hypotenuse, we have c2 62 112 36 121 157 c 157 12.5 (rounded) ◆◆◆ ◆◆◆ Example 10: A right triangle has a hypotenuse of 5 units in length and a leg of 3 units. Find the length of the other leg. Solution: a2 c2 b2 52 32 16 a 16 4 Common Error Remember that the Pythagorean theorem applies only to right triangles. Later we will use trigonometry to find the sides and angles of oblique triangles (triangles with no right angles). ◆◆◆ Section 6–2 ◆ 159 Triangles Some Special Triangles In a 30–60–90 right triangle [Fig. 6–29(a)], the side opposite the 30 angle is half the length of the hypotenuse. A 45° right triangle [Fig. 6–29(b)] is also isosceles, and the hypotenuse is 2 times the length of either side. A 3–4–5 triangle [Fig. 6–29(c)], is a right triangle in which the sides are in the ratio of 3 to 4 to 5. b= c 2 60˚ x 1 6 b 45˚ 30˚ a (a) 1 (b) FIGURE 6–29 ◆◆◆ 45˚ 公2 c A 8 (c) Some special right triangles. Example 11: In the 3–4–5 triangle of Fig. 6–29(c), side x is 10 cm. Exercise 2 ◆ ◆◆◆ Triangles 1. What is the cost, to the nearest dollar, of a triangular piece of land whose base is 828 m and altitude 412 m, at $2,925 per acre? Assume all the figures are exact. 2. The house shown in Fig. 6–30 is 12.0 m wide. The ridge is 4.50 m higher than the side walls, and the rafters project 0.750 m beyond the sides of the house. How long are the rafters? 3. What is the length of a handrail for a flight of stairs (Fig. 6–31) where each step is 30.5 cm wide and 22.3 cm high? il dra n Ha 16th step 30.5 cm 22.3 cm 0.750 m ? 4.50 m 1st step 12.0 m FIGURE 6–30 The 3–4–5 right triangle is a good one to remember, as it is in regular use. FIGURE 6–31 4. At $13.50 per square metre, find to the nearest dollar the cost of paving a triangular court, its base being 32.0 m and its altitude 21.0 m. 5. A vertical pole 15.0 m high is supported by three guy wires attached to the top and reaching the ground at distances of 20.0 m, 36.0 m, and 66.7 m from the foot of the pole. What are the lengths of the wires? 6. A ladder 13.0 m long reaches to the top of a building when its foot stands 5.00 m from the building. How high is the building? 7. Two streets, one 16.2 m and the other 31.5 m wide, cross at right angles. What is the diagonal distance between the opposite corners? For practice, do these problems using only geometry, even if you know some trigonometry. To help you solve each problem, draw a diagram and label it completely. Look for rectangles, special triangles, or right triangles contained in the diagram. Be sure to look up any word that may be unfamiliar, such as diagonal, radius, or diameter. 160 Sla nt h 21. eight 8 in . Chapter 6 18.4 in. FIGURE 6–32 We have not studied the hexagon as such, but you can solve problems 15 and 16 by dividing the hexagon into right triangles or a rectangle and right triangles. This also works for other regular polygons. ◆ Geometry 8. A room is 5.00 m long, 4.00 m wide, and 3.00 m high. What is the diagonal distance from one of the lower corners to the opposite upper corner? 9. What is the side of a square whose diagonal is 50.0 m? 10. A rectangular park 125 m long and 233 m wide has a straight walk running through it from opposite corners. What is the length of the walk? Height 11. A ladder 8.00 m long stands flat against the side of a building. How far must it be drawn out at the bottom to lower the top by 1.00 m? 12. The slant height of a cone (Fig. 6–32) is 21.8 in., and the diameter of the base is 18.4 in. How high is the cone? 13. A house (Fig. 6–33) that is 25.0 m long and 20.0 m wide has a pyramidal roof whose height is 7.50 m. Find the length PQ of a hip rafter that reaches from a corner of the building to the vertex of the roof. 14. An antenna, 125 m high, is supported by cables of length L that reach from the top of the tower to the ground at a distance D from the base of the antenna. If D is 43 the length of a cable, find L and D. 15. A hex head bolt (Fig. 6–34) measures 2.00 cm across the flats. Find the distance x across the corners. 16. A 1.00-m-long piece of steel hexagonal stock measuring 30.0 mm across the flats has a 10.0-mm-diameter hole running lengthwise from end to end. Find the mass of the bar (density 7.85 g/cm3). 17. Find the distance AB between the centres of the two rollers in Fig. 6–35. P 7.50 m 23.4 cm diameter Q A B .0 20 2.00 cm 15.8 cm diameter m 0m 25. x FIGURE 6–33 38.8 cm FIGURE 6–34 FIGURE 6–35 18. A 0.500-m-long hexagonal concrete shaft measuring 20.0 cm across the corners has a triangular hole running lengthwise from end to end (Fig. 6–36). The triangular hole is in the shape of an equilateral triangle with sides 10.0 cm long. What is the total surface area of the shaft? Side view 00 0.5 End view m 10.0 cm 20.0 cm FIGURE 6–36 Section 6–2 ◆ 161 Triangles 1874 ft. C ay hw ft. 77 28 g Hi A 2025 ft. 111.8° C B 100.4° A B FIGURE 6–37 right angle. Note: Angle B is not a FIGURE 6–38 19. A highway (Fig. 6–37) cuts a corner from a parcel of land. Find the number of acres in the triangular lot ABC. (1 acre 43 560 sq. ft.) 20. A surveyor starts at A in Fig. 6–38 and lays out lines AB, BC, and CA. Find the three interior angles of the triangle. 21. A beam AB is supported by two crossed beams (Fig. 6–39). Find distance x. 2.16 m A B x Parallel 5.16 m FIGURE 6–39 4.25 m 22. Find dimension x in Fig. 6–40. 23. Find dimension x on the beveled end of the shaft in Fig. 6–41. 8.0 mm 59 mm 45˚ 29 mm 150˚ 17 mm R 30° 92 mm 17 mm 27 mm 57 mm 35.0 mm 22 mm x x FIGURE 6–40 FIGURE 6–41 162 Chapter 6 6–3 ◆ Geometry Quadrilaterals A quadrilateral is a polygon having four sides. Quadrilaterals are the familiar figures shown in Fig. 6–42. The formula for the area of each region is given right on the figure. Lengths of line segments are measured in units of length (such as metres, centimetres, and inches), and areas of regions are measured in square units (such as square metres, square centimetres, and square inches). b a a (a) Quadrilateral Area = a2 a (b) Square Area = ab (c) Rectangle a Area = ah a h Area = bh a h a h Area = b (d) Parallelogram (e) Rhombus (a + b) h 2 b (f) Trapezoid FIGURE 6–42 Quadrilaterals. For the parallelogram, opposite sides are parallel and equal. Opposite angles are equal, and each diagonal cuts the other diagonal into two equal parts (they bisect each other). The rhombus is also a parallelogram, so the previous facts apply to it as well. In addition, its diagonals bisect each other at right angles and bisect the angles of the rhombus. The trapezoid has two parallel sides, which are called the bases, and the altitude is the distance between the bases. ◆◆◆ Example 12: A solar collector array consists of six rectangular panels, each 115 cm 235 cm, each with a blocked rectangular area (needed for connections) measuring 12.0 cm 22.5 cm. Find the total collecting area in square metres. Estimate: If the panels were 1.2 m by 2.4 m, or 2.9 m2 each, six of them would be 6 2.9, or 17.4 m2. Since we have estimated each dimension higher than it really is, and have also ignored the blocked areas, we would expect this answer to be higher than the actual area. Solution: area of each panel (115)(235) 27 000 cm2 bocked area (12.0)(22.5) 270 cm2 Subtracting yields collecting area per panel 27 000 270 26 700 cm2 There are six panels, so total collecting area (6)26 700 160 000 cm2 16.0 m2 since there are 100 100 104 square centimetres in a square metre. This is lower than our ◆◆◆ estimate, as expected.