a 501. The Altitude of a triangle is a line drawn perpendicular to the base from the angle opposite. 1. The dotted vertical lines in the figures represent the altitude. 2. Triangles are named from the relation both of their sides and angles. 502. An Equilateral Triangle has its three sides equal. 503. An Isosceles Triangle has only two of its sides equal. 504. A Scalene Triangle has all of its sides unequal. / FIG. 1. FIG. 2. FIG. 3. Scalene. Equilateral. Isosceles. 505. An Equiangular Triangle has three equal angles. (Fig. 1.) 506. An Acute-angled Triangle has three acute angles. (Fig. 2.) 507. An Obtuse-angled Triangle has one obtuse angle. (Fig. 3.) EXAMPLES. 508. The base and altitude of a triangle being given to find its area. 1. Find the area of a triangle whose base is 26 ft. and altitude 14.5 ft. OPERATION. = 14.5 14.5 x 26 = 2 = 1881 sq. ft. Or, 26 x 1887 sq. ft., area. 2 RULE. Divide the product of the base and altitude by 2, or multiply the base by one half the altitude. 2. What is the area of a triangle whose altitude is 10 yd. and base 40 ft. ? Aw.s. 600 sq. ft. Find the area of a triangle: 3. Whose base is 12 ft. 6 in. and altitude 6 ft. 9 in. Ans. 42 sq. ft. 4. Whose base is 25.01 ch. and altitude 18.14 ch. 5. What will be the cost of a triangular piece of land whose base is 15.48 ch. and altitude 9.67 ch., at $60 an acre? 509. The area and one dimension being given to find the other dimension. 1. What is the base of a triangle whose area is 189 sq. ft. and altitude 14 ft. ? OPERATION. 189 sq. ft. x 2 ÷ 14 27 ft., base. Double the area, then divide by the given di 2. Find the altitude of a triangle whose area is 20 sq. ft. and base 3 yards. Find the other dimension of the triangle: 3. When the area is 65 sq. in. and the altitude 10 in. Ans. 13 in. 4. When the base is 42 rd. and the area 588 sq. rd. 5. I paid $1050 for a piece of the rate of $5 per square rod. altitude? land in the form of a triangle, at If the base is 8 rd., what is its Ans. 50 rods. 510. The three sides of a triangle being given to find its area. 1. Find the area of a triangle whose sides are 30, 40, and 50 ft. OPERATION. (30+40 +50) ÷ 2 = 60; 603030; 60 - 40 = 20; 60 50 10. √60 x 30 x 20 x 10 = 600 sq. ft., area. RULE. = - From half the sum of the three sides subtract each side separately; multiply the half-sum and the three remainders together; the square root of the product is the area. 2. What is the area of an isosceles triangle whose base is 20 ft., and each of its equal sides 15 ft. ? Ans. 111.8 sq. ft. 3. How many acres are there in a field in the form of an equilateral triangle whose sides measure 70 rods ? Ans. 13 A. 41.76 P. 4. The roof of a house 30 ft. wide has the rafters on one side 20 ft. long, and on the other 18 ft. long. How many square feet of boards will be required to board up both gable ends ? 511. The following principles relating to right-angled triangles have been established by Geometry: 1. The square of the hypotenuse of a right-angled triangle is equal to the sum of the squares of the other two sides. 2. The square of the base, or of the perpendicular, of a right-angled triangle is equal to the square of the hypotenuse diminished by the square of the other side. 512. To find the hypotenuse. 1. The base of a right-angled triangle is 12, and the perpendicular 16. What is the length of the hypotenuse ? OPERATION. 122 + 162 = 400 (511, 1). V400 = 20, hypotenuse. RULE. Extract the square root of the sum of the squares of the base and the perpendicular; and the result is the hypotenuse. 2. The foot of a ladder is 15 ft. from the base of a building, and the top reaches a window 36 ft. above the base. What is the length of the ladder ? Ans. 39 ft. 3. If the gable end of a house 40 ft. wide is 16 ft. high, what is the length of the rafters ? 4. A park 25 ch. long and 23 ch. wide has a walk running through it from opposite corners in a straight line. What is the length of the walk ? Ans. 33.97 + ch. 5. A room is 20 ft. long, 16 ft. wide, and 12 ft. high. What is the distance from one of the lower corners to the opposite upper corner ? Ans. 28 ft. 3.36 in. = a a 513. To find the base or perpendicular. 1. The hypotenuse of a right-angled triangle is 35 ft., and the perpendicular 28 ft. What is the base ? OPERATION. = 352 — 282 = 441 (511, 2). V441 = 21 ft., base. RULE. - Extract the square root of the difference between the square of the hypotenuse and the square of the given side; and the result is the required side. 2. The hypotenuse of a right-angled triangle is 53 yd., and the base 84 ft. Find the perpendicular. 3. A line reaching from the top of a precipice 120 ft. high, on the bank of a river, to the opposite side is 380 ft. long. How wide is the river ? Ans. 360 ft. 6+ in. 4. A ladder 52 ft. long stands against the side of a building. How many feet must it be drawn out at the bottom that the top may be lowered 4 ft. ? Ans. 20 ft. QUADRILATERALS. 514, A Quadrilateral is a plane figure bounded by four straight lines, and having four angles. There are three kinds of quadrilaterals, the Parallelogram, Trapezoid, and Trapezium. 515. A Parallelogram is a quadrilateral which has its opposite sides parallel. There are four kinds of parallelograms, the Square, Rectangle, Rhomboid, and Rhombus. 516. A Rectangle is any parallelogram having its angles right angles. 517. A Square is a rectangle whose sides are equal. 518. A Rhomboid is a parallelogram whose opposite sides only are equal, but whose angles are not right angles. 519. A Rhombus is a parallelogram whose sides are all equal, but whose angles are not right angles. Rhomboid. Rhombus. Square. Rectangle. 520. A Trapezoid is a quadrilateral, two of whose sides are parallel and two oblique. 521. A Trapezium is a quadrilateral having no two sides parallel 522. The Altitude of a parallelogram or trapezoid is the perpendicular distance between its parallel sides. The vertical dotted lines in the figures represent the altitude. 523. A Diagonal of a plane figure is a straight line joining the vertices of two angles not adjacent. EXAMPLES. 524. To find the area of any parallelogram. 1. Find the area of a parallelogram whose base is 16.25 ft. and altitude 7.5 ft. OPERATION. 16.25 x 7.5 = 121.875 sq. ft., area. RULE. — Multiply the base by the altitude. 2. The base of a rhombus is 10 ft. 6 in., and its altitude 8 ft. What is its area ? 3. How inany acres are there in a piece of land in the form of a rhomboid, the base being 8.75 ch. and altitude 6 ch.? Ans. 54 A. |