XIII. TRIANGLE. 148. We have shewn in Art. 29 that a triangle is equivalent to half a rectangle having the same base and height: this is the reason of the rule now to be given. 149. To find the area of a triangle. RULE. Half the product of the base into the height will be the area. It is obvious that we may multiply together the base and half the height, or multiply together the height and half the base, or multiply together the base and the height and take half the product. 150. Examples: (1) The base of a triangle is 3 yards, and its height is 4 feet 6 inches. 3 yards 9 feet, 4 feet 6 inches=4 feet. Thus the area is 20 square feet. (2) The base of a triangle is 45 feet, and its height is 36 feet. Half of 36 is 18; 45 x 18-810. Thus the area is 810 square feet. 151. If we know the area of a triangle, and also one of the two dimensions, the base or the height, we can find the other. For if twice the number expressing the area be divided by the number expressing the height, the quotient is the base; and if twice the number expressing the area be divided by the number expressing the base, the quotient is the height. 152. The three sides of a triangle being given, to find the 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 will be the area. 153. Examples: (1) The sides of a triangle are 2 feet 2 inches, 2 feet 4 inches, and 2 feet 6 inches respectively. 2 feet 2 inches=26 inches, 2 feet 4 inches = 28 inches, 2 feet 6 inches 30 inches. 26+28+30=84, 42-26=16, of 84=42; 42-28=14, 42-30=12. 42 x 16 x 14 x 12=112896. The square root of 112896 is 336. Thus the area is 336 square inches. (2) The sides of a triangle are 24, 25, and 26 feet respectively. 24+25+26=75, of 75=37.5. 37.5-24-135, 37.5-25=12.5, 37.5-26-11.5. 37.5 × 13.5 × 125 × 11.5=72773-4375. The square root of 72773-4375 cannot be found exactly; if we proceed to three decimal places we obtain 269.766: so that the area is about 269.766 square feet. 154. We will now solve some exercises. (1) Find the area of the gable end of a house, the breadth being 24 feet, the distance of the eaves from the ground 30 feet, and the perpendicular height of the roof 10 feet. The figure is composed of a rectangle and a triangle. AB or EC is the breadth; the eaves are the junctions of the walls and the roof, as at E and C, so that AE or BC is the E A D B height of the eaves from the ground. The perpendicular height of the roof is the perpendicular from D on EC. The highest part of the roof is called the ridge, so that D is on the ridge. The triangle CDE is called the gable top. Here the area of the rectangle ABCE is 24 × 30 square feet, that is 720 square feet; and the area of the triangle is 24 × 5 square feet, that is 120 square feet. whole area is 840 square feet. Thus the (2) The side of an equilateral triangle is one foot; find the area. 3 area is therefore equal to the square root of that is 16' We may also obtain this result thus. It is shewn in Art. 65 that the height of the triangle is half of the square root of 3, and therefore by Art. 149, the area is one-fourth of the square root of 3. Thus the area is approximately 433 of a square foot; or to seven decimal places 4330127. (3) The sides of a right-angled triangle are 8 feet and 15 feet respectively: find the perpendicular from the right angle on the hypotenuse, and the two parts into which it divides the base. By Art. 149 the area of the triangle is 60 square feet. 120 By Art. 151 the perpendicular is feet, that is 7 feet. 17 Then, by Art. 60, the shorter of the two parts into which the perpendicular divides the base, in feet, is the square root of 8 × 8-X that is the square root of 17 17, 120 120 (4) Having given the sides of a triangle to find the diameter of the circle described round the triangle. The investigation which we shall now give is valuable not only for the result which will be obtained, but also for the illustration which it affords of the method by which geometrical truths are demonstrated. Let ABC be the triangle, AE a diameter of the circle described round the triangle, AD the perpendicular from A on the base BC. BA Join CE. By Art. 33, the angle ACE is a right angle; so that this angle is equal to the angle ADB. By Art. 32, the angle AEC is equal to the angle ABD. Therefore, by Art. 23, the angle BAD must be equal to the angle EAC. Therefore, by Art. 34, the triangles ABD and AEC are similar; so that AB is to AD as AE is to AC; and therefore AB× AC=AD × AE. Thus AE= ABX AC AB× AC× BC AD × BC Hence we have the following result: the diameter of the circle described round a triangle is equal to the product of the sides of the triangle divided by twice the area of the triangle. Suppose, for example, that the sides of the triangle are 26 inches, 28 inches, and 30 inches respectively. By Art. 153 the area is 336 square inches. Thus the diameter of the circle described round the triangle in inches EXAMPLES. XIII. Find the areas of the triangles having the following dimensions: 1. Base 18 feet, height 8 feet. 2. Base 8 yards 1 foot, height 5 yards 2 feet. 3. Base 10 yards 2 feet 6 inches, height 7 yards 1 foot 3 inches. 4. Base 14 chains 15 links, height 12 chains 24 links. Find the areas of the right-angled triangles having the following dimensions : 5. Hypotenuse 421, side 29. 6. Hypotenuse 730, side 152. Find approximately the areas of the right-angled triangles having the following dimensions: |