Calculate to three decimal places the areas of the triangles having the following sides:

27. The sides of a triangle are 11, 24, and 31: shew that the area is 66/3.

28. The sides of a triangle are 61, 62, and 63: shew that the area is 744/5.

29. The sides of a triangle are 68, 75, and 77; a straight line is drawn across the triangle parallel to the longest side, and dividing each of the other sides into two equal parts: find the area of the two parts into which the triangle is divided.

30. The sides of a triangle are 111, 175, and 176; two straight lines are drawn across the triangle parallel to the longest side, and dividing each of the other sides into three equal parts: find the areas of the three parts into which the triangle is divided.

31. The sides of a triangle are 13, 14, and 15 feet: find the perpendicular from the opposite angle on the side of 14 feet.

32. The sides of a triangle are 51, 52, and 53 feet: find the perpendicular from the opposite angle on the side of 52 feet, and find the area of the two triangles into which the original triangle is thus divided.

33. The side of a square is 100 feet; a point is taken inside the square which is distant 60 feet and 80 feet respectively from the two ends of a side: find the areas of the four triangles formed by joining the point to the four corners of the square.

34. ABC is a triangle, and AD is the perpendicular from A on BC. If AD=13 feet, and the lengths of the perpendiculars from D on AB and AC be 5 feet and 10-4 feet respectively, find the sides and the area of the triangle.

35. The base of a triangular field is 1166 links, and the height is 738 links; the field is let for £24 a year: find at what price per acre the field is let.

36. The sides of a triangular field are 350, 440, and 750 yards; the field is let for £26. 5s. a year: find at what price per acre the field is let.

37. Find to the nearest square inch the area of a triangle whose sides are 5, 6, and 7 feet.

38. A field is in the form of a right-angled triangle, the two sides which contain the right angle being 100 yards and 200 yards: find its area. If the triangle be divided into two parts by a straight line drawn from the right angle perpendicular to the opposite side, find the area of each part.

39. The sides of a triangle are in the proportion of the numbers 5, 12, and 13; and the perimeter is 50 yards: find the area.

40. The sides of a triangle are in the proportion of the numbers 13, 14, 15; and the perimeter is 70 yards: find the area.

41.. Find the cost of painting the gable end of a house at 1s. 9d. per square yard; the breadth being 27 feet, the distance of the eaves from the ground 33 feet, and the perpendicular height of the roof 12 feet.

Find the diameters of the circles described round the triangles having the following sides:

42. 293, 285, 68.

43. 136, 125, 99.

44. 123, 122, 49.

45. 267, 244, 161.

XIV. QUADRILATERALS.

155. A quadrilateral can be divided into two triangles by drawing a diagonal; then the area of each triangle can be found, and the sum of the areas of the triangles will be the area of the quadrilateral.

1

B

× 12 × 3 = 18;

the area of the triangle ACD = × 12×4=24;

18+24=42.

Thus the area of the quadrilateral is 42 square feet.

(2) A diagonal of a quadrilateral is 88 yards, and the perpendiculars on it from the opposite angles are 30 yards and 25 yards respectively.

1320+1100=2420. Thus the area of the quadrilateral is 2420 square yards, that is half an acre.

157. It is obvious that in Examples like those of the preceding Article, instead of calculating separately the areas of the two triangles, we may find the area of the quadrilateral by using the following rule: multiply the sum of the perpendiculars by the diagonal, and take half the product.

Thus in the first example of Art. 156 the sum of the perpendiculars is 7 feet; and therefore the area in square

1

feet × 12×7=42; in the second example the sum of

2

the perpendiculars is 55 yards, and therefore the area in

158. In the particular case in which the diagonals of a quadrilateral intersect at right angles the rule just given amounts to this: take half the product of the two diagonals.

By the aid of a figure the truth of this rule becomes self-evident. Let ABCD be a quadrilateral such that its diagonals AC and BD intersect at right angles; let E be the point of intersection. Through A and C draw straight lines parallel to BD; through

A

E

B and D draw straight lines parallel to AC. Thus a rectangle KLMN is formed. Now it is easy to see that, the triangle AEB is equal to the triangle BKA, the triangle BEC is equal to the triangle CLB, the triangle CED is equal to the triangle DMC, and the triangle DEA is equal to the triangle AND. Thus the quadrilateral ABCD is half of the rectangle KLMN; and therefore the area of the quadrilateral is equal to half the product of AC and BD.

159. The diagonals of a rhombus intersect at right angles; and therefore the rule of the preceding Article may always be applied to a rhombus.

T. M.

6

160. It is usual to give a special rule for finding the area of a trapezoid.

Let ABCD be a quadrilateral having the sides AB and CD parallel. From C draw CE perpendicular to AB; and from A draw AF perpendicular to CD. Then

1 2

the area of the triangle ABC=—AB×CE;

the area of the triangle ADC=CD× AF.

Now we may admit that AF=CE; and therefore the area of the quadrilateral is equal to the product of CE into half the sum of AB and CD. Thus we obtain the rule which will now be given.

161. To find the area of a trapezoid.

RULE. Multiply the sum of the two parallel sides by the perpendicular distance between them, and half the product will be the area.

162. Examples:

(1) The two parallel sides of a trapezoid are 2 feet 6 inches, and 3 feet 4 inches respectively; and the perpendicular distance between them is 1 foot 8 inches.

2 feet 6 inches=23 feet, 3 feet 4 inches=33 feet.

1 foot 8 inches=13 feet; 2+3)=5,

Thus the area of the trapezoid is 43 square feet.