Hot-water pipes and water cans being really cylinders, it is useful to know their surface areas as well as contents. To find the area of a curved surface of a circular cylinder (like a piece of piping), the rule is to multiply the circumference of the base by the height of the cylinder. It may be remarked that a cylinder is really a rectangle rolled up till its two long edges meet. Thus, if a piece of hot-water pipe is 4 in. in diameter and 3 ft. (36 in.) long, the area of its curved surface will be 4 x 3-1416 x 36 = 45239 sq. in. It is thus easy to find the heat surface of the piping in any greenhouse by means of this rule. The curved surface of a water can may be found in the same way.

Where, however, it is necessary to know not only the area of the curved surface, but also that of the two ends, the latter is found by the rule for finding the area of a circle (see above). The whole surface of a cylinder, therefore, equals the area of the two ends plus the length multiplied by the circumference.

Sphere Or Globe

The surface area is found by multiplying the square of the diameter by 3.1416. Thus a globular bush 5 ft. through has a superficial area of 5 x 5 x 31416 = 7854 sq. ft.


The ends of greenhouses and the gable ends of houses, sheds, etc, from the eaves up, are examples of triangles. To find the area of a triangle the rule is to multiply the base by half the perpendicular height; or the other way round - half the base by the perpendicular height - comes to the same thing. Thus if a triangle has a base of 9 ft. and a height of 4 1/2 ft, the area may be found either by (9x4 1/2)/2 =(9x9)/(2x2) = 20 1/2 sq„ ft; or 9/2 x 4 1/2 = 9/2 x 9/2=20 1/4 sq. ft. A somewhat more complicated problem is to find the area of a triangle from the given lengths of the three sides. The rule is: 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. Example: The sides of a triangle are 26 in., 28 in., and 30 in. respectively.

84 The sum of these is 26 + 28 + 30 = 84. Half the sum is 84/2 = 42.

From this subtract each side separately: 42 - 26 = 16; 42 - 28 = 14; 42 - 30 = 12. Multiply half the sum and the three remainders together, thus: 42x16x14x12 = 112,896. The square root of 112,896 is the area of the triangle, viz. 336 sq. in.

The area of an equilateral triangle is readily found by multiplying one side by .433.

From these examples it will be easy to find the total area of the end of a greenhouse or building by adding the area of the rectangular portion to the area of the triangular portion.

Other Figures

The area of a rhombus equals half the product of the two diagonals. The area of a trapezoid equals half the sum of two parallel sides by the perpendicular distance between them.

The area of a trapezium equals the longest diagonal by half the sum of the two perpendiculars falling upon it from the opposite angles.

The area of any irregular four-sided figure with straight sides may be found by dividing it into triangles. Find the area of each triangle separately and add together.