EXAMPLE II.

To find the area of a rectangle or right-angled parallelogram.

Rule.-Multiply the base into the altitude, and the product will give the area: thus the rectangle represented in Fig. 14, Plate 4, contains in the base AB 8 equal parts, and in the perpendicular AC, 4 of the same parts; if each of these parts represent a foot, the product of 8 by 4 = 5? will be the number of square feet contained in the given rectangle.

Again, let the base be in extent 33 feet 9 inches, and the altitude 18 feet 5 inches; these two quantities multiplied together, as before directed, (pages 189 and 190,) will give for the superficial content of the rectangle 621 square feet 6 inches, and 9 twelfth parts or lines.

EXAMPLE III.

To find the area of a rhombus or distorted square, whose sides are all equal but of which none of the angles are right, as represented in Fig. 11 of Plate 1.

Rule.-Multiply the base AB by the perpendicular alțitude CE, which must diminish in proportion to the acuteness of the angle at A, and the product will be the area.

Let the side of the rhombus be 36 feet 9 inches, and the perpendicular altitude be 32 feet 6 inches; then the product of these two dimensions, or 1194 square feet, is the area 4 inches.

Now, had the figure here measured been not only equalsided, like a square, but also equal-angled, the area would, as in Example I. have been found by squaring the side itself 36 feet 9 inches, giving an area of 1350 feet 61 inches, exceeding the area of the rhombus by 156 feet 2 inches.

EXAMPLE

EXAMPLE IV.

To measure the superficial contents of a rhomboides or parallelogram, as ACDB, (Fig. 15, Plate 4,) the base of which is 8 feet, and the inclined sides AC and BD each 5,657 feet, but the perpendicular altitude BF, or 4 C, is only 4 feet.

Rule.-Multiply the base into the perpendicular altitude, and the product will be the superficial contents, which in this case will be 32 feet; whereas, had the figure been a rectangle of 8 feet by 5,657 feet, the contents would have been 42 feet 3 inches.

These two last rules for calculating the superficial area of a rhombus and a rhomboides, are derived from the 14th Prop. of Geom. where it is shown that the area of any parallelogram (whether it be a rectangle or not) is equal to the product of the base multiplied into the perpendicular altitude. (See also Propositions 11 and 13 of Geom.)

EXAMPLE V.

To measure the area of a triangle, Fig. 6, 8, and 9, Plate 1. Rule. Multiply the base AC into the perpendicular altitude BD, and take one half of the product; or the whole base into one half of the perpendicular altitude; or one half of the base into the whole altitude: all these methods will give the same result for the area of the given triangle, ABC. Let the base be 18 inches, and the altitude be 16 inches, the area will be 144 inches.

This rule is founded on the 12th Prop, of Geometry, Fig. 27, Plate 1, where it is shown that a triangle is one half of a parallelogram, situated on the same base, and of the same altitude; the first method, above pointed out, gives the product of the base multiplied into the altitude for the area of the parallelogram, and its half is that of the given triangle. It is of no importance whether the triangle be acute, obtuse, or right-angled, as the perpendicular altitude must always be the same, whether it falls within the triangle, as in Fig. 6, Plate 1, or without it to the continuation of the base, as in Fig. 8, or coincides with one of the sides, as in Fig. 9.

EXAMPLE VI.

To find the contents of a trapezium (page 355), or figure of 4 sides, none of which are equal or parallel to any other side, such as appears in Fig. 14, Plate 1.

Rule. Multiply the diagonal of the figure into half the sum of the two perpendiculars let fall from the opposite angles to the diagonal, and the product will be the superficial

area.

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In the figure ACDB is a quadrilateral or trapezium, of which all the sides are unequal, and none is parallel to its opposite if the diagonal AD be drawn, the quadrilateral will be divided into two triangles, ADC and ADB, of which the common base AD may be measured, as also the altitude of each triangle, by letting fall on the diagonal perpendiculars from the angles at the vertex of each at C and B; consequently, as the area of the triangle is equal to the product of the base, multiplied by one half of its altitude, (Example 5,) the area of the trapezium must be equal to the product of the diagonal, multiplied by half the sum of the altitudes of both triangles.

It is evident that the result will be the same from which so ever angles the diagonal and perpendiculars to it are drawn ; for had it been drawn from C to B, instead of from A to D

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