Area of a Parallelogram.

The area of a parallelogram ABCD is equal to the area of the rectangle ABEF on the same base AB and between the same parallels AB, FC.

Now BE is the altitude of the parallelogram ABCD, if AB be taken as the base.

Hence area of

ABCD=rect. AB, BE.

If then the measure of the base be denoted by b,

and

altitude

h,

the measure of the area of the ☐ will be denoted by ōh. That is, when the base and altitude are commensurable, measure of area measure of base into measure of altitude.

Area of a Triangle.

If from one of the angular points A of a triangle ABC, a perpendicular AD be drawn to BC, fig. 1, or to BC produced,

and if, in both cases, a parallelogram ABCE be completed of which AB, BC are adjacent sides,

area of ▲ ABC= half of area of ABCE.

Now if the measure of BC be b,

Area of a Trapezium.

Let ABCD be the given trapezium, having the sides AB,

:: LBAO L CFO, and ▲ AOB= L FOC, and AB=CF;

=

.. A COF ▲ AОB.

=

Hence trapezium ABCD = ▲ ADF.

I. 26.

Now suppose the measures of AB, CD, AE to be m, n, p respectively;

.. measure of DF=m+n, :: CF= AB.

Then measure of area of trapezium

=(measure of DFx measure of AE)

= } (m + n) × p.

That is, the measure of the area of trapezium is found by multiplying half the measure of the sum of the parallel sides by the measure of the perpendicular distance between the parallel sides.

Area of an Irregular Polygon.

There are three methods of finding the area of an irregular polygon, which we shall here briefly notice.

I. The polygon may be divided into triangles, and the area of each of these triangles be found separately.

B

E

Thus the area of the irregular polygon ABCDE is equal to the sum of the areas of the triangles ABE, EBD, DBC.

II. The polygon may be converted into a single triangle of equal area.

If ABCDE be a pentagon we can convert it into an equivalent quadrilateral by the following process:

Join BD and draw CF parallel to BD, meeting ED produced in F, and join BF.

Then will quadrilateral ABFE=pentagon ABCDE.

For ABDF ▲ BCD, on same base BD and between same parallels.

If, then, from the pentagon we remove ▲ BCD, and add ▲ BDF to the remainder, we obtain a quadrilateral ABFE equivalent to the pentagon ABCDE.

The quadrilateral may then by a similar process be converted into an equivalent triangle, and thus a polygon of any number of sides may be gradually converted into an equivalent triangle.

The area of this triangle may then be found.

III. The third method is chiefly employed in practice by Surveyors.

Draw AE, the longest diagonal, and drop perpendiculars on AE from the other angular points of the polygon.

The polygon is thus divided into figures which are either right-angled triangles, rectangles, or trapeziums; and the areas of each of these figures may be readily calculated.

NOTE VII. On Projections.

The projection of a point B, on a straight line of unlimited length AE, is the point M at the foot of the perpendicular dropped from B on AE.

The projection of a straight line BC, on a straight line of unlimited length AE, is MN,-the part of AE intercepted between perpendiculars drawn from B and C.

When two lines, as AB and AE, form an angle, the projection of AB on AE is AM.

B

We might employ the term projection with advantage to shorten and make clearer the enunciations of Props. XII. and XIII. of Book II.

Thus the enunciation of Prop. XII. might be:

"In oblique-angled triangles, the square on the side subtending the obtuse angle is greater than the squares on the sides containing that angle, by twice the rectangle contained by one of these sides and the projection of the other on it."

The enunciation of Prop. XIII. might be altered in a similar

manner.