(114) For example suppose the figure EABCD is to be copied to scale. Place the piece of glass having squares ruled upon it so that the lines on the glass cross the figure as indicated.

Next draw lines horizontally and vertically at of the distance apart of those on the glass.

In the second figure at points in the squares corresponding to those in the first, mark a, b, c, d, e and draw in the figure.

If the figure EABCD had been given together with a scale shewing the lengths of the sides we could have proceeded as follows.

First draw a scale of the length of the first and make similar divisions upon it.

Measure EA, EB, AB from the first scale we find they are respectively 2', 4' 3", and 2' 7".

Construct the triangle cab of sides 2′, 4′ 3′′ and 2′ 7′′ taken from the second scale.

In the same way measure BC and EC of lengths 1' 3" and 5′ 4′′ and construct the triangle ebc with sides of the same length taken from the second scale.

The triangle ECD has sides EC = 5′ 4′′, CD = 2′ 7′′, ED = 4′ 2′′ to which the sides of ecd correspond.

CHAPTER XII.

ON AREAS.

THE area of any rectangular figure can be found by multiplying two of its adjacent edges: thus if a rectangle have sides 2 feet and 3 feet its area is 2 × 3=6 square feet.

It is necessary to distinguish between linear feet and square feet, thus twice three linear feet is a length of 6 feet, but 2 feet multiplied by 3 feet gives an area of 6 square feet.

If we multiply square feet by linear feet we obtain cubic feet so that 6 square feet multiplied by 2 linear feet gives the volume or space occupied by a block the area of the base of which is 6 square feet and of 2 feet thickness.

In the same way inches multiplied by inches give square inches, and square inches by linear give cubic inches.

In thinking then of the areas of any figures we compare them with rectangles and squares of equal area, because when we know the size of the rectangle, or square of equal area we only have to measure two adjacent sides and multiply them together to find the result required.

It will be found (Euclid, Bk. 1.) that a triangle is half the area of a parallelogram on the same base and between the same parallels.

To obtain the area of a triangle then we draw a perpendicular from the vertex to the base and multiply the altitude thus found by half the base, or else the whole base by half the altitude.

Again to find a rectangle equal in area to a triangle we make

its sides equal to the base and half the altitude of the triangle, or else to the altitude and half the base.

To make a triangle equal to a parallelogram we draw it on the same base but double its altitude, thus if ABCD be a parallelogram and if we produce BA to E making AE=AB then the triangle EBC is equal in area to ABCD.

It is proved in (Euclid, Bk. 1.) that triangles upon the same base and between the same parallels are equal to one another, by means of this proposition any rectilineal figure can be reduced to a triangle of equal area.

A triangle can be reduced to an isosceles triangle, or one having any given angle.

Take any triangle ABC and through its vertex draw AD parallel to the base BC.

Bisect BC at right angles by a straight line meeting AD in E, then EBC is an isosceles triangle equal to ABC.

If we draw BM making any given angle with BC meeting AD in M, the triangle MBC is also equal to ABC.

If on BC we make the segment of a circle to contain any given angle (32) and this segment meets AD in P or Q, the triangles PBC and QBC will each have the given vertical angle, and be equal in area to ABC.

When therefore the area and base of any triangle are given to construct it, make a rectangle on the given base of the given area; make a triangle of the same area by doubling the height of the rectangle; through the vertex of this triangle draw a parallel to the base. The vertex of the triangle required is somewhere on this parallel. For example suppose we want to describe a triangle on a base of 2 inches, the area of which is 4 square inches, and one side 5 inches long.

Draw the base of 2 inches, and a parallel to it at a distance of 4 inches, from one of the ends of the base as centre and at a distance of 5 inches make an arc to cut the parallel to the base, this point is the vertex of the triangle required.

When four straight lines are in proportion (Chap. III.) the product of the extremes is equal to the product of the means, thus 23 4: 6, then 2×6 the product of the means = 12, and 4× 3 the product of the means is also 12.