lelogram ABCD, is expressed by AB X BE; or in symbols, ABCD AB X BE. This, &c.

Thus, if AB were 6 times MN (the linear unit*), and EB 3 times the same, then the area of ABCD would be = 18MN2.

ART. 2. The area of a triangle is equal to half the product of any side taken as base, and the corresponding altitude.

In the triangle ABC, if the side ac be taken as base, and the perpendicular to it, BD, as altitude; then, the

B

D C

DEM. BY ART. 31, GEOM., the triangle ABC is equal to half the parallelogram on the same base and between the same parallels (i. e. having the same altitude). But by ART. preceding, a parallelogram on the base AC with the altitude BD would be = AC X BD. Hence, the triangle ABC is =

AC X BD

2

This, &c.

Thus if AC were 8 times MN, and BD 4 times the same, then the area of the triangle ABC would be 16мn2. ART. 3. The side of a square is to its diagonal as unity to the square-root of two.

Let ABCD be a square. Then, AB: BD:: D 1:√2.

DEM. Represent AB by s, and BD by d. Therefore, ART. 47, GEOM., as AB = AD

* In practice the linear unit is generally supposed to be a line of an inch in length. So that in this case the area of a parallelogram whose base was 6 inches, and altitude 3, would be = 18 squares of an inch line, i. e. 18 square-inches.

ART. 4. The altitude of an equilateral triangle is to the side as the square-root of three to two.

Let ABC be an equilateral triangle, BD its altitude. Then, BD AC:: √3:2.

DEM. Represent BD by p, and AC by s; therefore, as AC is divided at the point D into two equal parts (ART. 139, GEOM.) either of them AD, will be represented by A

B

D

S But, by ART. 47, GEOM., AB BD2+ AD, that is,

Observation. The altitude of an equilateral triangle is to half the side as the square-root of three to unity.

For, by the third equation above,

ART. 5. Given any two sides of a right-angled triangle, the third may be found.

Let any two sides of the triangle ABC, right-angled at в, be given, or known. Then, the third remaining side can be found.

DEM. Represent the three sides AB, BC, CA, by the letters s, s', s', respectively.

By ART. 47, GEOM.,

$2 = s2 + s'2;

A

therefore, any two of these quantities being given, the third will be determined. First let the sides, s, s', about the right angle be given; then,

that is, the third side ca is equal to the side equal to the sum of squares of AB and of BC. the side, s", opposite the right angle, and either sides, s, be given; then,

of a square Again, let of the other

that is, the third side BC is equal to the side of a square equal to the difference of the squares of CA and of AB. This, &c.

Thus, if the sides, s, s', about the right angle of a triangle were given respectively equal to 6 and 8 times the linear unit (an inch, for example, which we will represent by i), then,

that is, the third side in this case would be equal to ten times the linear unit, or 10 inches.

Again, if the side, s", opposite the right angle of a triangle were given equal to 5 times i, and one of the other sides, s', equal to 4 times i; then,

that is, the third side in this case would be equal to three times the linear unit, or 3 inches.

ART. 6. In a right-angled triangle, given one of the sides about the right angle, and the sum of the other two, the triangle itself may be found.

DEм. Represent AB by s, and PQ by a, and AC (which is unknown) by x. Consequently BC = PQ Therefore, by ART. 47, GEOM.

AC=α -X.

2ах

that is, if we take a right line (suppose GH represented by b) whose square= = AB2 + PQ2=s2 + a2, then

or, in other words, x is a third proportional to 2a and b. Having thus found ac, we may find BC, for BC = PQ

AC.

This, &c.

ART. 7. Given the base and altitude of a triangle, the inscribed square may be found.

Let ABC be a triangle, AC its base, BD its altitude. Then the area of EFGH, the inscribed square, may be found.

DEM. Represent the base AC by b, and the altitude BD by a, and GF the

B

side of the inscribed square (which is A ED H unknown) by x. Then, ART. 112, GEOM.

AC FG BC BG: BD BI, or, as BI = BD- - ID,

K

that is, a is a fourth proportional to (a+b), b, and a. Having thus found the side of the inscribed square, the square itself is determined. This, &c.

Obs. If the question proposed be numerical, that is to say, if the known quantities be given as so many inches, or

C

parts of an inch (an inch being supposed the linear unit), and if the sought quantity be required only in the same terms, this is immediately presented by the final equation. Thus, in the above example, if b were given = 12 inches,

and a 4 inches, then we should have

that is, the side of the inscribed square would be=3 inches, and.. the square itself would be = 9 square inches.

But if the question proposed be geometrical, that is to say, if the sought quantity be required in its geometrical magnitude, although its value is presented by the final equation, the quantity itself remains to be found by a construction. Thus, in the same example, if the square be required, not in numbers of square inches, but in actual geometric dimensions, we must, from the value found for x, scil.

ab
a+b

construct it. And a method of doing this is suggested by the equation. Draw BK parallel to AC, and equal to BD. Join AK, and draw GF parallel to AC, from the point G, where AK meets one side of the triangle, to the other. Then, GF will be a side of the inscribed square. For, GF BK AG: AK, by ART. 112, GEOM. But the triangles AGC, and BGK, being equiangular to each other, AG: GK AC: BK, by the same ART. Therefore, convertendo, AG AGGK: AC AC + BK, or AGAK¦¦ AC; AC + BD.

Consequently, GF BK:: AC AC + BD; that is, x; a:: b: a + b. Hence, x has been constructed a fourth-proportional to (a+b), b, and a; and therefore is the side of the square required, on which we have only to form a square that the question may be resolved.

However, not only the actual geometric dimensions, but the due position of the sought quantity, may be required. This also is frequently indicated by the equation. Thus, in the same example, not only the size of the inscribed square may be required, but its due position. In order to investigate its due position, it is not sufficient that we find a fourth-proportional to (a+b), b, and a, for that will only