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23. The student should compare the algebraical and the geometrical uses of the word square. In Geometry a square is a figure of a certain shape. In Algebra, a square is a product of two equal numbers. Now, by Art. 19, the measure of the area of a square=the product of the two equal measures of its adjacent sides. It might be supposed that this explains the identity of names in Algebra and Geometry, But this is only a partial explanation. In the above article, we had arbitrarily taken a square as our unit of area. If we had taken some other figure as unit of area, a figure of a corresponding shape would have been represented by the product of two equal numbers. If, e.g., the unit of area had been a triangle whose base and altitude were equal, then any other triangle whose base and altitude were equal would have been measured by the product of two equal numbers (and any triangle would be the product, (instead of į the product,) of its base and altitude]. It is, thus, purely arbitrary, but (as the student may see on reflection) highly convenient, to choose a square as the unit of area. Whatever unit of area had been chosen, any other area would have been proportional to the product of two lengths. This is equivalent to Euclid's statement that “ Similar rectilineal figures are to one another in the duplicate ratio of their homologous sides.”

24. The area of any rectilineal figure may be found by dividing it into triangles : and applying the formula,

triangle = 1 (base x altitude).

25. Since

triangle = 1 (base x altitude), :. triangles of equal altitudes are to one another as their bases.

This is Prop. 1 of Euc. Book VI. The student, who has not read Book VI. is advised, here, to read and learn Props. 2, 4, 5, 6 and 7 of that book. These props. are necessary for a great deal of what follows.

§ 3. MISCELLANEOUS PROPOSITIONS.

Equality in all respects of two triangles.

26. In Euc. I. (4, 8, 26) is proved the equality in all respects of two triangles, three elements of which in each case are given equal. Euclid omits one case, which may be now considered.

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Let ABC, DEF be two triangles, having the two sides AB, AC equal to the two sides DE, DF, each to each; and likewise the angle ABC, opposite to AC, equal to the angle DEF, opposite to DF. The angles ACB, DFE opposite the other equal sides AB, DE shall be either equal or supplementary.

(i) If EF and BC are equal, then, by Euc. I. 4, the triangles ABC, DEF are equal in all respects. But (ii) if not, one of them must be the greater. Let EF be the greater. Produce BC to C', so that BC' = EF, join C'A. Then, by Euc. I. 4, the triangles A BC", DEF are equal in all respects.

But :: AC= AC", ... ACC" = AC'C. [Euc. I. 5.]
And the angles ACB, ACC" are supplementary,
.. the angles ACB, AC'C are supplementary,
i.e. the angles ACB, DFE are supplementary,
.. in case (i) ACB, DFE are equal,
and in case (ii) ACB, DFE are supplementary. Q.E.D.

Cor. 1. Since of two supplementary angles, not equal to one another, one must be acute and the other obtuse, therefore if, in the above proposition, we know either that both the angles ACB, DFE are acute,

that both the angles ACB, DFE are obtuse,

that one of the angles ACB, DFE is right, then it follows that ACB, DFE are necessarily equal without any other alternative.

Cor. 2. If AC is equal to or greater than AB, then ABC " is equal to or greater than ACB.

or

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But two angles of a triangle are together less than two right-angles, therefore 2 ACB is acute.

Similarly DF being equal to AC, and DE to AB, the angle DFE is acute.

This brings us to Cor. 1, showing that the angles ACB, DFE are necessarily equal.

27. The proposition of the last article is called The Ambiguous Case of the equality of two triangles. The ambiguity is removed by the corollaries. Further consideration of the matter is postponed to the chapter on Solution of Triangles.

28. We see from Art. 26, that there are four cases in which the exact equality of two triangles may be shown. But the last case is Ambiguous.

In each case, a side is given equal.
The three elements given equal are in
Case I. Two angles and a side. Euc. I. 26.
Case II. Three sides. Euc. I. 8.

Case III. Two sides and the angle included by them. Euc. I. 4.

Case IV. Two sides and the angle opposite one of them. Art. 27. .

These four cases correspond, respectively, to the four propositions 4, 5, 6, 7 of Book VI. in which the similarity of two triangles is shown.

Case I. Two angles given equal. Euc. VI. 4.
Case II. Three sides given proportional. Euc. VI. 5.

Case III. Two sides given proportional, and the angle included by them given equal. Euc. VI. 6.

Case IV. Two sides given proportional, and the angle opposite one of them given equal. Euc. VI, 7.

29. By Euclid I. 32, the angles of a triangle are together equal to two right-angles.

But 1 right-angle = 90 degrees or 90°,

:. 2 right-angles = 180 degrees or 180°,
.. the angles of a triangle are together equal to 180°.

Example. The angles of a triangle are as 2 : 3:4. Find them.

We have to divide 180° into 3 parts, which are in the proportion 2:3 : 4.

The required parts are

2
(1)

of 180o =
2+3+4
3

of 180°
2+3+4

4 (3)

of 180° 2+3+4

2

of 180° = 40°,
9
3

of 180° = 60°,

4

of 180° = 80°. 9

30. By Euclid I. 47, the square on the hypothenuse of a right-angle is equal to the sum of the squares on the two sides containing the right-angle.

Let ABC be a right-angled triangle, right-angle at A.

If a, b, c are the measures of the 6 sides opposite to A, B, C, respec

tively; then a, 67, ca are the mea

sures of the squares on these sides, A respectively,

.. a= + c.

a

B

с

By this equation, we can find any one side of a right-angled triangle when the other two are given.

Example 1. The sides of a right-angled triangle are 5 ft. and 12 ft. respectively, find the hypothenuse.

Take a foot as the unit of length.
Then 5 and 12 are the measures of the two sides.

.: (measure of hypothenuse)2=52 + 122=25 +144=169,
. measure of hypothenuse = 169=13,
.: hypothenuse = 13 ft.

Example 2. The hypothenuse and base of a right-angled triangle are 2 ft. 5 in. and 1 ft. 9 in. respectively. Find the altitude.

Take an inch as unit of length. Let x measure altitude. Then, we have,

292=212 + x2.
.: x2=292 – 212=841 – 441=400,
.: x=20,
.: altitude=20 in.=1 ft. 8 in.

EXAMPLES I.

1. Express in grades, minutes and seconds the following decimals of a right-angle.

(1) 000003, (2) .0025, (3) 04, (4) 356, (5) 1.888, (6) 234.25789, (7) 52-04065042003.

2. Express the above angles in degrees, minutes, seconds. 3. Express as decimals of a right-angle:

(1) 4", (2) 70', (3) •048, (4) 58 2 45", (5) 268 469 35":62, (6) 32":4, (7) 3":24, (8) 0”:324, (9) 11° 4, 12", (10) 31° 3', (11) 61° 1' 12".

4. Express in the centesimal method :-
(1) 1° 8' 8":8, (2) 3° 4' 7":2, (3) 61° 1' 12".
5. Express in the sexagesimal method :-
(1) 38 33' 33":3, (2) 458 5' 3", (3) 78 879".

6. Show that the number of degrees contained in any angle is o of the number of grades contained in the same angle.

7. If an angle is expressed as a decimal of a degree, show that it may be expressed as a decimal of a grade by removing the decimal point one place to the right and dividing by 9.

8. If an angle is expressed as a decimal of a grade, show that it may be expressed as a decimal of a degree by subtracting from the given decimal the decimal formed from it by removing the point one place to the left.

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