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CP'=s- b=CP”, BP=s-C=BP". Similarly, AR=s-b=AR", BR'=s-a=BR", etc. Again, CD=CE=b-AE=b-AF=b-(c-BF)
CE=CD=s-C=BP". Similarly, AE=AF=s-a=BR", etc.
These relations are frequently useful.
If we put Ai to denote the distance of the vertex A from the adjacent points of contact of the in-circle, and Ab, Ac to denote its distances from the points of contact of the ex-circles upon the sides b and c respectively, we have
1. In testing the straightness of a “rule" three rules are
virtually tested. How? 2. To construct a rectangle, and also a square. 3. To place a given line-segment between two given lines
so as to be parallel to a given line. 4. On a given line to find a point such that the lines joining
it to two given points may make equal angles with the
given line. 5. To find a point equidistant from three given points. 6. To find a line equidistant from three given points. How
many lines ?
7. A is a point on line L and B is not on L. To find a point
P such that PA+PB may be equal to a given segment. 8. On a given line to find a point equidistant from two
given points. 9. Through a given point to draw a line which shall form an
isosceles triangle with two given lines. How many solutions ?
10. Through two given points on two parallel lines to draw
two lines so as to form a rhombus. 11. To construct a square having one of its vertices at a
given point, and two other vertices lying on two given
parallel lines. 12. Through a given point to draw a line so that the intercept
between two given parallels may be of a given length. 13. To construct a triangle when the basal angles and the
altitude are given. 14. To construct a right-angled triangle when the hypothen
use and the sum of the sides are given. 15. To divide a line-segment into any number of equal parts. 16. To construct a triangle when the middle points of its
sides are given. 17. To construct a parallelogram when the diagonals and
one side are given. 18. Through a given point to draw a secant so that the chord
intercepted by a given circle may have a given length. 19. Draw a line to touch a given circle and be parallel to a
given line. To be perpendicular to a given line. 20. Describe a circle of given radius to touch two given lines. 21. Describe a circle of given radius to touch a given circle
and a given line. 22. Describe a circle of given radius to pass through a given
point and touch a given circle. 23. Describe a circle of given radius to touch two given circles. 24. To inscribe a regular octagon in a circle. 25. To inscribe a regular dodecagon in a circle. 26. A, B, C, D, are consecutive vertices of a regular
octagon, and A, B', C', D', ..., of a regular dodecagon in the same circle. Find the angles between AC and
B'C'; between BE' and B’E. (Use 108°.) 27. Show that the plane can be filled by
(a) Equilateral triangles and regular dodecagons.
136°. Def. 1.—The area of a plane closed figure is the portion of the plane contained within the figure, this portion being considered with respect to its extent only, and without respect form.
A closed figure of any form may contain an area of any given extent, and closed figures of different forms may contain areas of the same extent, or equal areas.
Def. 2.—Closed figures are equal to one another when they include equal areas.
This is the definition of the term "equal” when comparing closed figures.
Congruent figures are necessarily equal, but equal figures are not necessarily congruent. Thus, a A and a may have equal areas and therefore be equal, although necessarily having different forms.
137o. Areas are compared by superposition. If one area can be superimposed upon another so as exactly to cover it, the areas are equal and the figures containing the areas are equal. If such superposition can be shown to be impossible the figures are not equal.
In comparing areas we may suppose one of them to be divided into any requisite number of parts, and these parts to be afterwards disposed in any convenient order, since the whole area is equal to the sum of all its parts.
Illustration.—ABCD is a square.
Now, if AD and DE be equal and in line, the As ADC and EDC are congruent and equal.
Therefore the AABC may be taken
Ē from its present position and be put into the position of CDE. And the square ABCD is thus transformed into the AACE without any change of area ;
DABCD=AACE. It is evident that a plane closed figure may be considered from two points of view.
1. With respect to the character and disposition of the lines which form it. When thus considered, figures group themselves into triangles, squares, circles, etc., where the members of each group, if not of the same form, have at least some community of form and character.
2. With respect to the areas enclosed.
When compared from the first point of view, the capability of superposition is expressed by saying that the figures are congruent. When compared from the second point of view, it is expressed by saying that the figures are equal.
Therefore congruence is a kind of higher or double equality, that is, an equality in both form and extent of area. This is properly indicated by the triple lines ( =) for congruence, and the double lines (=) for equality.
138°. Def.—The altitude of a figure is the line-segment which measures the distance of the farthest point of a figure from a side taken as base.
The terms base and altitude are thus correlative. A triangle may have three different bases and as many corresponding altitudes.
(87) In the rectangle (82°, Def. 2) two adjacent sides being perpendicular to one another, either one may be taken as the
base and the adjacent one as the altitude. The rectangle having two given segments as its base and altitude is called the rectangle on these segments.
Notation.—The synbol stands for the word rectangle and o for parallelogram.
Rectangles and parallelograms are commonly indicated by naming a pair of their opposite vertices.
COMPARISON OF AREAS--RECTANGLES,
139o. Theorem.-1. Rectangles with equal bases and equal altitudes are equal.
2. Equal rectangles with equal bases have equal altitudes. 3. Equal rectangles with equal altitudes have equal bases. 1. In the Os BD and FH, if
OBD=OFH. Proof.-Place E at A and EH along E AD. Then, as ZFEH=LBAD= 7, EF will lie along AB.
And because EH=AD and EF=AB, therefore H falls at D and F at B, and the two os are congruent and therefore equal.
9.e.d. 2. If OBD=OFH and AD=EH, then AB=EF.
Proof.-If EF is not equal to AB, let AB be > EF.
OPD=OFH, by the first part,