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SECTION V.

CONSTRUCTIVE GEOMETRY.

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180°. Problem.-AB being a given segment, to construct the segment ABJ2.

Constr. - Draw BC I to AB and equal to it.
Then AC is the segment AB/2.
Proof.-Since ABC is right-angled at B,

AC2=AB2 + BC2=2AB2, (169, Cor. 1)

AC=AB2. Cor. The square on the diagonal of a given square is equal to twice the given square.

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181°. Problem.To construct AB/3.

Constr.— Take BC in line with AB and equal to it, and on AC construct an equilateral triangle ADC.

(124", Cor. 1) BD is the segment AB/3.

Proof.ABD is a 7, and AD=AC=2AB. Also

AD2=AB2+BD2=4AB2. (169°, Cor. 1)

BD2=3AB, and BD=AB/3. Cor. Since BD is the altitude of an equilateral triangle and AB is one-half the side,

... the square on the altitude of an equilateral triangle is equal to three times the square on the half side.

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182°. Problem.—To construct AB/5.

Constr. —Draw BC I to AB and equal to twice
AB. Then AC is the segment AB/5.
Proof.—Since LB is a right angle,

AC?=AB2+ BC?.
But

BC2=4AB%;

ACP=5AB', and

AC=AB/5.

183o. The three foregoing problems furnish elements of construction which are often convenient. A few examples are given.

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Ex. I. AB being a given segment, to find a point C in its

line such that ÁC=AB. CB.
Analysis - ACP=AB. CB=AB(AB-AC),

AC? + AC. AB=AB2. Considering this an algebraic form and solving as a quadratic in AC, we have AC=(AB/5-AB), and this is to be constructed. Constr.Construct AD=ABJ5 (by 182') as in the figure,

and let E be the
middle point of BD.

Take DF=DE.
Then
AF=AB./5-AB;
.. bisecting AF in G,
AG=AC

=}(AB/5-AB), and the point C is found.

Again, since 5 has

two signs + or

Yf' take its negative sign and we have AC'= -}(ABJ5+AB).

Therefore, for the point C', on AD produced take DF=DE, and bisect AF' in G'. Then

AG'= }(AB/5+AB); and since AC' is negative we set off AG' from A to C', and C' is a second point. The points C and C satisfy the conditions,

AC=AB.CB and AC'2=AB. C'B. A construction effected in this way requires no proof other than the equation which it represents.

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It is readily proved however. For AD2=5AB%, and also ADP=(AF+FD):=(2AC+AB)", whence

AC=AB(AB - AC)=AB. CB. It will be noticed that the constructions for finding the two points differ only by some of the segments being taken in different senses. Thus, for C, DE is taken from DA, 'and for C, added to DA; and for C, AC is taken in a positive sense equal to AG, and for C', AC' is taken in a negative sense equal to AG'.

In connection with the present example we'remark :

1. Where the analysis of a problem involves the solution of. a quadratic equation, the problem has two solutions corresponding to the roots of the equation.

2. Both of the solutions may be applicable to the wording of the problem or only one may be.

3. The cause of the inapplicability of one of the solutions is commonly due to the fact that a mathematical symbol is more general in its significance than the words of a spoken language.

4. Both solutions may usually be made applicable by some change in the wording of the problem so as to generalize it.

The preceding problem may be stated as follows, but whether both solutions apply to it, or only one, will depend upon our definition of the word “part.” See Art. 23°.

To divide a given segment so that the square upon one of the parts is equal to the rectangle on the whole segment and the other part.

Def.-A segment thus divided is said to be divided into extreme and mean ratio, or in median section.

Ex. 2. To describe a square when the sum of its side and diagonal is given.

Analysis.-If AB is the side of a square, ABJ2 is its diagonal,

(180°)

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.. AB(I + N2) is a given segment =S, say. Then

AB=S(2-1).
Constr:Let EF be the given segment S.

Draw FG I and to EF, and with centre G and radius GF describe a cutting EG in H and H'.

EH is the side of the square; whence the square is easily constructed.

If we enquire what EH' means, we find it to be the side of the square in which the difference between the side and diagonal is the given segment S. The double solution here is very suggestive, but we leave its discussion to the reader.

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184°. Problem.To find a segment such that the rectangle on it and a given segment shall be equal to a given rectangle.

Constr.--Let S be the given segment, F and AC the given rectangle.

On DA produced make AP=S, and

draw PBQ to cut DC produced in Q.
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CQ is the segment required.
Proof.-Complete the Os PEQD, PGBA, and BCOF.
Then DAC=OGF=GB. BF=PA.CQ,

S.CQ=DAC. Def.The segments AP and CQ are reciprocals of one another with respect to the DAC as unit.

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185°. Problem.To find the side of a square which is

equal to a given rectangle.

Constr.-Let AC be the rectangle. Make BE=BC and in line with BA. E On AE describe a semicircle, and produce CB to meet it in F.

BF is the side of the required square. Proof.-Since AE is a diameter and FB a half chord I to it,

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Cor. This is identical with the problem, “To find a geometric mean between two given segments,” and it furnishes the means of constructing the segment a, when a=Nbc, b and c being given.

(165°) Ex. I. To construct an equilateral triangle equal to a given rectangle.

Let AC be the given rectangle, and suppose PQR to be the required triangle. Then

AB. BC={PR. QT

=PT.QT. But

QT=PT/3, (181°, Cor.)

PT.QT=PT3 whence PT2=AB/3.BC. And PT is the side of a square equal to the rectangle whose sides are ABJ3 and }BC, and is found by means of 181°, 127°, and 185o.

Thence the triangle is readily constructed.

Ex. 2. To bisect the area of a triangle by a line parallel to its base.

Let ABC be the triangle, and assume PQ as the required line, and complete the parallelograms AEBC, KFBC, and let BD be the altitude to AC. Because PQ is || to AC, BD is I to PQ. Now EP=OPC,

(145°) JFC=DEQ, or PQ.BD=AC. BG. (153°, 1) But 2DFQ=DEC, or 2PQ. BG=AC.BD; .. dividing one equation by the other, and reducing to one line,

BD’=2BG2; and therefore BG is one-half the diagon

of the square of which BD is the side, and the position of PQ is determined.

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