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C

Let ac be the proposed square, and put the side вc, or cd,

OC.

A

Then, if the difference of BD and Bс be put d, the hypothenuse BD will be = x + d.

B

But since, as in the former problem. BC2 + CD2, or 2BC2 = BD2, we shall have

2x2

a2

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D

x2 Which equation, being resolved according to the rule laid down for quadratics in the preceding part of the work, gives d + d √2,

X

Which is the value of the side вc, as was required.

x2+2dx + d2, or

2dx

d2;

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PROBLEM 3.—The diagonal of a rectangle ABCD, and the perimeter, or sum of all its four sides, being given, to find the sides.

a

Let the diagonal AC === d, half the perimeter AB + BC= α and the base BC= x; then will the altitude AB = a

And since, as in the former problem, AB2 + BC2 shall have

2ax + x2 + x2
d2

x2

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D

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da, or

a2

OC.

Ac2, we

C

ax

2

Which last equation, being resolved, as in the former instance, gives

1 x = α = √( 2d2 — a2). a ± ± Where a must be taken greater than d and less than d√2.

PROBLEM 4.-The base and perpendicular of any plane triangle ABC being given to find the side of its inscribed square.

B FD G

Let EG be the inscribed square; and put BC= and the side of the square EH or EF = x.

P,

Then, because the triangles ABC, AEH, are similar, we shall have

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AD: BC :: AI: EH, or

p: b :: (p
(p - x): x.

Whence, taking the products of the means and extremes, there will arise

px = bp
= bp - bx,

Which, by transposition and division, gives

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bp

b + p

Where b and p may be any numbers whatever, either whole or fractional.

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PROBLEM 5. Having the lengths of three perpendiculars, EF, EG, EH, drawn from a certain point E, within an equilateral triangle ABC, to its three sides, to determine the sides.

A

H

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H

B

DF

C

Draw the perpendicular AD, and having joined EA, EB, and EC, put EF = a, EG = b, EH = c, and BD (which is BC) = x. Then, since AB, BC, or ca, are each by Euc. 1, 47,

2x, we shall have,

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√(4x2 - x2)

AD = √(AB √ (AB2 — BD2) x2) = √ 3x2 * √ 3. And because the area of any plane triangle is equal to half the rectangle of its base and perpendicular, it follows that A ABC 1 BC X AD = x × ∞ √ 3 = x2 √ 3, X x

A BEC

A AEC

A AEB

ABX EH x X c

= cx.

But the last three triangles BEC, AEC, AEB, are together equal

to the whole triangle ABC; whence

T G

b, AD

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BC XEF x X a

AC X EG = x x b

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= ax,

bx,

x2 √3 = ax + bx + cx,

And consequently. if each side of this equation be divided by x, we shall have

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Which is, therefore, half the length of either of the three equal sides of the triangle.

COR.-Since, from what is above shown, Ad is = x√3, it follows, that the sum of all the perpendiculars, drawn from any point in an equilateral triangle to each of its sides, is equal to the whole perpendicular of the triangle.

PROBLEM 6.-Through a given point P, in a given circle ACBD, to draw a chord CD, of a given length.

A

Draw the diameter APB; and Cpx; then will PD

But, by the property of the circle (Euc. 111. 35,) cp X PD = APX PB; whence

B

and put CD =
and put CD = ά, AP = b, PB = c,

a

XC.

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ax

bc.

x (a — x) = bc, or x2 Which equation, being resolved in the usual way, gives x = 1⁄2 a ± √ (1 a2 — bc); Where x has two values, both of which are positive.

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PROBLEM 7.-Through a given point P, without a giver circle ABDC, to draw a right line so that the part CD, intercepted by the circumference, shall be of a given length.

B

D

Draw PAB through the centre o; and put CD=α, RÀ PB = c, and PC = x; then will PD = x + a.

But, by the property of the circle, (Euc. III, 36, cor.,) PCX PD PA X PB; whence

x (x -|- a) = bc, or
x2 + ax = bc.

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b,

Which equation being resolved, as in the former problem, gives

XC 1⁄2 a ± √ (1 a2 + bc);

Where one value of x is positive and the other negative.*

PROBLEM 8.-The base of BC, of any plane triangle ABC, the sum of the sides AB, AC, and the line AD, drawn from the vertex to the middle of the base, being given, to determine the triangle.

B

D

Put BD or DC= a, AD= b, AB + ACs, and AB = x; then will AC = S

x.

2

But, by my Geometry, B. 11, Prop. 19, AB+ AC2 = 2BD +2AD2; whence

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x2 + (s —x)2 = 2a2 + 2b3, or
x2 SI · a2 + b2 — — s2.

Which last equation, being resolved as in the former instances, gives

#t 1⁄2 s ± s√ (a2 + b2 for the values of the two sides AB and taking the sign + for one of them, and observing that a2 + b2 must be greater than 1s2.

B

1 s2),

4

AC of the triangle; for the other, and

PROBLEM 9.-The two sides AB, AC, and the line AD, bisecting the vertical angle of any plane triangle ABC, being given, to find the base BC.

The two last problems, with a few slight alterations, may be readily employed for finding the roots of quadratic equations by construction; but this, as well as the methods frequently given for constructing cubic and some of the higher order of equations, is a matter of little importance in the present state of mathematical science; analysis, in these cases, being generally thought a more commodions instrument than geometry.

b, ADC, and BC

Put AB α, AC = VI., 3, we shall have

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AB (a): Ac (b):: BD: DC. And consequently, by the compositions of ratios, (Euc. v, 18,)

a+b: a :: x; BD =

and

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c, and BC= x; then, by Euc.

a+bb::x : DC =

a+b

But, by Euc. VI., 13, BD X DC + AD3 = AB X AC; wherefore, also,

abx2
(a + b) a

+c2 = ab, or

abx2 = (a+b)2 × (ab — c2).

From which last equation, we have

ab

c2

x = (a + b) √

ab

Which is the value of the base BC, as required.

PROBLEM 10.-Having given the lengths of two lines AD, BE, drawn from the acute angles of a right-angled triangle ABC, to the middle of the opposite sides, it is required to determine the triangle.

x2+4y2

y2+ 4x2

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ax

a+b2

bx

B

D

2

Put AD α, BE = α, b, CD or ‡CB = x, and CE or à ca = y; then, since (Euc. 1, 47) CD2 + CA2, = AD2, and CE2 — CB3 BE2, we shall have

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Whence, taking the second of these equations from four times the first, there will arise

15y2 = 4a2
4a2 b2

15

A.

72.

E

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y = √

15

And, in like manner, taking the first of the same equations from four times the second, there will arise

15x2 - 462 a2, or

462

a2

62, or

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