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Where the base (b) and the given difference (d) may

be any numbers as before, provided 6 be greater than d.

PROBLEM II.

The difference between the diagonal of a square and one of its sides being given, to determine the square,

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Let ac be the proposed square, and put the side Bc, or CD, X.

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

But since, as in the former problem, Bca 400%, or 2BC2 =BD2, we shall have

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2.0=x? +2dx+da, or

22 - 2dx=d2.

Which equation being resolved according to the rule laid down for quadratics, in the preceđing part of the work, gives

x=dtd/2. Which is the value of the side BC, as was required.

PROBLEM III.

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

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Let the diagonal ac=d, half the perimeter AB+BC sa, and the base bc=x ; then will the altitude Ab=

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And since, as in the former problem, AB? +BC=AC, we shall have

al - 2ax+x+x=da, or

d? ax=

2

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

x=aiv(2d2-23). Where a must be taken greater than d and less than d./2.

PROBLEM Iy.

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

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A

B F D G Let eg be the inscribed square ; and put bc=b, AD=P,

i and the side of the square Eh or EF=%.

BCE

Then, because the triangles ABC, AEH, are simila VI, 4,) we shall have

AD : BC :: AI : EH, or

p:b1: (2-2): 2. Whence, taking the products of the means and extremes, there will arise

px=bp-bx.
Which, by transposition and division, gives

bp

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

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PROBLEM v.

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

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Draw the perpendicular AD, and having joined EA, EB, and ec, put eF=d, Eg=b, eH=c, and BD (which is {BC)

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Then, since AB, BC, or'ca, are each =2x, we shall have, by Euc. I, 47,

AD=V (ABPBD)=(4x2 - **)=3x2=*/3.

And because the area of any plane triangle is equal to half the rectangle of its bage and perpendicular, it follows, that

A

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