x2+ 2xy+y2 = 2s(x+y)+h2, or (x+y)' - 28(x+y) = h', which is a quadratic in terms of x+y: hence we find x+y= 8+√s2+h3. [3] [4] Substituting this value of x + y in [1], and we shall obtain [5] From the square of [4] subtracting four times [5], we Extracting the square root of [6], we have [6] [7] xy = s2+s√s2+h3. h2-2s - 2s√s2+h3. x—y = √h2. 2 s2 — 2 s√/s2+h2. Taking half the sum of [4] and [7], and also half their s + √ s3 +h3 — (h3 — 2 s2 — 2 s√s*+h3)3 ̧ PROBLEM X. To determine a right-angled triangle; having given the hypothenuse, and the radius of the inscribed circle. r = radius of the inscribed circle; AB =x, and AC = Then BF = BE = x−r, BCBF + CF = x+y-2r=h, or y-r; therefore In [2], for x+y substitute its value given by [1], and it PROBLEM XI. To determine a right-angled triangle; having given the lengths of the lines drawn from the acute angles, to the middle points of the opposite sides. Equations [1] and [2], when cleared = a2, or = a2. [1] [2] A B D [3] [4] A A PROBLEM XII. To determine a triangle; having given the base, the perpendicular, and the difference of the two other sides. Then from the right-angled triangles ADC, BDC, we have Expanding [1] and [2], and then taking half their sum and one-fourth of their difference, we obtain PROBLEM XIII. To determine the radii of three equal circles, described in a given circle, to touch each other, and also the circumference of the given circle. Let D be the centre of the given circle, whose radius we will denote by R; also let A, B, C be the centres of the three equal circles, whose common radius we will denote by r. Then joining A, B and C, we have the triangle ABC equilateral, each of whose sides is 2 r. Drawing CK perpendicular to AB, the right-angled triangle AKC gives CK2 = AC2 — AK2, or in symbols CK' 432; consequently, CK r√3. Now [B. IV, Prop. XIV] CD = 3 CK = r√3. But ED EC + CD; that is, R = r + r√3: from which we readily find [1] [2] [3] PROBLEM XIV. Given the radius r of a circle, to find the sides of the inscribed and circumscribed pentagons and decagons. D N M E 1. The inscribed pentagon. Let ADBCE be the pentagon inscribed in the circle (B. V, Prop. vII), and let O be the centre of the circumscribing circle. Join AB, AC, and draw AF perpendicular to BC; then, by known properties, BC is bisected in G, and the line AF passes through the centre O of the circle. Since the angle ADX P V G H K H is measured by half the arc AEC (B. III, Prop. VIII), and the angle AXD is measured by half the sum of the arcs AD and BC (B. III, Prop. x), it follows that these angles are equal, and the triangle DAX is isosceles. Again, the angleș DAX and BAC are equal, being measured by halves of the equal arcs DB, BC: hence the triangles ABC and DAX are similar. But the triangle DAX is obviously equal to BCX; therefore |
