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Chapter 4
r 2r a
Problem 9 Draw in a diagonal from the southwest to the northeast corner of the large square a in figure 4.8. Then we see that 2a = 2 2r + r + 2r , or, solving for r, r =
2a 3 2 +1
.
Problem 10 By inspection of figure 4.9 one easily sees that r = 3b + 4c, R = b + 2a, R = 5b + 4c, and a + b = 2b + 4c. Solving these equations simultaneously yields b = 2c, a = 6c, and r = 10c. Problem 11 We notice that S = 6 × 1652 and c = 5 × 165. Thus, we have a right triangle with sides 3, 4, and 5 (in units of 165). Hence a = 495 and b = 660. By similar triangles, we have b:a = (b − t):t, which gives t = ab/(a + b) = 282.85. By inspection of figure 4.10 n = t / 2 = 200.01, and because d is an altitude, d = 2S/c = 396.
d c
a t n b
Problem 12 Since the side of the rhombus is equal to the distance between the two horizontal lines, each half of the rhombus consists of an equilateral triangle. Then b = 2 3R and with the Pythagorean theorem, R = 3r. Hence 2R = 6r = 106.5 and b = 184.4. The Pythagorean theorem also gives directly a = 3b = 6 R = 18r = 319.5 and thus aπ = 1009.6. Also, d = (a − b)/2 = 67.5.
d r b a
R
d
B
s a t C
A
Problem 13 The angles of the square and equilateral triangle in figure 4.12 force triangle ABC to be a 30- 60-90 triangle with ∠A = 30° and ∠B = 60°. Hence, AB = 2a. Examining the other interior angles shows AB = 2a = 3t + s + s / 3 . Also we see that BC = a = 2s / 3 + s . Eliminating s quickly yields t = ( 3 − 1)a. Problem 14 Draw the auxiliary lines shown in figure 4.50 to get an equilateral triangle. Thus r red is the radius of the inscribed circle. By inspection rred = rwhite + r blue and by the Pythagorean theorem rred = 3r blue (see problem 12). Therefore rwhite = 2r blue and r blue = rwhite/2.