Question

4. (a) Start with a square of sides 2, on the xy-plane, with corners at (0,0), (0, 2), (2,0), and (2,2). Construct the length V5 by connecting the point (1,0) to a suitable point in the rectangle, and use it construct a rectangle of sides 1 + V5 and 2. make sure this rectangle extends the original square in such a way that one side in on the y-axis extended between (0,0) and (2,0). Such rectangle is known as golden rectangle. Your construction is to be minimal, precise, and precisely documented by refering to the Greek Construction axioms. Call this golden rectangle R. Keep this drawing clean and clear, and then use different colors to draw the constructions of parts (b) and (d) in the same drawing.

Answer 1

No Construck golden rectangle R . - we at have, (0,0) square of (0,2), plane side (2,0) 2 4 on xy (2,2 with corner. y2- 2 10 (2,2) 2 0 (0,0) (20) -175 a first construct 2 unit x Now, take an aoc point. . Now join AC golden rectangle join unit from Take & R. square point mid - line P BC P point to of gives A side of OQCB & B I a square as (6) Construct line - 1+15 H-N- 0. -I+5 o o first now construct take on golden are OR rectangle. K with 2x2 Square. from s with joining of poins length (+1+85/2. We have,

o Poove a a - atb ba ' abitb footatha - q² - ab-b² = 0 New, using Qudratic formula a bt 6² + 4x1 X6² bt 156² bt 5b 2 Therefore - bt 5b blir $5) 2 => a = 1+ 5 - Golden Ratio Also 02-0h = b? Now. b a-b a²_ab – b² = 0 Using Qudratic tamula bt 6² + 4 x 186² 2 1t5 G = bi 5b Therefore Golden ratio. 2 a This more golden rectangle with golden rectangle Ri- i OQ BC is golden rectangle R. Here OB - BC Ох . It = OM BZ QM 5 unit ont

1. Noe I+ I+ ... derni A= /1+ it... Hence, 1+ A A2 – ITA A²-A-1=0 Qudratic formula Then using A = 1 I 1 + 4 It 15 Hence JAF lt 5 2 Any
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