Mathematics Question and Solution
Sir Mike Ambrose is the author of the question.
Let the side length of the ascribed square be 1 unit.
Area square is;
1²
= 1 square units.
a = ½(90-60)
a = 15°
tan15 = b/1
b = 0.2679491924 units.
cos15 = 1/c
c = 1.0352761804 units.
Let d be the radius of the inscribed circle.
Calculating d.
sin30 = d/e
e = 2d units.
f² = 2d²
f = √(2)d units.
It implies;
e+f = √(1²+1²)
2d+√(2)d = √(2)
d(2+√(2)) = √(2)
d = ½(2√(2)-2)
d = (√(2)-1) units.
d = 0.4142135624 units.
Again, d is the radius of the inscribed square.
g = 90-15
g = 75°
h = ½(180-75)
h = ½(105)
h = 52.5°
j = 90-52.5
j = 37.5°
tan37.5 = k/0.4142135624
k = 0.3178372452 units.
Shaded Area is;
2(area triangle with height and base 0.3178372452 units and 0.4142135624 units respectively - area sector with radius 0.4142135624 units and angle 37.5°)
= 2((0.5*0.4142135624*0.3178372452)-(37.5π*0.4142135624*0.4142135624÷360))
= 0.01935831332 square units.
Therefore;
Shaded Area ÷ Area Square to 3 significant figures is;
0.01935831332÷1
= 0.01935831332
≈ 0.019
