Mathematics Question and Solution
Let AB be 2 units.
tan(atan½) = a/1
a = ½ unit.
b = (atan(⅔))°
tan(atan⅔) = c/1
c = ⅔ units.
d = (180-atan⅔-atan2)°
(e/sin(atan2)) = (1/sin (180-atan⅔-atan2))
e = ¼√(13) unit.
It implies;
Area Blue is;
2(area triangle with height ⅔ unit and base 1 unit - area triangle with height ¼√(13) unit and base sin(atan⅔) units.
= 2(½*⅔*1- ½*¼*√(13)*sin(atan⅔))
= 2(⅓ - ¼)
= 2(1/12)
= ⅙ square units.
Calculating Area Shaded.
f² = (½)²+(⅓)²
f = ⅙√(13) unit.
g = f = ⅙√(13) unit.
h² = 2(½)²
h = ½√(2) unit.
i = h = ½√(2) unit.
Therefore;
Area quadrilateral ascribing the circle is:
Area triangle with height ⅙√(13) units and base ⅙√(13)sin(2atan(3/2)) units + Area triangle with height and base ½√(2) unit respectively.
= ½(⅙√(13))²*sin(2atan(3/2)) + ½(½√(2))²
= ⅙ + ¼
= 5/12 square units.
Radius, r of the inscribed circle is;
2(½(½√(2)r)) + 2(½(⅙√(13)r)) = 5/12
6√(2)r + 2√(13)r = 5
r = 5/(6√(2)+2√(13)) unit.
It implies;
Area Shaded is;
(5/12) - Area circle with radius 5/(6√(2)+2√(13)) unit.
= (5/12) - π(5/(6√(2)+2√(13)))²
= (5/12) - 0.3187796987
= 0.09788696797 square units.
Therefore;
Area Shaded ÷ Area Blue to 2 decimal places is;
0.09788696797 ÷ (1/6)
= 0.58732180782
≈ 0.59
