Mathematics Question and Solution
Sir Mike Ambrose is the author of the question.
Let CD = 1 units.
It implies;
DE = 1 units.
AD = 2 units.
AE = 2 units.
1² = 2²+2²-2*2*2cosa
1 = 8-8cosa
cosa = 7/8
a = acos(⅞)°
b = ½(180-acos(⅞))
b = (90-0.5acos(⅞))°
sinb = c/2
c = 2sin(90-0.5acos(⅞))
c = 1.9364916731 units.
c = CF.
cosb = d/2
d = 2cos(90-0.5acos(⅞))
d = ½ units.
d = EF.
AF = AE-EF
AF = 2-½
AF = ½(3) units.
tan(acos(⅞)) = e/AF
e = 1.5tan(acos(⅞))
e = 0.82992500276 units.
e = BF
Therefore;
BC = CF-BF
BC = 1.9364916731-0.82992500276
BC = 1.10656667034 units.
(a) BF÷BC is;
0.82992500276÷1.10656667034
= 0.75
= ¾ exactly.
Area ADE is;
½*2*2sin(acos(7/8))
= 2sin(acos(7/8))
= 0.96824583655 square units.
Area yellow is;
Area ADE - Area ABF
= 2sin(acos(7/8))-(½*½(3)*0.82992500276)
= 0.34580208448 square units.
(b) Area yellow ÷ Area ADE is;
0.34580208448 ÷ 0.96824583655
= 0.35714285714 exactly in decimal.
Therefore;
(a) BF÷BC = ¾ exactly.
(b) Area yellow ÷ Area ADE
= 0.35714285714 exactly in decimal.
