Mathematics Question and Solution
Sir Mike Ambrose is the author of the question.
12²=10²+10²-2*10*10cosa
cosa = (200-144)/200
a = acos(56/200)°
a = acos(7/25)°
a = 73.73979529169°
b = c = ½(180-73.73979529169)
b = c = 53.13010235416°
d² = 10²+10²
d = 10√(2) cm
e = ½d
e = 5√(2) cm.
Let PR = f.
f² = 2(5√(2))²-2(5√(2))²cos(90+acos(7/25))
f = PR = 14 cm.
g² = 6²+6²
g = 6√(2) cm.
Let PQ = h = QR
h² = 50+72-2*60cos(90+53.13010235416)
h = PQ = QR = 14.76482306023 cm.
14²=2(14.76482306023)²-2(14.76482306023)²cosi
cosi = (2(14.76482306023)²-14²)/2(14.76482306023)²
cosi = 0.5504587156
i = 56.60151153176°
(a) Calculating tan theta. Let theta be x.
x = ½(180- 56.60151153176)
x = 61.69924423412°
tanx = tan61.69924423412
= 1.85714285715 in decimal.
= (4181913939717259/2251799813685248) or 13/7 exactly in fraction.
(b) Calculating Area Red ÷ Area Yellow exactly.
Area Yellow is;
½*14*14.76482306023sin61.69924423412
= 91 cm².
Area Red is;
(j/sin73.73979529169) = (k/sin53.13010235416)
j = (ksin73.73979529169)/(sin53.13010235416)
j = 1.2k cm.
l = ½(14-1.2k)
l = (7-0.6k) cm.
m = 180-53.13010235416
m = 126.86989764584°
It implies;
(7-0.6k)/(sin45) = (5√(2)/sin126.86989764584)
k = (7sin126.86989764584-5√(2)sin45)/(0.6sin126.86989764584)
k = 1.25 cm.
Area Red is;
½*(1.25)²sin73.73979529169
= 0.75 cm².
= ¾ cm².
Therefore;
Area Red ÷ Area Yellow exactly is;
(¾) ÷ 91
= 3/364
