Let the inscribed square side be a.
tan60 = a/b
a = √(3)b cm.
Where b is BS = CR
It implies;
b+√(3)b+b = 10
(2+√(3))b = 10
b = 10(2-√(3)) cm.
Therefore;
a = √(3)*10(2-√(3))
a = 10(2√(3)-3) cm....
Let a be the side of the inscribed square.
Calculating a.
tan60 = a/b
b = a/(tan60) cm.
Where b is BS = CR.
Therefore;
2b+a = 10
2a/(tan60)+a = 10
a((2/tan60)+1) = 10
a = 10(2√(3)-3) cm.
a = 4.6...
a² = 6²-1²
a = √(35) units.
b = atan(√(35))
cos(atan(√(35))) = c/2
c = ⅓ unit.
Where c is CE.
d = atan(√(35))-60
d = 20.40593177314°
Where d is angle FCE.
tan20.40593177314 = e/(1/3)
e = 0.12400...
Sir Mike Ambrose is the author of the question.
Let the side of the regular pentagon be 1 unit.
It implies;
Area red is;
½*sin54*cos54
= 0.23776412908 square units.
Area orange is;
½*((sin54-cos...
Radius of the inscribed small semi circles, r is;
(3+r)² = 3²+(6-r)²
36 = 18r
r = 2 cm
Therefore;
Area blue is;
Area semi circle with radius 6 cm - Area semi circle 2 cm - Area circle with radiu...
tan60 = a/5
a = 5√(3) cm.
Where a is AD.
b = ½(a)
b = ½(5√(3)) cm
c = atan((½(5√(3))/5)
c = atan(½√(3))
Where c is angle AGB
d = atan(5√(3)/5)
d = atan(√(3))
Where d is angle CAG.
e = 180-c-d
e...
Let the square side be a.
tan60 = a/b
b = a/tan60 cm.
Where b is BS.
Calculating a.
2b+a = 6
2(a/tan60)+a = 6
a((2/tan60)+1) = 6
a = 6/((2/tan60)+1)
a = 6(2√(3)-3) cm.
Area Shaded STR is;
½(a²)...
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a² = 2*6²
a² = 2*36
a = √(2*36)
a= 6√(2) units.
Let the radius of the small inscribed circle be r.
Calculating r.
b² =2r²
b = √(2)r units.
It implies;
b+r+6 = a
√(2)r+r+6 = 6√(2)
(√(2)+1)r = 6√...
a = atan(2)°
b = 180-2a
b = (180-2atan(2))°
Where b is angle FBP.
Notice;
CQ = 5 cm.
Calculating FP using sine rule.
Let FP be c.
(c/sin(180-2atan(2))) = (5/sin(atan(2)))
c = 4.472135955 cm.
T...
