r² = (r-1)²+(r-2)²
Where r is the radius of the circle.
r² = r²-2r+1+r²-4r+4
r²-6r+5 = 0
r²-r-5r+5 = 0
r(r-1)-5(r-1) = 0
Therefore;
r ≠ 1 unit.
r = 5 units.
r-1
5-1= 4 units.
Where 4 units is half...
a =1
b = 0.5
Cross Multiply.
b = 0.5a ----- (1).
½(a)+½(a)+½(1*0.5)+½(1-b) = 1
2a+0.5+1-b = 2
2a-b = 0.5
b = 2a-0.5 ---- (2).
Equating (1) and (2).
2a-0.5 = 0.5a
1.5a = 0.5
a = (1/2)/(3/2)
a = ⅓...
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Let the two congruent angles be a.
Let the base of the ascribed triangle be b.
tana = 8/b
b = 8/(tana) ----- (1).
tan2a = 18/b
b = 18/(tan2a) ----- (2).
Equating (1) and (2).
8/(tana) = 18/(tan...
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Using 1 unit base.
tan70 = 1/a
a = 1/tan70
a = 0.36397023427 units.
tan80 = 1/b
b = 1/tan80
b = 0.17632698071 units.
tan40 = c/a
c = 0.36397023427tan40
c = 0.30540728934 units.
Therefore, the re...
Let the radius of the ascribed quarter circle be a.
a² = 2(a-1)²
a² = 2(a²-2a+1)
a² = 2a²-4a+2
a²-4a+2 = 0
(a-2)² = -2+4
a = 2±√(2)
a ≠ 2-√(2)
a = 2+√(2) units.
a = 3.41421356237 units.
Where a is...
Let DE = BF = a.
Let EG = GH = HF = b.
(3b)² = 12²+(12-2a)
9b² = 144+144-48a+4a²
b² = ⅑(288-48a+4a²) ----- (1).
CF = √(144+a²)
(CG)² = (12-a)²-b²
CG = √((12-a)²-b²)
Therefore;
(144+a²) = 4b²+(...
Let r be the radius of the inscribed circle.
Calculating r.
tan30 = r/(6-r)
√(3)r = 6-r
r = 6/(√(3)+1) cm.
r = 3(√(3)-1) cm.
sin30 = r/a
½ = 3(√(3)-1)/a
a = 6(√(3)-1) cm.
Where a is BO, O is the...
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