Let r be the radius of the inscribed circle.
Calculating r.
√(5)² = (2r)²+r²
5 = 5r²
r = 1 unit.
Notice!
The square side is 4r.
Let a be the square side.
a = 4r
And r = 1 unit.
a = 4(1)
a = 4...
sin30 = a/4
Where a is the red length.
a = 4sin30
a = 2 units.
Calculating PQ.
Let r be the radius of the smaller circle.
7 units is the radius of the bigger circle.
sin60 = a/7
√(3)/2 = a/7
a = ½(7√(3)) units.
a = 6.0621778265 units.
a is the radius, r of t...
Let r be the radius of the two equal inscribed circles.
tana = 21/28
a = atan(3/4)°
b = 90-a
b = 90-atan(3/4)°
b = atan(4/3)°
c² = 28²+21²
c = √(28²+21²)
c = 35 units.
tan(0.5atan(...
Observing cosine rule.
a² = 12²+4²-2*12*4cos60
a = 10.5830052443 units.
Observing cosine rule.
b² = 3²+12²-2*3*12cos120
b = 13.7477270849 units.
Observing cosine rule.
10.5830052443² = 13.74772708...
Let c3 = a = 6 units.
Let c2 = b = 8 units.
Therefore;
c1 = c = √(6²+8²)
c = √(100)
c = 10 units.
Calculating Area S3.
It is;
Area equilateral triangle with side 6 units - 2(area equilateral tria...
Calculating area blue.
6 - a
b - 2
Cross Multiply.
ab = 12 cm²
c = ½(ab)
c = ½(12)
c = 6 cm²
Therefore, calculating a.
6+½(2a) = ½(2+b)*a
6+a = ½*2a*ab
And
ab = 12 cm²
6+a = ½*2a*12
6+a = 12a
6...
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cos60 = a/4
½ = a/4
a = 2 units.
b = a+4
b = 2+4
b = 6 units.
tan60 = c/6
c = 6√(3) units.
d = b+4
d = 6+4
d = 10 units.
It implies, area red is;
½(cd)
= ½*6√(3)*10
= 30√(3) square units.
Let r be the radius of the circle.
It implies, calculating r.
r² = (r-1)²+(r-0.5)²
r² = r²-2r+1+r²-r+(¼)
0 = r²-3r+(5/4)
4r²-12r+5 = 0
Resolving the above quadratic equation via completi...
