Let AB be 5 units.
Radius, r of the inscribed circle is;
5*(2/5)
r = 2 units.
a = ⅛(180*6)
a = 135°
2b² = 25
b = ½(5√(2)) units.
c = 180-0.5(135)-45
c = 67.5°
sin22.5 = ½(5√(2))/d
d = 9.238795...
AD = √(40²-20²)
AD = 20√(3) units.
AE = DE = ½(AD)
= 10√(3) units.
Angle BED is;
= atan(20/(10√(3)))
= atan(⅔√(3))°
= 49.10660535087°
Angle DAC is;
= 30°
(EF)² = 300+625-20√(3)*25cos30
EF = 5√(7...
Let AD = 2AB.
For AB = 2 units, AD = BC = 4 units.
Angle APD is;
= atan(4)°
PD = √(4²+1)
PD = √(17) units.
Considering similar triangles APD and QCD ratios.
1 = a
√(17) = 2
Crossing Multiply....
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Calculating R.
(5+R)² = (10-R)²+(5-R)²
25+10R+R² = 100-20R+R²+25-10R+R²
20R = 100-20R+R²
R²-40R+100 = 0
Therefore;
R ≠ 20+10√(3) units.
R = 20-10√(3) units.
R = 2.67949192431 units.
Calculating r....
b² = 12²+12²
b² = 144+144
b² =288
b = 12√(2) cm.
Where b is AE = AC.
c² = 12²+6²
c = 6√(5) cm.
Where c is AM.
d = b-c
d = (12√(2)-6√(5))
d = 3.55415488348 cm.
Where d is EM.
e = 45-atan(1/2)
e =...
Let a be the radius of the inscribed circle.
Calculating a.
tan30 = a/6
a = 2√(3) cm.
b = inscribed circle diameter.
b = 4√(3) cm.
12² = c²+6²
c = 144-36
c = √(108)
c = 6√(3) cm.
d = 6√(3)-4√(3)...
The single interior angle of the ascribed regular hexagon is;
⅙(180*4)
= 120°
a = 360-90-120-(120-53)
a = 150-120+53
a = 83°
b = 360-120-53-53
b = 240-106
b = 134°
It implies;
The required angl...
Let the small inscribed half circle be r.
Big inscribed half circle's radius is 0.5 cm.
Calculating r.
(0.5+r)² = 0.5²+(1-r)²
0.25+r+r² = 0.25+1-2r+r²
3r = 1
r = ⅓ cm.
1² = 2a²
a = 1/√(2)
a = 0.7...
18² = 14²+16²-2*14*16cosa
448cosa = 14²+16²-18²
a = acos((14²+16²-18²)/448)
a = 73.39845040098°
(18/sin73.39845040098) = (14/sinb)
b = 48.18968510422°
c = 180-48.18968510422-73.39845040098
c = 58....
