a² = 11*11+10*10-2*11*10cos50
a = 8.92113926968 cm.
(8.92113926968/sin50) = (10/sinb)
b = 59.16920318624°
c = 180-50-59.16920318624
c = 70.83079681376°
d = 0.5a
d = 4.46056963484 cm.
e² = 4.460...
Sir Mike Ambrose is the author of the question.
Radius of the semicircle is 2 units.
Therefore;
Shaded area exactly in decimal square units is;
Area trapezoid with parallel sides 2 units...
Radius, r1 of the inscribed semicircle is;
sin60 = r1/6
r1 = 3√(3) cm.
Radius, r2 of the inscribed circle is;
tan60 = 3/r2
r2 = 3/√(3)
r2 = √(3) cm.
It implies;
Area of the coloured inscribe...
Sir Mike Ambrose is the author of the question.
Let the base of the parallelogram be 2 units.
Therefore the area of the parallelogram is;
2(area triangle with height (sin70/sin35) unit and base 2s...
Sir Mike Ambrose is the author of the question.
Let the side of the regular hexagon be 1 unit.
Therefore;
Area green is;
Area triangle with height ½ units and base √(3) units + Area trian...
Area yellow is;
2(area triangle with height (4.5/cos15) cm and base (4.5/cos15)sin60 cm + Area square with side (4.5/cos15) cm.
= 2*½*(4.5/cos15)²(√(3)/2) + (4.5/cos15)²
= (4.5/cos15)²(√(3)/2) +...
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...
Sir Mike Ambrose is the author of the question.
Area green exactly in decimal is;
Square side is (16/√(17)) units.
Area trapezoid with parallel sides (16/√(17)) units and (4/√(17)) units, an...
Let the bigger inscribed side of the square be 2 units.
Therefore, the big inscribed square side is 1 unit.
The marked angle is;
½(atan(½)+atan(2))
= ½(90°)
= 45°
Area blue is;
Area square with side 10 cm - Area quarter circle with radius 10 cm - Area square with side 5 cm + Area quarter circle with radius 5 cm.
= (10*10) - (¼*10*10*π) - (5*5) + (¼*5*5*π)...
