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)...
Area pink is;
4(area semicircle with radius 4 cm - area quarter circle with radius 4√(2) cm + area triangle with height and base 4√(2) cm) respectively.
It implies;
4(½(π(4²)) - ¼(π(4√(2))²) + ½(...
Sir Mike Ambrose is the author of the question.
Yellow area exactly in decimal cm² is;
Area triangle with height 7.49903313849 cm and base 8sin(32.2356103172) cm + Area triangle with height 7.49...
Sir Mike Ambrose is the author of the question.
Let c be 4 units.
Therefore;
b is;
2cos(25)+2cos(25)tan(65/2)
b = 2.96737905059 units.
a is;
2sin(65)-2sin(65)tan(65/2)
a = 0.65...
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.
a...
Sir Mike Ambrose is the author of the question.
Area ABC (yellow area) exactly is;
Area triangle with height ⅓√(265) cm and base ((5√(26)/6)sin84.3306169587) cm.
= ½*⅓√(265)*((5√(26)/6)sin84.33061...
