Notice!
AC = 7+7
AC = 14 units.
It implies;
14² = 2a²
a² = ½(196)
a = √(98)
a = 7√(2) units.
a is the side length of square ABCD.
Therefore, shaded area yellow is;
Area square w...
sin30 = 10/a
a = 20 units.
a is the side length of the square.
sin30 = b/10
b = 5 units.
Therefore;
Area Yellow is;
Area triangle with height 20 units and base 5 units.
= ½*20*5...
a² = 3²+5²-2*3*5cos60
a = √(19) units.
a = 4.3588989435 units.
a is AC.
b² = 2²+1²-2*2*1cos60
b = √(3) units.
b = 1.7320508076 units.
b is CD.
(√(19)/sin120) = (√(3)/sinc)
c = 20.12830...
Calculating x.
(a+30)*x = (30+15)(30+15)
x = 45²/(a+30) --- (1).
a*x = 15*15
a = 225/x --- (2).
Therefore, Substituting (2) in (1) to get x.
x = 45²/((225/x)+30)
x = 2025÷(225+30x)/x...
5² = 2a²
a² = ½(25)
a = √(½(25)) cm.
a = √(12.5) cm.
b = 5-a
b = (5-√(½(25))) cm.
b = (5-√(12.5)) cm.
It implies, blue area is;
½(b²)
= ½(5-√(½(25)))²
= ½(5-√(12.5))²
= ½(25-10√(12...
Sir Mike Ambrose is the author of the question.
Let the square side be x unit.
Calculating x.
x² = (x-4)² + ((x/2)+√((x²/4)-16)²
320 = 32x
x = 10 units.
Therefore;
Shaded Area is;...
a² = 3²+4²
a = √(25)
a = 5 cm.
a is the radius of the quarter circle.
Let r be the radius of the inscribed blue circle.
Calculating r.
b = (3+r) cm.
c = (5-r) cm.
It implies;
(...
Area red is;
Area semi circle with radius 5 cm - Area sector with radius 5 cm and angle (180-2atan(2))° - Area kite with diagonal lengths √(10²+5²) cm and √(200-200cos(2atan(½))) cm + Area secto...
Let the two equal lengths, AB and CD be 1 unit.
a = 180-50-15
a = 115°
a is angle ADB.
b = 180-a
b = 65°
b is angle ADC.
(1/sin115) = (c/sin50)
c = 0.8452365235 units.
c is AD.
d²...
Calculating x, length CD.
4² = 6²+6²-2*6*6cosa
72cosa = 72-16
cosa = 56/72
a = acos(7/9)
a = 38.942441269°
a is angle ABC.
b = ½(180-a)
b = 70.5287793655°
b is angle ADB = angle BAD....
