Let a be the radius of the inscribed blue circle.
b = (24-a) units.
c = ½(24)
c = 12 units.
Therefore, Calculating a.
a²+12² = (24-a)²
a²+144 = 576-48a+a²
48a = 432
a = 9 units.
Ag...
Calculating shaded area (triangle BCE).
Let a be the side length of the square.
5² = a²+3²-2*3*acosb
25 = a²+9-6acosb
16 = a²-6acosb --- (1).
8² = a²+3²-2*3*acos(90-b)
64 = a²+9-6a(cos9...
Sir Mike Ambrose is the author of the question.
Notice;
The single side length of the square is 12 units.
Area blue is;
Area triangle with height and base 8 units each - Area triangle wit...
Please, move the above question left/right one time to review the solution.
Thank you.
Please, move the above question left/right one time to review the solution.
Thank you.
Sir Mike Ambrose is the author of the question.
Please, move the above question left/right one time to review the solution.
Thank you.
1²+(1-x)²=(1+x)²
1+1-2x+x²=1+2x+x²
1 = 4x
x = ¼ units.
Therefore;
Area Orange/Shaded is;
Area triangle with height 1 unit and base (1-x) units.
= ½(1*(1-x)), and x = ¼ units.
= ½(1-...
a = (24-R) units.
It implies;
R² = (24-R)²+12²
R² = 576-48R+R²+144
48R = 576+144
48R = 720
4R = 60
R = 15 units.
Calculating x, side length of the inscribed regular hexagon.
sin60 = a/6
2a = 6√(3)
a = 3√(3) units.
It implies;
Calculating x.
(2x)² = 6²+(3√(3))²
4x² = 36+27
4x² = 63
x² = 63/4...
Let the square side be 3 units.
Area of the plane shape is;
Area square with side 3 units + Area triangle with height 3 units and base ½(3) units.
= 3²+(½*½(3)*3)
= 9+(9/4)
= 45/4 space...
