Radius of the circle is;
5-(√(18-18cos(180-2atan(4/3)))
= 5-3.6
= 1.4 units
= (7/5) units.
Therefore area of the circle is;
π(7/5)²
= (49/25)π square units.
Red shaded area will be;
Area square with single side length 3 inches
= 3 * 3
= 9 square inches.
Sir Mike Ambrose is the author of the question.
Calculating Area Orange.
Radius of the circle, r is;
(5-r)²=18-18cos(180-2atan(4/3))
(5-r)²=12.96
5-r=√(12.96)
5-r=3.6
Therefore;
r =...
Let BD = x
Calculating x;
5²= √(9+x²)²+√(9+(x+5)²)²-2√(9+x²)*√(9+(x+5)²)cos45
It implies;
x = 1 unit.
Calculating AC.
AC = √(9+(x+5)²)
And x = 1
It implies;
AC = √(9+(1+5)²...
Let the radius of the quarter circle be 2 units.
Radius, r of the small inscribed semi circle is;
(2+r)²=1+(2-r)²
r = ⅔ units.
Calculating angle x;
½(atan(3/4))
= 18.43494882292°
Let Alpha be x.
Calculating x;
½(180-atan2)-atan(0.5)
x = ½(90-atan(0.5))
It implies;
tan x exactly is;
= tan(½(90-atan(0.5)))
Calculating Area Orange.
Notice, Area Orange is...
a = ½(360-90)
a = ½(270)
a = 135°
It implies;
b² = 1²+√(2)²-2√(2)cos135
b = √(5) units.
Calculating c, radius of the quarter circle.
2c² = b²
2c² = √(5)²
c = √(5/2) units.
Again,...
Sir Mike Ambrose is the author of the question.
Let the single side length of the regular nonagon be 2 units.
Therefore;
Area purple is;
Area triangle with two equal length 2 units each a...
½ - 1
½√(3) - x.
Where x is the base of the blue shaded trapezium.
Cross multiply
½(x) = ½(√(3))
Therefore x = √(3) units.
Therefore area blue (area trapezium) is;
½*½(½√(3)+√(3))...
Let a be the radius of the inscribed circle.
Calculating a.
b = 3a units.
c = b+a
c = 4a units.
It implies;
3a²+4a²+5a = 3a*4a
7a²+5a = 12a²
5a² = 5a
a = 1 unit.
Again, a is the...
