Notice.
The ascribed right-angled triangle is not drawn to scale.
Let a be the side length of the inscribed square.
Calculating Area Inscribed Square.
tanb = 3√(3)/(9√(3))
b = atan(⅓)°...
Calculating Length x (AD).
Let a be the diameter of the green inscribed circle.
Therefore;
x ~ a
(b+x) ~ 2a
Cross Multiply.
2ax = ab+ax
2ax-ax = ab
ax = ab
x = b units.
Theref...
Let OF be r, radius of the ascribed circle.
a²+r² = (2+3)²
a = √(25-r²) units.
b = a+r
b = (r+√(25-r²)) units.
c = r-a
c = (r-√(25-r²)) units.
Therefore;
4*(2+3) = (r-√(25-r²))(r+...
Total area of the ascribed composite plane shape is;
(¼*12*12π) + (½*6*6π)
= 36π+18π
= 54π cm²
Calculating the inscribed shaded area.
Let re be the radius of the inscribed shaded half ci...
a² = 3²+4²
a = √(25)
a = 5 units.
a is the hypotenuse of the ascribed right-angled triangle.
tanb = 3/4
b = atan(3/4)°
tanc = 4/3
c = atan(4/3)°
Let d be the radius of the inscribed c...
Let BH be a.
It implies;
25 ~ a
a ~ 4
Cross Multiply.
a² = 100
a = 10 units.
b² = 25²+10²
b = √(725) units.
b is AB, the side length of the ascribed square.
c² = 10²+4²
c = √...
Notice.
The angle formed at the meet point (vertex) of length 4 units and x units is equal the angle formed at the meet point (vertex) 9 units and x units.
Therefore, as a result, two similar...
Sir Mike Ambrose is the author of the question.
Let the square side be 1 unit.
Therefore;
Area B is;
Area triangle with height 1 unit and base sin(53.13010235416) units - Area triangle wi...
Calculating Area Blue.
(2/sin120) = (1/sina)
a = 25.6589062733°
b = 180-120-a
b = 60-25.6589062733
b = 34.3410937267°
Therefore, shaded blue area is;
Area sector with radius 2 units...
a = ½(4+6)
a = 5 units.
a is radius.
10² = x²+b²
b = √(100-x²) units.
c = 10-½(4)
c = 8 units.
x² = c²+d²
x² = 8²+d²
d = √(x²-64) units.
Calculating x.
Therefore;
8 ~ √(x²-6...
