Let the side length of the ascribed square be 1 unit.
Therefore, the side length of the inscribed green regular triangle 1 unit.
a = 90-60°
a = 30°
b² = 2-2cos30
b = 0.51763809021 units....
sina = 4/(4+1)
a = asin(4/5)°
b² = 1²+4²-2*1*4cos(asin(4/5))
b = ⅕√(305) units.
b = 3.49284983931 units.
(3.49284983931/sin(asin(4/5))) = (4/sinc)
c = asin(0.9161573349)
c = 113.62937773...
a² = (3+5)²+x²
a² = a²-64
a = √(64+x²) cm.
b = (8-x) cm.
It implies;
5 ~ 8
(8-x) ~ √(64+x²)
Cross Multiply.
8(8-x) = 5√(64+x²)
(64-8x)² = 25(64+x²)
4096-1024x+64x² = 1600+25x²
2...
Let single side of the square be x.
Notice;
PB=¼(AB)=¼(x).
Calculating single side length of square ABCD.
It will be;
Area triangle ADP of height x and base (x-¼(x)) + area triangle...
Sir Mike Ambrose is the author of the question.
Let the bigger inscribed square be 2 units.
BC = 1+2+x = (3+x) units.
Calculating x.
(3+x)²=1²+8-4√(2)cos135
3+x=√(13)
x = (√(13)-3)...
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...
