Let the radius of the inscribed circle be r.
The square side is 8 cm.
b = 8-r
c = 2+r
Therefore;
(2+r)² = 2²+(8-r)²
4+4r+r² = 4+64-16r+r²
4r = 64-16r
20r = 64
5r = 16
r = (16/5) cm.
Area circle...
Sir Mike Ambrose is the author of the question.
Let the side let of the square be 2 units.
Therefore;
Area square is;
2²
= 4 square units.
It implies;
Area R is;
Area trapezoid with parallel sid...
a = ½(4√(6)-2√(6))
a = √(6) cm.
tan60 = b/√(6)
b = 3√(2).
b is the height of the trapezoid.
c² = (√(6))²+(3√(2))²
c² = 6+18
c = √(24)
c = 2√(6) cm.
Or
cos60 = √(6)/c
c = √(6)/(½)
c = 2√(6) cm.
c i...
Notice.
64 cm is the perimeter of the square.
Therefore, calculate a, the side length of the square.
a = ¼(64)
a = 16 cm.
b = 16-11.5
b = 4.5 cm.
c = 16-10
c = 6 cm.
It implies;...
3A = ¼(9*9)π
Where;
3A is the sum of the inscribed three equal areas.
¼(9*9)π is the area of the ascribed quarter circle.
It implies;
12A = 9*9π
A = (27/4)π cm²
A is an area of the three equal are...
Let r be the radius of the inscribed green semi circle.
Notice.
The ascribed triangle is equilateral.
Therefore;
sin60 = r/(0.5*7)
r = 0.5*7sin60
r = 3.0310889132 units.
Again, r is the radius of...
Notice.
Rectangle width is 4 units.
a = 4-1
a = 3 units.
tanb = ½
b = atan(⅓)°
c = 2b
c = 2atan(⅓)°
d = 90-c
d = (90-2atan(⅓))°
tan(½d) = 2/e
tan(0.5(90-2atan(⅓))) = 2/e
e = 2/tan(0.5(90-2atan(⅓...
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(2√(2)/sin135) = (√(2)/sina)
a = 20.70481105464°
b = 180-20.70481105464-135
b = 24.29518894536°
(c/sin24.29518894536) = (2√(2)/sin135)
c = 1.64575131106 units.
Where c is the height of the yellow...
