Let AB be 5 units.
Radius, r of the inscribed circle is;
5*(2/5)
r = 2 units.
a = ⅛(180*6)
a = 135°
2b² = 25
b = ½(5√(2)) units.
c = 180-0.5(135)-45
c = 67.5°
sin22.5 = ½(5√(2...
Shaded inscribed area is;
Area regular hexagon with side 1 unit - 6(Area sector with radius 1 unit and angle 60°) + 6(Area equilateral triangle with side 1 unit).
= 6/(4tan(180÷6)) - 6(60π÷36...
Sir Mike Ambrose is the author of the question.
Let the single side length of the regular octagon be 2 units.
Therefore;
Area green is;
Area triangle with height and base √(2) units + Are...
Sir Mike Ambrose is the author of the question.
18² = 14²+16²-2*14*16cosa
448cosa = 14²+16²-18²
a = acos((14²+16²-18²)/448)
a = 73.39845040098°
(18/sin73.39845040098) = (14/sinb)
b = 48.189...
Green Area is;
2(area square with side 2 cm - area quarter circle with radius 2 cm) + 3(area sector with radius 1 unit and angle 300°) + 4(area equilateral triangle with side 1 unit)
= 2(4 -...
Sir Mike Ambrose is the author of the question.
a = 6-2
a = 4 units.
b² = 6²-4²
b = 2√(5) units.
c = 6-b
c = (6-2√(5)) units.
d = asin(⅔)°
tan(asin⅔) = e/(6-2√(5))
e = ⅕(12√(5)-20)...
½(xy)=24
xy = 48 ------ (1)
x²+y²=10² ------- (2)
And;
x²+y²=(x+y)²-2xy
Therefore;
(x+y)²-2xy=100
And;
xy = 48
(x+y)²-2(48)=100
(x+y)²=196
x+y=14 ----- (3)
x = 14-y --...
Let the radius, r of the circle be 7 units.
Therefore perimeter of the circle is;
2πr
= 44 units.
Calculating the length of the square.
14² = (14-2x)²+(14-4x)²
196 = 196-56x+4x²+196-1...
Let r be the radius of the semi circle.
Therefore;
A right-angled triangle with hypotenuse 6 units, and adjacent height and base, (r-7) units and √(r²-49) units respectively is derived.
It...
Let the side length of the regular pentagon be 1 unit.
tan36 = a/0.5
a = 0.363271264 units.
b² = 0.363271264²+0.5²
b = 0.61803398875 units.
tan72 = c/0.5
c = 1.53884176859 units.
Are...
