Let the side length of the regular pentagon be 4 units.
a = ⅕*180(5-2)
a = 108°
a is the single interior angle of the regular pentagon.
b = ½(180-a)
b = ½(180-108)
b = 36°
c = 2 units....
Notice.
7 square units is the are of the small pink half circle.
½*a²π = 7
And π = 22/7
½*a²(22/7) = 7
11a² = 49
√(11)a = 7
a = 7√(11)/11 units.
a = 2.110579412 units.
a is the radius...
a² = 4²+3²
a = √(25)
a = 5 units.
a is MN, the side length of the inscribed square.
b² = 2(5²)
b = 5√(2) units.
b is MK, the diagonal of the inscribed square.
tanc = 4/3
c = atan(4/3)°...
Notice.
7 square units is the are of the small pink half circle.
½*a²π = 7
And π = 22/7
½*a²(22/7) = 7
11a² = 49
√(11)a = 7
a = 7√(11)/11 units.
a = 2.110579412 units.
a is the radius...
tana = 5/10
a = atan(½)°
b = 2a
b = 2atan(½)°
c = b-45
c = 8.1301023542°
d² =10²+10²-2*10*10cos8.1301023542
d = 1.4177804018 units.
e = ½(180-8.1301023542)
e = 85.9349488229°
f²...
Sir Mike Ambrose is the author of the question.
Let length AB be 3 units.
Therefore;
AC = 20 units.
It implies;
Area large square is;
20²
= 40 square units.
Area orange is;
A...
Let EB be a.
Let AE be b.
It implies;
a/b = √(2) --- (1).
a+b = 1 --- (2).
From (1).
a = √(2)b --- (3).
Substituting (3) in (2) to get b.
√(2)b+b = 1
b(√(2)+1) = 1
b = (√...
Sir Mike Ambrose is the author of the question.
Let the side of the square be 2 units.
Therefore;
Area square is;
2²
= 4 square units.
Area S is;
Area triangle with two side 0.9892...
Let the square side be 2 units.
tan30 = a/2
a = ⅓(2√(3)) unit.
sin60 = 2/b
b = ⅓(4√(3)) units.
c = 2-a
c = ⅓(6-2√(3)) units.
It implies;
Area Shaded is;
2(½*(⅓(6-2√(3)))²sin60)...
Sir Mike Ambrose is the author of the question.
Let the side of the inscribed square be 2 units.
Therefore;
Area S is;
Area triangle with two side 1.03528 units and 0.53589838486 units, a...
