Side length of the square is;
Let it be x.
x = √((16*34)-144)
x = √(400)
x = 20 cm.
Therefore;
Area Orange exactly in decimal cm² is;
Half area square with side 20 cm - Area triangle with height...
4^(2x)+2 = 3*4^(x)
Let 4^(x) = p
It implies;
p²+2 = 3p
p²-3p+2 = 0
Resolving the above quadratic equation.
p²-p-2p+2 = 0
p(p-1)-2(p-1) = 0
(p-2)(p-1) = 0
Therefore;
p = 1 or p = 2
And p = 4^(x...
Let the blue square side length be x.
2² = a²+b²
b² = 4-a²
b = √(4-a²)
4² = a²+(√(4-a²)+x)²
16 = a²+((4-a²)+2x√(4-a²)+x²
12 = 2x√(4-a²)+x² --- (1).
5² = (x+a)²+(4-a²)
25 = x²+2xa+a²+4-a²
21 = x²+...
a = ½(8+2)
a = 5 cm.
a is the radius of the ascribed semi circle.
b = 8
2 = b. Cross Multiply.
b² = 16
b = 4 cm.
Therefore, area Blue is;
½(8*4)
= ½(32)
= 16 cm²
Calculating Area Red.
tanc = 4/3...
Sir Mike Ambrose is the author of the question.
Let the side of the regular pentagon be 1 unit.
a = 360-2(108)-90
a = 54°
tan72 = b/0.5
b = 1.53884176859 units.
tan54 = 1.53884176859/c
c = 1.1180...
Notice.
a = 18+55 = 73°
b = 28+45 = 73°
It implies;
The ascribed triangle is isosceles.
Calculating angle x.
Let the shorter length of the ascribed isosceles triangle be 1 unit.
c = 180-45-55-1...
Notice.
ABCD is a square.
Let AB = 1 unit.
cos50 = a/1
a = 0.6427876097 units.
a is CE.
b = ½√(1²+1²)
b = ½√(2) units.
b = 0.7071067812 units.
b is OC.
c² = 0.7071067812²+0.6427876097²-2*0.70710...
tana = (10/5)
a = atan(2)°
b = ½(180-a)
b = ½(180-atan(2))°
cosb = c/10
cos(½(180-atan(2))) = c/10
c = 5.2573111212 cm.
sinb = d/10
sin(½(180-atan(2))) = d/10
d = 8.5065080835 cm.
It implies, bl...
Let the side of the regular pentagon be 1 unit.
Notice;
The regular pentagon side is equal the regular heptagon side.
a = ⅐(180*5)
a = ⅐(900)°, the single interior angle of the regular heptagon....
It implies;
tan(0.5acos(3/5)) = r/3
Where r is the radius of the inscribed semi circle.
r = 3tan(0.5acos(3/5))
r = 1.5 units.
r = 1½ units.
