Calculating Shaded Area.
a² = 1²-(4k)²
a = √(1-16k²) units.
1 ~ √(1-16k²)
4k ~ b
Cross Multiply.
b = 4k√(1-16k²) units.
Again.
1 ~ √(1-16k²)
√(1-16k²) ~ c
Cross Multiply.
c = (1-16k²) units....
a = x°
b = (90-x)°
R² = 6²+16²-2*6*16cosx
R² = 292-192cosx
cosx = (292-R²)/192 --- (1).
Notice.
At (1).
(292-R²) is adjacent.
192 is hypotenuse.
Let c be opposite.
c² = 1...
Let the radius of the ascribed circle be 2 units.
Therefore, radius of the bigger inscribed circle is 1 unit.
a = (1+y) units
b = (2-y) units.
c = (1-y) units.
Calculating y, radius of the big...
Notice.
The ascribed quadrilateral is a square.
Let it's side length be x.
a²+x² = 3²
a = √(9-x²) units.
2 ~ 3
b ~ x
Cross Multiply.
b = ⅓(2x) units.
It implies;
Calculating x.
x²-½(x*√(...
Let OA be 1 unit.
Area Triangle AOB is;
½*1*1
= ½ square units.
Calculating Area Triangle ODG.
a² =2(1)²-2(1)²cos30
a = √(2-√(3)) units.
a = 0.51763809021 units.
a is AC, the side length of the...
Let a = 1 unit.
It implies;
1 ~ b
b ~ e
Cross Multiply.
e = b²
f = (1-b²) units.
cos60 = g/(1-b)
g = ½(1-b) units.
h = 2g+1
h = (2-b) units.
It implies;
Calculating b....
Calculating length RT.
a = ½(10)
a = 5 units.
a is half the diagonal of the rectangle.
b = 8+a
b = 13 units.
It implies;
(RT)²+5² = 13²
RT = √(169-25)
RT = √(144)
RT = 12 units.
Calculating x, length BE.
a = 90-75
a = 15°
a is angle CBE.
sin15 = b/x
b = (xsin15) units.
b is CE.
c = 5+b
c = (5+(xsin15)) units.
c is CD.
cos15 = d/x
d = (xcos15) units.
d is BC.
Notice.
c...
Let the length of the rectangle be a.
Let the width of the rectangle be (6+b) units.
½(2a-4)(6+b)-70 = ½(6a)+½(a-4)b
12a+2ab-24-4b-140 = 6a+ab-4b
6a+ab = 164
Therefore;
a(6+b) = 164
Notice....
Let the side length of the square be x.
a = (x-4) units.
Calculating x.
½(x²) = ½*x(x-4)-10+30
x² = x(x-4)+40
x² = x²-4x+40
4x = 40
x = 10 units.
Again, x is the side length of the square.
A...
