Calculating the required angle, let it be x.
Let a be the side length of the ascribed square.
½(ab) = 1
b = 2/a --- (1).
1+1+c = ½(a²)
c = ½(a²)-2
c = ½(a²-4) square units.
d = a-b...
Calculating x.
Let a be the height of the inscribed isosceles triangle.
b = ½(7)+7
b = 10.5 units.
c² = 10.5²+a²
c = √(a²+110.25) units.
Let d be one of the two equal inscribed angles.
cosd = 1...
Let a be the side length of the red inscribed square.
b = (4-a) cm.
Therefore;
4² = 2b²
4² = 2(4-a)²
16 = 2(16-8a+a²)
8 = 16-8a+a²
a²-8a+8 = 0
(a-4)² = -8+(-4)²
(a-4)² = 16-8
a = 4±√(8)
a ≠ (4+2√...
Calculating x.
sina = 3.5/12.5
a = asin(3.5/12.5)°
b = 2a
b = 2asin(3.5/12.5)°
sinc = 10/12.5
c = asin(10/12.5)°
d = 2c
d = 2asin(10/12.5)°
e = 180-b-d
e = 180-2asin(3.5/12.5)-2asin(10/12.5)
e...
cosa = (2√(5))/(3√(5))
a = acos(⅔)°
a is angle PRT.
tana = 25/b
b = 25/(tan(acos(⅔)))
b = 22.360679775 units.
b = 10√(5) units.
Therefore, area triangle TRP is;
½*25b
= ½*25*10√(5)
= 125√(5) squa...
Let the radius of the ascribed half circle be 1 unit.
a² = 2x²
a = √(2)x units.
b²+1² = a²
b = √(2x²-1) units.
c = 1-b
c = (1-√(2x²-1)) units.
d = ½(c)
d = ½(1-√(2x²-1)) units.
e...
Let a be the radius of the circle.
b² = 2(2)²
b = √(8)
b = 2√(2) cm.
c = (2√(2)-a) cm.
It implies;
2a² = c²
2a² = (2√(2)-a)²
2a² = 8-4√(2)a+a²
a²+4√(2)a-8 = 0
(a+2√(2))² = 8+(2√(2...
Let the bigger inscribed semi circle radius be 2 units.
Let the big inscribed semi circle radius be r.
Let the inscribed circle radius be q.
Calculating r.
(2+r)² = 2²+(4-r)²
4+4r+r²...
Let a be the radius of the circle.
2πa = 14π
a = 7 cm.
Calculating x.
cos49 = b/7
b = 7cos49
b = 4.59241320293 cm.
c = 2b
c = 9.18482640587 cm.
c is CD
d = 180-49-37.5
d = 93....
Calculating Total Shaded Area, Area Grey, Purple and Blue.
Notice.
KLMN is a square.
Calculating area grey.
a = 6-4
a = 2 units.
sinb = 2/6
b = asin(⅓)
b = 19.4712206345°
c²+2² = 6²
c = 4√(2) u...
