(½(r))²+(3/2)² = r²
(9/4) = r²-¼(r²)
(9/4) = ¾(r²)
r² = 3
r = √(3) units.
Where r is the radius of the circle.
Let the square side be a.
b = 45-20
b = 25°
c² = 2a²
c = √(2)a
Calculating a.
a² = 1²+(√(2)a)²-2*1*√(2)acos25
a² = 1+2a²-2.5634255282a
a²-2.5634255282a+1 = 0
It implies;
a = 2.0834...
a² = 2²+2²
a = √(8) cm.
a = 2√(2) cm.
b = 4+2
b = 6 cm.
c = 2+6-4
c = 4 cm.
d² = 6²+4²
d = √(52) cm.
d = 2√(13) cm.
Therefore, shaded portion perimeter is;
4+4+2√(2)+6+2√(13)...
tana = 16/8
a = atan(2)°
b² = 16²+8²
b = 8√(5) cm.
Let BC = c.
Therefore;
c+3c = b
4c = 8√(5)
c = 2√(5) units.
Again, c is BC.
AB = 3c
AB = 3*2√(5)
AB = 6√(5) cm.
d = 180-30-at...
Let the diameter of the inscribed half circle be 2 units.
It radius is 1 unit.
Calculating x, angle alpha.
tana = 1/0.5
a = atan(2)°
b = 2a
b = 2atan(2)°
c = 180-b
c = (180-2atan(...
Notice!
The inscribed blue area is a square.
Let the inscribed blue area side length be a.
Calculating a.
tan63 = a/b
b = a/tan63 --- (1).
tan43 = a/c
c = a/tan43 --- (2).
It implie...
a = (10-b) cm.
tan30 = (10-b)/10
10 = 10√(3)-√(3)b
b = (10√(3)-10)/√(3)
b = ⅓(30-10√(3)) cm.
tan30 = ⅓(30-10√(3))/c
3c = 30√(3)-30
c = (10√(3)-10) cm.
d = 10-c
d = 10-(10√(3)-10)
d...
sin45 = 12/a
a = 12√(2) cm.
a is AC.
Notice !
AB = 2BC
Let BC = b
Therefore;
AB = 2b cm.
AC = AB+BC
AC = 2b+b
AC = 3b cm.
And AC = 12√(2) cm.
It implies;
12√(2) = 3b
b = 4...
Calculating r, radius of the circle.
a² = 4
a= 2 cm.
a is the side length of the small square.
b = (r-2) cm.
It implies;
r² = 4²+(r-2)²
r² = 16+r²-4r+4
4r = 20
r = 5 cm.
Again, r...
c² = 7²+24²
c = 25 units.
a² = (7+25)²+24²
a = 40 units.
b = ½(a)
b = ½√(40) units.
b = 20 units.
d²+20² = 25²
d = √(625-400)
d = √(225)
d = 15 units.
It implies, R radius of the...
