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...
a*15 = 540
a = 108/3
a = 36 cm.
a is AB.
b = a-R
b = (36-R) cm.
b is OB.
It implies;
Calculating R.
R² = 15²+(36-R)²
R² = 225+1296-72R+R²
72R = 1521
R = 21.125 cm.
Calculating the area of the blue inscribed rectangle and the area of the ascribed regular pentagon.
πr² = 36π
r = 6 cm.
r is the radius of the inscribed circle.
sin(0.5*108) = 6/a
a = 7.41...
a = (1-r) units.
b²+r² = (1-r)²
b = √(1-2r) units.
c = (0.5+r)
It implies;
(0.5+r)² = r²+√(1-2r)²
¼+r+r² = r²+1-2r
3r = 1-¼
3r = ¾
r = ¼ units.
Or
(½+r) = (1-r)
Where;
AB =...
Let a be the radius of the inscribed sector.
b = (2+a) units.
Therefore;
(2+a)² = a²+6²
4+4a+a² = a²+36
4a = 32
a = 8 units.
Again, a is the radius of the inscribed sector.
b = (2+a...
Let AD = CD = a.
Calculating a.
b² = 2a²
b = √(2)a units.
b is AC.
c²+3² = a²
c = √(a²-9) units.
c is DE.
a² = d²+7²
d = √(a²-49) units.
e = c-d
e = (√(a²-9)-√(a²-49)) units....
