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....
Calculating blue area.
It is;
Area square with side length 3 cm + Area semi circle with radius 1 cm + Area semi circle with radius 2 cm.
= (3*3)+(π*½*1²)+(π*½*2²)
= 9+½(π)+2π
= ½(18+π+4π...
cos30 = a/16
a = 8√(3) cm.
cos30 = 10/b
b = ⅓(20√(3)) cm.
c = a-b
c = 8√(3)-⅓(20√(3))
c = ⅓(4√(3)) units.
d = 2c
d = ⅓(2√(3)) cm.
e = b+d
e = ⅓(20√(3))+⅓(2√(3))
e = ⅓(22√(3)) cm...
Notice!
Radius of the half circle is 9 units.
Let the three equal lengths be b.
5² = b²+a²-2bacos45
25 = b²+a²-√(2)ab --- (1).
9² = b²+a²-2abcos135
81 = b²+a²+√(2)ab --- (2).
Subtra...
Sir Mike Ambrose is the author of the question.
Calculating Yellow Area.
It is;
Area triangle with height 4sin(atan(⅔)) units and base 8√(13)/9 units - Area triangle with height 1 unit and bas...
a²+2² = 8²
a² = 64-4
a = √(60)
a= 2√(15) units.
a is BD
b = ½(a)
b = √(15) units.
c = ½(AB)
c = 4 units.
c is the radius of the half circle.
cosd = √(15)/4
d = acos(¼√(15))°
d is a...
