30² = 18²+a²
a² = 900-324
a = √(576)
a = 24 units.
b = ½(a)
b = 12 units.
Therefore, the required length, x is;
x² = 18²+b²
x² = 18²+12²
x² = 324+144
x = √(468)
x = 6√(13) units....
tana = 7/4
a = atan(7/4)°
b = 180-a
b = (180-atan(7/4))°
c² = 7²+4²
c = √(49+16)
c = √(65) cm.
It implies, area blue triangle is;
Area triangle with height √(65) cm and base 4sin(18...
Let AB be x cm.
BC = 3x cm.
AC = AB+BC
AC = 4x cm.
sin60 = a/16
a = 8√(3) cm.
a is AC.
Therefore;
4x = 8√(3)
x = 2√(3) cm.
Again, x is AB.
BC = 6√(3) cm.
CE = CD = 8 cm....
tan30 = 8/a
a = 8√(3) cm.
b = ½(a)
b = 4√(3) cm.
Area Blue is;
½*8√(3)*10-½*10*4√(3)
= 40√(3)-20√(3)
= 20√(3) cm²
Area triangle ABC is;
½*10*8√(3)+½*8√(3)*8
= 40√(3)+32√(3)
= 72√(3) cm²
Theref...
a = (10-x) cm.
10 cm is the radius of the ascribed semi circle.
12² = a²+b²
b² = 144-(100-20x+x²)
b² = 44+20x-x²
b = √(44+20x-x²) cm.
10² = x²+b²
100 = x²+44+20x-x²
56 = 20x
14 = 5x
x...
Let a be the radius of the inscribed orange circle.
(3+a)² = a²+b²
b² = 9+6a+a²-a²
b = √(9+6a) cm.
c = c-3
c = (√(9+6a)-3) cm.
It implies;
(6-a)² = a²+c²
(6-a)² = a²+(√(9+6a)-3)²
36-12a+a² = a²...
a² = 4²+3²
a = √(25)
a = 5 units.
a is the radius of the semi circle.
tanb = (3/4)
b = atan(3/4)°
c = 90-b
c = (atan(4/3))°
tan(atan(3/4)) = d/5
(3/4) = d/5
d = 3.75 units.
d = ¼(15) units.
tan(...
Calculating x (radius of the small inscribed circle)
28² + x² = (56-x)²
784 + x² = 3136 - 112x + x²
112x = 2351
x = 21 units.
a = 108-90
a = 18°
cos18 = b/1
b = 0.9510565163 units.
c = 1+0.9510565163
c = 1.9510565163 units.
c is the side length of the regular ascribed pentagon.
sin18 = d/1
d = 0.3090169944 units.
e² =...
a = 120π*2*30÷360
a = ⅓(60)π
a = 20π cm.
a = 62.8318530718 cm.
a is the length of arc of the sector and also, the base circumference of the cone.
20π = 2πb
b = 10 cm.
b is the radius of th...
