a = ⅕*180(5-2)
a = 108°
a is the single interior angle of the regular pentagon.
b = 180-a-42
b = 180-150
b = 30°
(12/sin108) = (c/sin30)
c = 6.30877334543 cm.
c is the side length of th...
Let r be the radius of the half circle.
a²+2² = r²
a = √(r²-4) units.
b = 2a
b = 2√(r²-4) units.
c = (r-2) units.
Calculating r.
√(r²-4) ~ 2√(r²-4)
(r-2) ~ 9
Cross Multiply....
2πr = 13π
r = ½(13) cm.
r is the radius of the ascribed large circle.
a = 2r
a = 2*½(13)
a = 13 cm.
a is the diameter of the ascribed large circle.
It implies;
c = a-3b
c = (13-3b) c...
Calculating R, radius of the circle.
R² = (½(9))²+a²
a² = R²-(81/4)
a = √(R²-(81/4)) units.
b = ½(9)+3
b = ½(15) units.
c = a-4
c = (√(R²-(81/4))-4) units.
There, R is;
R² = b²+c...
a² = 2²+2²
Where;
a is radius of the ascribed half circle.
2 is the side length of the inscribed blue square.
a² = 8
a = 2√(2) cm.
Again, a is the radius of the circle.
Therefore, leng...
Notice.
R-r = 10 cm.
It implies;
R = (10+r) cm.
R is the side length of the ascribed square.
Calculating length AB.
a = R-r
a = 10 cm.
b = ½(R) units.
It implies;
R² = 10²+...
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 b...
Sir Mike Ambrose is the author of the question.
Let the square aide length be 1 unit.
Square Area is;
1²
= 1 square unit.
Calculating Green Area.
a² = 1²+1²
a = √(2) units.
a is AB....
Let a be the side length of the big inscribed square.
Let b be the side length of the bigger inscribed square.
Therefore;
27 ~ b
a ~ 64
Cross Multiply.
ab = 27*64
b = (1728)/a unit...
Sir Mike Ambrose is the author of the question.
Let a be the equal lengths inscribing triangle ABC.
Calculating a.
½(a)² = 15
a = √(30) units.
b = 180-60-45
b = 75°
b is angle BAC....
