a² = 3²+4²
a = √(25)
a = 5 units.
a is the hypotenuse of the ascribed right-angled triangle.
tanb = 3/4
b = atan(3/4)°
tanc = 4/3
c = atan(4/3)°
Let d be the radius of the inscribed c...
Let BH be a.
It implies;
25 ~ a
a ~ 4
Cross Multiply.
a² = 100
a = 10 units.
b² = 25²+10²
b = √(725) units.
b is AB, the side length of the ascribed square.
c² = 10²+4²
c = √...
Notice.
The angle formed at the meet point (vertex) of length 4 units and x units is equal the angle formed at the meet point (vertex) 9 units and x units.
Therefore, as a result, two similar...
Sir Mike Ambrose is the author of the question.
Let the square side be 1 unit.
Therefore;
Area B is;
Area triangle with height 1 unit and base sin(53.13010235416) units - Area triangle wi...
Calculating Area Blue.
(2/sin120) = (1/sina)
a = 25.6589062733°
b = 180-120-a
b = 60-25.6589062733
b = 34.3410937267°
Therefore, shaded blue area is;
Area sector with radius 2 units...
a = ½(4+6)
a = 5 units.
a is radius.
10² = x²+b²
b = √(100-x²) units.
c = 10-½(4)
c = 8 units.
x² = c²+d²
x² = 8²+d²
d = √(x²-64) units.
Calculating x.
Therefore;
8 ~ √(x²-6...
Notice.
r, radius of the ascribed quarter circle is (25/√(2)) units.
r = ½(25√(2)) units.
(AD)² = 2(½(25√(2)))²
(AD)² = 2*¼*625*2
AD = √(625)
AD = 25 units.
It implies;
AB+BC+CD = AD
10+BC+4 =...
Calculating Area Blue.
a = 4+8
a = 12 cm.
a is the radius of the ascribed quarter circle.
b² = 12²+4²
b = √(144+16)
b = √(160)
b = 4√(10) units.
tanc = 12/4
c = atan(3)°
(12/sin(60+atan(3))) =...
Let FC=x.
Let CE=y
Therefore;
5+x=y+4
x=y-1 ------- (1).
Notice;
Triangle AEF is similar to triangleEFC.
It implies;
5+x=y
4=x, cross multiply.
x²+5x=4y ----- (2).
Subs...
a = (r+9) units.
b = (r-9) units.
a² = b²+c²
(r+9)² = (r-9)²+c²
c² = r²+18r+81-r²+18r-81
c² = 36r
c = 6√(r) units.
d = (r+4) units.
e = (r-4) units.
f²+e² = d²
f² =...
