Sir Mike Ambrose is the author of the question.
tan30 = a/3
a = √(3) units.
b = a+3
b = (3+√(3)) units.
tan30 = c/b
⅓√(3) = c/(3+√(3))
c = ⅓(3√(3)+3) units.
c = (1+√(3)) units.
It im...
Let a be the blue length.
b² = 4²+3²
b = √(25)
b = 5 units.
Let alpha be c.
cos(2c) = 4/5
cos(c+c) =4/5
(cosc)²-(sinc)² = 4/5 --- (1).
cosc = 3/a --- (2).
a² = 3²+d²
d = √(a²-9)...
Let a be the radius of the circle.
b = 7-3
b = 4 cm.
c = (7-a) cm.
d = (3-a) cm.
e = c+d
e = 7-a+3-a
e = (10-2a) cm.
Calculating a.
e² = (2a)²+4²
(10-2a)² = 4a²+16
100-40a+4...
Notice.
Triangle ACD and Triangle BCD are right-angled triangles.
Let AD = 2 units.
AB = 6 units.
It implies;
BD = 6-2
BD = 4 units.
a² = 2(0.5*4)²
a = 2√(2) units.
a is BE = C...
Notice.
The inscribed square area is 3 square units.
Therefore the single side length of the inscribed square is √(3) units.
The radius of the ascribed quarter circle will be;
√((√(3))²+(...
Let AG be y.
Calculating y.
a² = 8²+y²
a = √(y²+64) cm.
a is AD.
It implies;
Considering area ADG including the inscribed circle with radius 1 cm.
½(1*8)+½(1*y)+½(1*√(y²+64)) = ½(8*y)
8+y+√(y²...
Let the side length of the regular pentagon be 1 unit.
tan36 = a/0.5
a = 0.363271264 units.
b² = 0.363271264²+0.5²
b = 0.61803398875 units.
tan72 = c/0.5
c = 1.53884176859 units.
Are...
Let the single side length of the square be x.
Therefore;
Area rectangle of length x unit and width 1 cm + area triangle of height x cm and width (x-2) cm + 2(area triangle of height (x-1) cm...
Let a be the side length of the regular hexagon.
3² = 1²+a²-2acosb
2acosb = a²-8
cosb = (a²-8)/(2a) --- (1).
c²+(a²-8)² = (2a)²
c² = 4a²-(a⁴-16a²+64)
c = √(-a⁴+20a²-64) units.
sinb = c...
Let the radius of the semi circle be 3 units.
Calculating area Red.
Area red will be;
2(area triangle of two length 3 units and 1 unit, and angle (cos–¹(⅓)), angle they form) + area sector...
