Sir Mike Ambrose is the author of the question.
Let the side of the regular hexagon be 1 unit.
Therefore;
Area blue is;
2(area triangle with height 1 unit and base (⅓√(3)) units).
= 2(½*...
8 = 4
a = (8-a) Cross multiply.
64-8a = 4a
64 = 12a
a = ⅓(16) cm.
b = 8-a
b = ⅓(8) cm.
c = atan(⅔)°
d = atan(3/2)°
⅔ = 8/e
e = 12 cm
⅔ = f/8
f = ⅓(16) cm.
g = 8-f
g = ⅓(...
Sir Mike Ambrose is the author of the question.
Radius of the inscribed circle, r is;
2r² = 4
r² = 2
r = √(2) units.
Therefore;
Area R in square unit to 3 d. p. is;
Area sector wit...
The radius of the inscribed circle is 2 units.
a = 2atan(⅓)°
b = 2atan(⅕)°
c = 180-a-b
c = (180-2atan(⅓)-2atan(⅕))°
sin(2atan(⅓)) = d/16
d = (48/5) units.
cos(2atan(⅓)) = e/16
e =...
Sir Mike Ambrose is the author of the question.
Let the side length of the three congruent, inscribed square be x.
Calculating x.
x + 2 = 2√(3)
x = (2√(3)-2) units.
Therefore;
Area gr...
Let a be the side of the square.
cos30 = b/8
b = 4√(3) cm.
sin30 = c/8
c = 4 cm.
a = b+12
a = (12+4√(3)) cm.
Area Square is;
a² = (12+4√(3))²
a² = 358.27687752661 cm²
d² = 12...
Let the diameter of the inscribed white semicircle be x.
Therefore the diameter of the ascribed semicircle is 2x.
Calculating x.
(3x/2)² = x²+36
(9x²/4)-x² = 36
(5x²/4) = 36
√(5)x = 12...
Sir Mike Ambrose is the author of the question.
Area purple is;
Area trapezoid with two parallel sides 4 cm and (50/9) cm, and height (7/2) cm - Area triangle with height 4 cm and base 3 cm....
Let the height of the yellow area trapezoid be a.
Calculating a.
tan72 = a/b
b = (a/tan72) cm.
c = 2b
c = (2a/tan72) cm.
d = 8+c
d = (8+(2a/tan72)) cm.
It implies;
½*a(8+(8+(2...
Sir Mike Ambrose is the author of the question.
Let AB be 2 unit.
Therefore CD = 1 unit.
It implies;
Area Blue is;
Area triangle with height 0.72111025509 unit and base sin(atan(3/2)) un...
