Sir Mike Ambrose is the author of the question.
Let AC be 1 unit.
Area smaller pentagon is;
½(5)*(1/(2tan(36)))
= 1.72047740059 square units.
a² = 1²+0.5²-cos108
a = 1.24860602048 uni...
Sir Mike Ambrose is the author of the question.
Area yellow is;
Area triangle with two side 8√(2) cm and 2 cm, and angle 45° - Area triangle with height (20/3) cm and base ⅓(20tan(45-2atan(⅓)...
Sir Mike Ambrose is the author of the question.
Area red is;
Area triangle with two side 4 cm and 8 cm, and angle 110°.
= 0.5*4*8sin110
= 15.0350819326 cm²
Area blue is;
Area triang...
Sir Mike Ambrose is the author of the question.
The side of the square is;
6.065425 cm
Therefore;
Area red in cm² to 2 d. p is;
Area triangle with height (2.1tan29.11) cm and base 2.1...
Sir Mike Ambrose is the author of the question.
Point P coordinate is;
P(4, -3)
Point R coordinate is;
R(0, ⅓(37))
Point Q (maximum turning point) coordinate is;
Q((-3√(259)/14),...
ac+1 = a ----- (1)
bc+1 = 4c ----- (2)
3ab+3 = 7b ---- (3)
From (1)
ac+1 = a
a = 1/(1-c) ----- (4)
Sub. (4) in (3)
3ab+3 = 7b
3(1/(1-c))b+3 = 7b
3b/(1-c) = (7b-3)
3b = (1-c)(7b-3)
3b = 7...
Sir Mike Ambrose is the author of the question.
Shaded area is;
Area under the curve at points (-1, 0) and (7, 0) - Area triangle with height 4 units and base 8 units.
= ⅓(208) - ½(4*8)...
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...
