Area ABCD is;
½((5√(3)) square units.
Please, move the above question left/right to review the solution.
Thank you.
Sir Mike Ambrose is the author of the question.
Let the single side length of triangle ABC be 2 units.
Calculating Area ABC.
½(4sin60)
= √(3) square units.
Calculating Area Yellow.
It...
a = 180-60
a = 120°
b = ½(8)
b = 4 units.
It implies;
(8/sin120) = (4/sinc)
c = 25.6589062733°
d = 690-c
d = 34.3410937267°
Therefore, yellow area is;
Area sector with radius...
Let a be the radius of the inscribed circle.
b = ½(2a+1+4)
b = ½(2a+5) units.
b is the radius of the ascribed semi circle.
c = b-4
c = ½(2a+5)-4
c = ½(2a-3) units.
d = a-c
d = a-½(2a-...
Sir Mike Ambrose is the author of the question.
Let the side length of the ascribed square be 1 unit.
Area square is;
1²
= 1 square units.
a = ½(90-60)
a = 15°
tan15 = b/1
b = 0.267...
Sir Mike Ambrose is the author of the question.
Calculating k.
3k² = 108
k = 6 units.
The equation of the curve is;
y = 4x-⅓(x²)
Area R is;
Area under the curve at x=0 and x=(6-3√(2...
Sir Mike Ambrose is the author of the question.
Let the side of the regular quadrilateral (square) be 4 units.
Area square is;
4²
= 16 square units.
Area Red is;
Area triangle with he...
Let the radius of yellow circle be r.
Let the radius of the blue semi circle be R.
Calculating r.
(4r+1)x1=5x1
4r+1=5
4r = 4
r = 1 unit.
Calculating R.
2²+3²=R²
R = √(4+9)
R = √...
Calculating r, radius of the inscribed circle.
r²=(r-4)²+(r-2)²
r²=r²-8r+16+r²-4r+4
r²-12r+20=0
(r-6)²=-20+36
r = 6±√(16)
r = 6±4
Therefore;
r ≠ 2 units.
r = 10 units.
Therefore, area...
Area Blue (Star) is;
Area circle of radius 20 cm - 5(area triangle with two length 14.5308505601 cm and angle 108°, the angle they form) - 5(Area sector of radius 20 cm and angle 72°) + 5(area t...
