Calculating Area Yellow.
Let a be the radius of the ascribed semi circle.
b²+6² = a²
b = √(a²-36) units.
b is the radius of the inscribed half circle.
Therefore, area yellow is;
½(a²)...
Let the side length of the regular hexagon be 1 unit.
a = ⅙*180(6-2)
a = 120°
a is the single interior angle of the regular hexagon.
sin30 = b/1
b = ½ units.
cos30 = c/1
c = ½√(3) uni...
Sir Mike Ambrose is the author of the question.
a² = 10²+16²-2*10*16cos120
a = 2√(129) units.
Where a is length AC.
(3√(129)/sin120) = (16/sinb)
b = 37.58908946897°
c = 90-b
c = 52.410...
Sir Mike Ambrose is the author of the question.
a² = 10²+16²-2*10*16cos120
a = 2√(129) units.
Where a is length AC.
(3√(129)/sin120) = (16/sinb)
b = 37.58908946897°
c = 90-b
c = 52.410...
Let the side length of the regular hexagon be 1 unit.
Therefore;
a = 1 unit.
Calculating b.
Cos30=a/c
And a = 1
c = 1/cos30
c = ⅔√(3) units.
Where c is the hypotenuse of the r...
Sir Mike Ambrose is the author of the question.
Area Blue, in cm² to 1 d. p. is;
Half area regular pentagon with side 12 cm - area triangle with height 6 cm and base 6tan24 cm - area triangle w...
Calculating yellow area.
a = ½(10)
a = 5 km.
a is the radius of the inscribed circle.
tanb = 5/10
b = atan(½)°
c = 180-2b
c = (180-2atan(½))°
d² = 2(5²)-2(5²)cos(180-2atan(½))
d =...
Let a be the radius of the inscribed circle.
Let b be the radius of the ascribed circle.
It implies;
πb²-πa² = 20π
b²-a² = 20
b = √(20+a²) units.
Again, b is the radius of the ascribed...
Sir Mike Ambrose is the author of the question.
Let the side length of the large square be x.
Calculating x.
25=x²+64-(16xcos20)
Therefore;
x = 11.70244 cm.
It implies;
Area large sq...
Sir Mike Ambrose is the author of the question.
Let the side length of the regular pentagon be 2 units.
a² = 2²-1²
a = √(3) units.
b² = 5-4cos108
b = 2.49721204096 units.
(2.49721204096/...
