Sir Mike Ambrose is the author of the question.
Calculating a.
½((0, 6√(3)) (-3, 3√(3)) (a, (16a/3)) (0, 6√(3))) = ¼(9+20√(3))
Therefore;
a = x = ½√(3) units.
y = 16a/3 = (16/3)(√(3)/2...
Let the radius of the ascribed semi circle be 2 units.
Let the radius of the inscribed circle be r.
Calculating r.
(2-r)² = r² + 1
Therefore;
r = ¾ units.
It implies;
OH = r =...
Shaded area is;
Area triangle with height 4√(3) cm and base 8 cm + 2(area sector with radius 8 cm and angle 60° - area triangle with height 4√(3) cm and base 8 cm) - Area circle with radius 2 cm...
Sir Mike Ambrose is the author of the question.
Area green is;
½*3*6sin120
= ½(9√(3)) cm²
Radius, r of the circle is;
r = √(63)/√(3)
r = √(21) cm.
Therefore;
Area circle is;...
Sir Mike Ambrose is the author of the question.
Let the side of the large square be 2 units.
Therefore;
Area large square is;
2²
= 4 square units.
Area gold is;
Area triangle wit...
Total coloured area is;
2(area triangle with height 6.69991708075 cm and base √(65)sin(atan(10)-atan(7/4)) cm.
= 2*½*6.69991708075*√(65)sin(atan(10)-atan(7/4))
= 6.69991708075*3.2836227276...
Sir Mike Ambrose is the author of the question.
x = 16
The sequence is;
16, 12, 9, (27/4), ...
a = 16
r = ¾
Calculating n.
(2187/16383) = (¾)^(n)
n = 7.
Therefore;
T...
Shaded area is;
Area trapezoid with parallel sides 58 cm and 28 cm, and height 30 cm - Area trapezoid with parallel sides 9 cm and 30 cm, and height 28 cm - Area triangle with height 12 cm and b...
Sir Mike Ambrose is the author of the question.
Radius, r of the inscribed circle is;
r = (3-√(3)) cm.
Therefore;
Area purple exactly in square cm is;
Area trapezium with parallel side...
Sir Mike Ambrose is the author of the question.
P(2, (16√(5)/5))
Gradient of the curve at the tangent is;
= 24/5√(5)
Therefore;
Gradient of the curve at the normal is;
= -5√(5)/2...
