Let the radius of the inscribed circle be 2 unit.
Therefore;
AB = 8 units.
BC = 6 units.
Area triangle ABC is;
½*6*8
= 24 square units.
It implies;
OG = ⅓(10) units.
GE =...
Sir Mike Ambrose is the author of the question.
Area orange exactly in square units is;
Area triangle with height ⅓(10√(2)) units and base 5sin45 units - Area triangle with height (8/3) units a...
Let the side of the regular pentagon be 1 unit.
a = 360-2(108)-90
a = 54°
tan72 = b/0.5
b = 1.53884176859 units.
tan54 = 1.53884176859/c
c = 1.11803398875 units.
d = c-0.5
d = 0.618...
(x/√(3))+x+2x+3x+(3x/√(3)) = 1
(4√(3)x/3)+6x = 1
(4√(3)x)+18x=3
(4√(3)+18)x = 3
x = 3/(4√(3)+18)
x = 3(18-4√(3))/(324-48)
x = 3(15-4√(3))/276
x = (18-4√(3))/92
x = (9-2√(3))...
Sir Mike Ambrose is the author of the question.
Area yellow in square units to 2 decimal places is;
Area triangle with height (tan70+tan(450/7)) units and base (tan70+tan54)sin80 units.
= ½(...
Sir Mike Ambrose is the author of the question.
At t = 2, x = 3 and y = 3/4
dx/dt = 2/3
dy/dt = 13/16
Therefore;
Gradient at the tangent and of the curve at t = 2 is;
dy/dx = dy/dt÷...
Sir Mike Ambrose is the author of the question.
At x = 0, y = 2.
At y = 16, x = 1.
Gradient of the curve at x = 1 or at the tangent is;
dy/dx = 3In2*2^(3x+1)
For x = 1
dy/dx = 48I...
#Sir Mike Ambrose is the author of the question.
Let the side length of the regular Pentagon be 1 unit.
Therefore Area Pentagon is;
(5*1)÷(4tan(180/5))
= 5/(4tan36) space units.
= 1.72047740...
Sir Mike Ambrose is the author of the question.
Let the side length of the square be 1 unit.
Therefore;
Area square is;
1² = 1 square units.
It implies;
Area R is;
Area square -...
Sir Mike Ambrose is the author of the question.
Let the side length of the be 2 units.
Therefore;
BC = 2 units.
It implies;
Area orange is;
½(1*2) = 1 square units.
Area green is;...
