Notice.
R-r = 10 cm.
It implies;
R = (10+r) cm.
R is the side length of the ascribed square.
Calculating length AB.
a = R-r
a = 10 cm.
b = ½(R) units.
It implies;
R² = 10²+...
Sir Mike Ambrose is the author of the question.
Calculating Yellow Area.
It is;
Area triangle with height 4sin(atan(⅔)) units and base 8√(13)/9 units - Area triangle with height 1 unit and b...
Sir Mike Ambrose is the author of the question.
Let the square aide length be 1 unit.
Square Area is;
1²
= 1 square unit.
Calculating Green Area.
a² = 1²+1²
a = √(2) units.
a is AB....
Let a be the side length of the big inscribed square.
Let b be the side length of the bigger inscribed square.
Therefore;
27 ~ b
a ~ 64
Cross Multiply.
ab = 27*64
b = (1728)/a unit...
Sir Mike Ambrose is the author of the question.
Let a be the equal lengths inscribing triangle ABC.
Calculating a.
½(a)² = 15
a = √(30) units.
b = 180-60-45
b = 75°
b is angle BAC....
Let R=the bigger inscribed circle radius.
Let r=the smaller inscribed circle radius.
It implies;
R+r=6√(3)-6=4.4
R+r=4.4 -------(1).
And the radius of the ascribed semi circle is 6...
Notice.
Radius rod the half circle is 5 units.
a = (5-x) units.
Calculating x.
5² = 3²+(5-x)²
16 = 25-10x+x²
x²-10x+9 = 0
x = 1 unit.
It implies;
a = 5-x
And x = 1 unit.
a = 4...
a = (2r-14) units.
b = ½(a)
b = (r-7) units.
Calculating r, radius of the half circle.
Therefore;
2r ~ 6
6 ~ (r-7)
Cross Multiply.
2r(r-7) = 36
r²-7r-18 = 0
Resolving the qu...
Sir Mike Ambrose is the author of the question.
Equation of the curve is;
y = ⅔(8x-x²)
Calculating Area E.
(Area under the curve at x = (3/2) and x = 4) - Area trapezium with parallel sid...
