Notice.
3 units is the radius of the circle.
6² = (6-2a)²+(6-a)²
36 = 36-24a+4a²+36-12a+a²
5a²-36a+36 = 0
5a²-30a-6a+36 = 0
5a(a-6)-6(a-6) = 0
(5a-6)(a-6) = 0
It implies;
a ≠ 6
a = ⅕(6)...
Calculating area of the rectangle.
a² = 5²+2²
a = √(29) units.
a is the width of the rectangle.
b = 2+6
b = 8 units.
Observing similar plane shape (right-angled) side length ratios to g...
Calculating r, radius of the circle.
a = 6+4
a = 10 units.
a is AD = AC.
b²+6² = 10²
b = √(100-36)
b = √(64)
b = 8 units.
c = b-r
c = (8-r) units.
Therefore, r, radius of the circle i...
Calculating R, radius of the inscribed quarter circle (side length of the square).
a = (R-3) units.
b = (R+1) units.
Therefore;
(R+1)² = R²+c²
c² = R²+2R+1-R²
c = √(2R+1) units.
d...
Calculating x.
Notice.
x is the radius of the big inscribed half circle.
Let y be the radius of the small inscribed half circle.
It implies;
2x+2y = 16
2y = 16-2x
y = (8-x) units.
A...
a = 12+1
a = 13 units.
a is the radius of the quarter circle.
13² = 12²+b²
b² = 169-144
b = √(25)
b = 5 units.
R is the radius of the inscribed circle.
c = b+R
c = (5+R) units.
d...
Sir Mike Ambrose is the author of the question.
The equation of the curve is;
y = ½(3x²)+4
Point P is;
P(4, 28)
Calculating OQ (x)
y = ½(3x²)+4
dy/dx = 3x, and x = 4.
Therefore;
dy...
Sir Mike Ambrose is the author of the question.
Let the two equal lengths of the isosceles triangle be 2 units.
Therefore;
Area triangle ABC is
= 2sin36 square units.
= 1.17557050458 squ...
Sir Mike Ambrose is the author of the question.
Let the single side length of the regular heptagon be 1 unit.
Therefore;
Area Regular Heptagon is;
7/(4tan(180/7))
= 3.633912444 square un...
Calculating a.
b² = a²+1
b = √(a²+1) units.
b is AC.
Therefore;
a - 2a
1 - √(a²+1)
Cross Multiply.
a√(a²+1) = 2a
√(a²+1) = 2
a²+1 = 4
a² = 3
a = √(3) units.
a is BC.
Therefo...
