Calculating length PQ.
tana = 12/5
a = atan(12/5)
a = 67.380135052°
b = 180-a
b = 180-67.380135052
b = 112.619864948°
c = ½(b)
c = ½(112.619864948)
c = 56.309932474°
tan56.309932474 = r/d
d = r/...
Sir Mike Ambrose is the author of the question.
Let AC = 2 units.
Therefore;
AB = ⅖ unit.
BC = (8/5) unit.
It implies;
Area Blue is;
Area triangle with height ⅖ unit and base 2sin120 units + Ar...
(7/sin60) = (8/sina)
a = 81.7867892983°
b = 120-a
b = 120-81.7867892983
b = 38.2132107017°
(7/sin60) = (c/sin38.2132107017)
c = 5 units.
c is the side length of the regular hexagon.
cos60 = c/d
½...
Observing similar plane shape (right-angled triangle) side length ratios.
3 - 5
a - (4-a)
Cross Multiply.
5a = 3(4-a)
5a = 12-3a
8a = 12
a = (3/2) units.
b = 4-a
b = 4-(3/2)
b = (5/2) units.
c...
a² = 2²+4²
a = √(20) cm.
a is radius OA = OB.
tanb = 4/2
b = atan(2)°
c = atan(½)°
d = 180-b-c
d = 180-atan(2)-atan(½)
d = 90°
d is the angle subtended by chord AB and the radius (OA and OB) of t...
Notice!
Radius of the circle is 4 cm.
a² = 4²+8²
a² = 16+64
a = √(80) cm.
a = 4√(5) cm.
a is length AC.
tanb = 2/4
b = atan(½)°
c = 180-2b
c = (180-2atan(½))°
It implies;
d² = 4²+4²-2*4*4cos(1...
Calculating x.
(2+x)² = (2x)²+x²
4+4x+x² = 5x²
4x²-4x-4 = 0
x²-x-1 = 0
(x-½)² = 1+¼
x = ½±½√(5)
It implies;
x ≠ ½(1-√(5))
x = ½(1+√(5)) units.
x = 1.6180339887 units.
Shaded Area is;
¼(x²)π
= ¼(½(...
Sir Mike Ambrose is the radius of the question.
Area R is;
Area triangle with height 4 cm and base (4tan52.5) cm - Area sector with angle 52.5° and radius 4 cm.
= ½*4(4tan52.5) - (52.5*4*4π)/360
=...
Let the inscribed bigger quarter circle radius be a.
Let the inscribed big quarter circle radius be b.
c = 5 cm.
c is the radius of the inscribed semi circle.
It implies;
a+b = 10 --- (1).
a+5...
