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...
Let a be the side length of the square and the diameter of the biggest inscribed semi circle.
Let b = diameter of the bigger inscribed semi circle.
Let c = diameter of the big inscribed semi circl...
a = (√(3)-1) units.
b = (1-c) units.
Calculating c.
Observing similar plane shape (right-angled triangle) side length ratios.
(1-c) - (√(3)-1)
(√(3)-1) - 1
Cross Multiply.
1-c = (√(3)-1)²
c = 1-...
Let the square side length be 2 units.
It implies;
R = 1 unit.
a = (2+2x) units.
a is diameter of the bigger circle.
b = ½(a)
b = ½(2+2x)
b = (1+x) units.
b is the radius of the bigger circle.
c...
a = (x+17) cm.
a is the length of the the ascribed rectangle.
b = a-9
b = (x+17)-9
b = (x+8) cm.
b is the width of the ascribed rectangle and also, the diameter of the inscribed semi circle...
a² = 2²+2²-2*2*2cos120
a = √(12)
a = 2√(3) units.
a is the side length of each of the congruent 4 inscribed regular hexagon.
Calculating the side length of the ascribed regular hexagon.
sin30 = c/...
