Radius of the inscribed circle, r is;
2r² = 4
r² = 2
r = √(2) unit.
Therefore;
Area R in square unit to 3 d. p. is;
Area sector with radius √(2) unit and angle atan(0.5√(2))° - Area triangle with...
tan60 = a/10
a = 10√(3) cm.
a is the height of the equilateral triangle.
sin(atan(2)) = b/(10√(3))
b = 15.4919333848 cm.
b is a side length of the square.
It implies;
Area Square is;
b...
Let the base of ABCD be 1 unit.
a = 180-24-38-48
a = 180-110
a = 70°
b = 38+48
b = 86°
(1/sin70) = (c/sin86)
c = 1.0615854889 units.
d = 180-26-24-38
d = 180-88
d = 92°
(1/sin92) = (e/sin38)
e...
Let a be the side of the square.
cos30 = b/8
b = 4√(3) cm.
sin30 = c/8
c = 4 cm.
a = b+12
a = (12+4√(3)) cm.
Area Square is;
a²
= (12+4√(3))²
= 358.27687752661 cm²
d² = 12²+4²-2*4*12cos30
d =...
Sir Mike Ambrose is the author of the question.
Let the two equal side length of the ascribed isosceles triangle be 1 unit.
Therefore;
Area triangle ABC is;
½(1*1sin36)
= ½(sin36) space unit.
= 0.2...
Area green is;
Area square with side 6 unit + 4(Area sector with radius 6 unit and angle 60° - Area equilateral triangle with side 6 unit) - Area square with diagonal (6√(3)-6) unit - 4(area secto...
6a = 4*17
a = ⅓(34) units.
b = ½(4+17)
b = ½(21) units.
c = b-4
c = ½(21)-4
c = ½(21-8)
c = ½(13) units.
d = ½(a+6)
d = ½(⅓(34)+6)
d = ½*⅓(34+18)
d = ½*⅓(52)
d = ⅓(26) units.
It implies, radius...
Let the square side be a.
2(a-14)² = 14²
(a-14)² = 98
a = 14+7√(2) units.
Green area length and width is;
Length = 14+7√(2)-18
= 7√(2)-4 units.
Width = 14+7√(2)-10
= 4+7√(2) units.
= (4+7√(2))(7√...
Let AB be 2 unit.
Therefore CD = 1 unit.
It implies;
Area Blue is;
Area triangle with height 0.72111025509 unit and base sin(atan(3/2)) unit.
= 0.5*0.72111025509sin(atan(3/2))
= 3/10 space unit....
Let the square side be x.
x²+(x-4)² = a²
a = √(2x²-8x+16)
b² = 2²+(x-2)²
b = √(x²-4x+8)
It implies;
(½(a))²+6² = b² --- (1).
And;
a = √(2x²-8x+16)
b = √(x²-4x+8)
It implies, subs...
