Shaded area is;
Area under the curve at points (-1, 0) and (7, 0) - Area triangle with height 4 unit and base 8 unit.
= ⅓(208) - ½(4*8)
= ⅓(160) space unit.
Area S is;
Area under the curve at p...
Let the side of the regular hexagon be 1 unit.
Therefore;
Area blue is;
2(area triangle with height 1 unit and base (⅓√(3)) unit).
= 2(½*1*(⅓√(3)))
= ⅓√(3) space unit.
Area orange is;...
Area blue is:
Area triangle with height 1 cm and base (3sin(109.471220634)) cm.
= 0.5*1*3sin(109.471220634)
= 1.41421356238 cm²
= √(2) cm²
Let the radius of the inscribed white semi circle be r.
Therefore;
r²+(1+r)²=(2+r)²
r²+1+2r+r²=4+4r+r²
r²-2r-3 = 0
(r-1)² = 3+(-1)²
(r-1)² = 4
r = 1±2
r ≠ -1
r = 3 unit.
It implies;
Area blue is;...
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...
