Let AG be y.
Calculating y.
a² = 8²+y²
a = √(y²+64) cm.
a is AD.
It implies;
Considering area ADG including the inscribed circle with radius 1 cm.
½(1*8)+½(1*y)+½(1*√(y²+64)) = ½(8*y)
8+y+√(y²...
Let the side length of the regular pentagon be 1 unit.
tan36 = a/0.5
a = 0.363271264 units.
b² = 0.363271264²+0.5²
b = 0.61803398875 units.
tan72 = c/0.5
c = 1.53884176859 units.
Are...
Let the single side length of the square be x.
Therefore;
Area rectangle of length x unit and width 1 cm + area triangle of height x cm and width (x-2) cm + 2(area triangle of height (x-1) cm...
Let a be the side length of the regular hexagon.
3² = 1²+a²-2acosb
2acosb = a²-8
cosb = (a²-8)/(2a) --- (1).
c²+(a²-8)² = (2a)²
c² = 4a²-(a⁴-16a²+64)
c = √(-a⁴+20a²-64) units.
sinb = c...
Let the radius of the semi circle be 3 units.
Calculating area Red.
Area red will be;
2(area triangle of two length 3 units and 1 unit, and angle (cos–¹(⅓)), angle they form) + area sector...
a = 180-75-27
a = 78°
b = 1+½(√(5)+1)
b = 2.61803398875 units.
(2.61803398875/sin78) = (c/sin27)
c = 1.21511575349 units.
sin75 = d/1.21511575349
d = 1.17371168822 units.
cos75 = e...
Sir Mike Ambrose is the author of the question.
Area W to 3 decimal places is;
Area triangle of two length 5 units and 7.5 units, and angle 86.4166783015° - Area triangle of two length 5 units...
Sir Mike Ambrose is the author of the question.
Let the length of the inscribed rectangle be 2 units.
Let the width of the inscribed rectangle be √(3) units.
Therefore the single side length...
Let r be the radius of the quarter circle.
Let x be the interior angle facing 1 cm.
a = (x+45)°
a is alternate to y, the angle formed by the meet point of r and 2 cm at the circumference of...
Let a be the radius of the inscribed half circle.
b²+8² = (2a)²
b = √(4a²-64) units.
It implies;
Calculating a.
2a ~ 8
a ~ √(4a²-64)
Cross Multiply.
8a = 2a√(4a²-64)
4 = √(4a²-...
