Let the length of the rectangle be (a+3) cm.
Let the width of the rectangle be (b+2) cm.
Therefore, Calculating a and b.
Notice!
The area of the two inscribed triangles and quadrilateral are equal...
a² = 85²-68²
a = 51 units.
a is the height of the ascribed triangle.
Calculating r, radius of the inscribed circle.
85r+51r+68r = 51*68
204r = 3468
r = 17 units.
b = 51-17
b = 34 units.
tanc = (...
Notice!
Length 3 units and length 6 units are both tangent to the green inscribed circle.
It implies;
Radius r, of the ascribed semi circle is;
r = ½(3+6)
r = ½(9)
r = 4.5 units.
Where 9 units is...
Let the side length of the regular pentagon be a.
b = ⅕(180(5-2))
b = 108°
b is the single interior angle of the regular pentagon.
c = 108-90
c = 18°
d = (180-2(18))
d = (180-36)
d = 144°
e = ½(...
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Let AE = CE = BE = b.
Let OE = a.
4² = a²+b²-2abcos45
16 = a²+b²-√(2)ab --- (1).
6² = a²+b²-2abcos135
36 = a²+b²+√(2)ab --- (2).
Solving (1) and (2) simultaneously.
Subtracting (1) from...
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Area of coloured region is;
Area triangle with height 53.3479501083 m and base 60sin(102.042575143) m + Area triangle with height 100.027406452 m and base √(5924)sin(12.5245961777) m
= (½*53.34795...
Observing Pythagoras Rule.
(n³)² = (n²)²+n²
n⁶ = n⁴+n²
n⁶-n⁴-n² = 0
n²(n⁴-n²-1) = 0
Let n² = p.
It implies;
p(p²-p-1) = 0
p²-p-1 = 0
Resolving the above quadratic equation, applying c...
Sir Mike Ambrose is the author of the question.
Let beta be x.
x = (½*atan(5/4))
Let alpha be y.
y = (½*atan(4/5))°
Therefore;
Area blue exactly in cm² decimal is;
Area triangle with hei...
