Shaded Area is;
4(area quarter circle with radius 2 units - area triangle with height and base 2 units each) + 2(area square with side 2 √(2) units - area semi circle with radius 2 units)
= 4(¼*4π...
Let r be the radius of the ascribed semi circle.
It implies;
r² = 14²+a²
a² = r²-196
a = √(r²-196) cm.
a is OA.
b = a+r
b = (√(r²-196)+r) cm.
b is AB.
Therefore, calculating r.
½*14*(√(r²-196)+r...
Notice, the triangle is isosceles.
It implies;
½(12x+22)+(6y-5) = 90
6x+11+6y-5 = 90
6x+6y = 84
x+y = 14 --- (1).
½(12x+22)+(10y-41) = 90
6x+11+10y-41 = 90
6x+10y = 120
3x+5y = 60 --- (2).
Solvi...
Calculating Area Blue Plus Area Orange.
Radius of the congruent four semi circles is 1 unit each.
(a-1)²+1² = 2²
a²-2a+2 = 4
a²-2a-2 = 0
Resolving the above quadratic equation via completing the...
Calculating x, side length of the regular pentagon.
a = ⅕(180(5-2))
a = ⅕(180*3)
a = 108°
a is the single interior angle of the regular pentagon.
b = ½(180-108)
b = 36°
c = 108-36
c = 72°
d = 4...
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Let the radius of the red inscribed circle be r.
Calculating r.
2a² = 6²
a = √(18)
a = 3√(2) units.
2b² = 8²
b = √(32)
b = 4√(2) units.
c² = 6²+8²
c = √(36+64)
c = 10 units.
10²...
Calculating yellow area.
2² = 2a²-2a²cos120
Where a = regular hexagon side length.
120° = single interior angle of the regular hexagon.
It implies;
4 = 3a²
a = (2/√(3)) units.
a = 1.1547005384 u...
Calculating r, radius of the quarter circle.
a = (r-2) units.
Therefore;
r² = b²+(r-2)²
r²-(r²-4r+4) = b²
b² = 4r-4
b = √(4r-4) units.
c = (r-4) units.
Therefore;
r² = d²+(r-4)²
r²-(r²-8r+16) = d²...
Notice, the ascribed quadrilateral is regular (square).
Calculating Area Blue.
a² = 4²+8²
a = √(16+64)
a = √(80)
a = 4√(5) units.
b = ½(a)
b = ½(4√(5))
b = 2√(5) units.
tanc = 4/8
c...
