Mathematics Question and Solution
Let c3 = a = 6 units.
Let c2 = b = 8 units.
Therefore;
c1 = c = √(6²+8²)
c = √(100)
c = 10 units.
Calculating Area S3.
It is;
Area equilateral triangle with side 6 units - 2(area equilateral triangle with side 3 units) - Area sector with radius 3 units and angle 60°
= ½*6*6sin60-2(½*3*3sin60)-(60π*3*3/360)
= 9√(3)-½(9√(3))-½(3π)
= ½(18√(3)-9√(3)-3π)
= ½(9√(3)-3π) square units.
Calculating Area S2.
It is;
Area equilateral triangle with side 8 units - 2(area equilateral triangle with side 4 units) - Area sector with radius 4 units and angle 60°
= ½*8*8sin60-2(½*4*4sin60)-(60π*4*4/360)
= 16√(3)-8√(3)-⅓(8π)
= ⅓(48√(3)-24√(3)-8π)
= ⅓(24√(3)-8π) square units.
Calculating Area S1.
It is;
Area equilateral triangle with side 10 units - 2(area equilateral triangle with side 5 units) - Area sector with radius 5 units and angle 60°
= ½*10*10sin60-2(½*5*5sin60)-(60π*5*5/360)
= 25√(3)-½(25√(3))-⅙(25π)
= ⅙(150√(3)-75√(3)-25π)
= ⅙(75√(3)-25π) square units.
Area S2 plus S3 is;
S2+S3 = ½(9√(3)-3π)+⅓(24√(3)-8π)
S2+S3 = ⅙(27√(3)-9π+48√(3)-16π)
S2+S3 = ⅙(75√(3)-25π) square units.
Notice!
S1 = ⅙(75√(3)-25π) square units.
It implies;
S1 = S2+S3
Proved!
