Shaded blue area will be;
5(area square of single side length 2 cm) - area semi circle of radius 2 cm.
=5(2x2) - ½(4π)
= 20-2π
= 2(10-π) cm²
Green area will be;
½(area quarter circle with radius (2+√(2)) unit) - area isosceles right-angled triangle with height and base √(2) unit respectively - area sector with radius √(2) unit and angl...
Sir Beg is the author of the question.
Let the square side length be 1 unit.
1² = 2a²
a = √(1/2)
a = ½√(2) units.
a is AD = CD.
Let the radius of the inscribed circle be b.
Therefore;...
Sir Beg is the author of the question.
Let AB be a cm, the side length of the ascribed equilateral triangle ABC.
b² = 9
b = 3 cm.
b is the side length of the inscribed square and the base of t...
Let the single side length of the square be x.
Therefore;
½(x*(3x/4))+½(4√(x²+(3x/4)²)+½(8x)+½(¼(x)(x-4))=x²
(3x²)/8+2√(x²+(9x²/16))+4x+(x²/8)-(x/2)=x²
3x²+16√(x²+(9x²/16))+32x+x²-4x=8x²
1...
Exact enclosed area R will be;
Area sector with radius 2 unit and angle (90-2tan–¹(½)) - Area sector with radius 1 unit and angle (180-4tan–¹(½)) + Area isosceles triangle with two equal lengths...
tan30 = 6/a
a = 6√(3) units.
a is AB.
cos30 = 6/b
b = 4√(3) units.
b is CR.
c = a-b
c = 2√(3) units.
c is DR.
d = 180-30
d = 150°
d is angle DRQ.
It implies, shaded area is;...
a² = 6²+6²-2*6*6cos120
a = 6√(3) units.
tanb = 3/(6√(3))
b = atan(2/(2√(3)))
b = 16.102113752°
c = 90-16.102113752
c = 73.897886248°
d = 30+16.102113752
d = 46.102113752°
It implie...
Let the radius of the inscribed circle be a.
tanb = 4/3
b = atan(4/3)°
c = ½(b)
c = ½(atan(4/3))°
d = (3-a) units.
It implies;
tan(½(atan(4/3))) = a/(3-a)
½ = a/(3-a)
2a = 3-a
a...
Let the inscribed square side length be a.
2² = 1²+b²
b² = 4-1
b = √(3) units.
c = 2b+a
c = (a+2√(3)) units.
c is BC.
d = (a+1) units.
sin60 = (a+1)/e
½√(3) = (a+1)/e
√(3)e = 2a+2...
