Calculating x.
a²+(x-3)² = 10²
a² = 100-x²+6x-9
a² = 91+6x-x²
a = √(91+6x-x²) cm.
It implies;
(a+x)² = (x-3)²+(x+4)²
(√(91+6x-x²)+x)² = x²-6x+9+x²+8x+16
91+6x-x²+2x√(91+6x-x²)+x² = 2x²+2x+25
2x√(...
Calculating Length MN.
Let BD = a.
b = 180-30-75
b = 75°
b is angle BDC.
Notice!
Area ABCD = 10 square units.
Calculating a.
It implies
0.5*3*a*sin60 + 0.5*a²sin30 = 10
¼(3√(...
Notice!
The composite plane shape consist of two half circles with equal diameter, 10 units each.
Radius = 5 units each.
Calculating shaded area.
a = 180-2*30
a = 120°
Shaded Area i...
Let the single side length of the equilateral triangle be 4 unit.
Therefore;
Shaded area ÷ Area circle exactly in decimal is;
(Area sector with radius √(3) unit and angle (180-2(sin–¹(sin60...
sin15 = a/6
a = 1.5529142706 units.
a is CD.
cos15 = b/6
b = 5.7955549577 units.
b is BD.
c² = 4²+6²-2*4*6cos30
c = 3.2296719057 units.
c is AC.
(3.2296719057/sin30) = (4/sind)
d =...
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...
