7² = 3²+5²-2*3*5cosa
49 = 34-30cosa
cosa = (34-49)/30
a = acos(-15/30
a = 120°
a is angle ADC.
b = 180-a
b = 60°
b is angle ADB.
Therefore, the required angle, angle ABC is;
Let it...
Shaded Area exactly in square cm is;
Area rectangle with length 6 cm and width 2√(3) cm - Area triangle with height (10-4√(3)) cm and base (10-4√(3))sin60.
= 6*2√(3) - ½(10-4√(3))²sin60
= 12...
Let the single side length of the inscribed equilateral triangle be 1 unit.
Therefore;
Green Area is;
¼(2√(3)+2-π) square unit.
Large Semi Circle Area is;
½(π(1+(√(3)/2))²
= (π(7+4√...
Calculating angle x.
Let the hypotenuse of the inscribed right-angled triangle of the ascribed quadrilateral be 1 unit.
2a² = 1²
a = √(1/2) units.
a = 0.7071067812 units.
c = (b+0.707106...
Let a be the white inscribed circle radius.
a² = (0.5*8)²+b²
b = √(a²-16) units.
Let c be the radius of the ascribed.
c² = (0.5*8+4)²+b²
c² = 64+√(a²-16)²
c² = 64-16+a²
c = √(a²+48) un...
a²+8² = 12²
a = √(144-64)
a = √(80)
a = 4√(5) units.
tanb = 8/(4√(5))
b = atan(2/√(5))°
c = 90-b
c = atan(√(5)/2)°
cos(atan(√(5)/2)) = 4√(5)/d
d = 13.416407865 units.
Therefore,...
Let a be the radius of the half circle.
(2a)² = 8²+3²
4a² = 73
a = ½√(73) units.
b = 90-atan(3/8)
b = atan(8/3)°
Therefore, area blue triangle is;
0.5*½√(73)*(8+3)sin(atan(8/3))
= 0...
a = ½(180-45)
a = 67.5°
sin67.5 = (0.5*18)/b
b = 9.7415298026 units.
b is the radius of the half circle and the width of the rectangle.
2c² = b²
2c² = 9.7415298026²
c = 6.8883017826 unit...
a*6 = 12*8
a = 16 cm.
a is DM.
b² = 16²+12²
b = 20 cm.
b is BD.
c = ½(16+6)
c = 11 cm.
d = ½(8+12)
d = 10 cm.
e = d-8
e = 2 cm.
f² = 11²+2²
f = 5√(5) cm.
f is OD, radius of...
2a² = 2²
a² = 2
a = √(2) units.
a is the radius of the inscribed blue quarter circle.
Calculating R, radius of the inscribed circle.
b = (√(2)+R) units.
c = (√(2)-R) units.
It implie...
