Let the square side be 2a cm.
It implies;
a is the radius of the inscribed semi circle.
Let b equal the radius of the big inscribed quarter circle.
Therefore;
(a+b)² = a²+(2a)²
a²+2ab...
Let the two equal lengths of the isosceles triangle be 1 unit each.
a = 180-2(12+36)
a = 180-96
a = 84°
b² = 2-2cos84
b = 1.3382612127 units.
c = 180-12-30
c = 138°
(d/sin30) = (1.3...
Let the side length of the regular pentagon be 1 unit.
a² = 2-2cos108
108° is the single interior angle of the regular pentagon.
a = 1.6180339887 units.
a is BE.
b = 108-60
b = 48°
c =...
a² = 3²+1²
a = √(10) units.
a is AB.
It implies;
AC = ½√(10) units.
tanb = 1/3
b = atan(1/3)°
tanc = ½√(10)/√(10)
c = atan(½)°
d² = (½√(10))²+√(10)²
d² = ½(5)+10
d = √(25/2)
d...
x+50+180-a = 180
a = x+50 --- (1).
x+30+180-b = 180
b = x+30 --- (2).
3x+180-2a+180-2b = 180
3x+180-2(a+b) = 0
2(a+b) = 3x+180 --- (3).
Therefore, substituting (1) and (2) in (3) to get...
tan30 = a/12
a = 12tan30
a = 4√(3) cm.
a is the side length of the square.
b = ½(12)+a
b = (6+4√(3)) cm.
c = ½(a)
c = 2√(3) cm.
Observing similar right-angled triangle side length rat...
Let x be the single interior angle of the regular polygon.
a = 180-x-20
a = (160-x)°
b = 180-x-30
b = (150-x)°
c = 180(5-2)
c = 540°
d = 2(360-x)°
It implies;.
a+b+d+86 = c
(16...
Notice, triangle ABC is isosceles.
Let BC, the base of triangle ABC = 1 unit.
a = ½(180-45)
a = ½(135)
a = 67.5°
a is angle ABC = angle ACB.
b = a-45
b = 22.5°
b is angle DBE.
c =...
Area Blue is;
Area semi circle with radius √(2) units - Area quarter circle with radius 2 units + Area triangle with height and base 2 units respectively.
= ½(√(2)²)π - ¼(2²)π + ½(2²)
= π -...
Area Blue is;
Area triangle with height 4 cm and base (4sin150) cm
= ½*4*(4sin150)
= 2(4sin150)
= 8sin150
= ½(8)
= 4 cm²
