Notice!
AB = AC
It implies, triangle ABC is isosceles.
a = 180-45-30
a = 105°
a is angle ADC.
b = ½(180-45)
b = ½(135)
b = 67.5°
b = angle ABC = angle ACB
c = 67.5-45
c = 22.5°
c is angle DBE....
Area green is;
6(2(area sector with radius 4 units and angle 60°-area triangle with height 4 units and base 4sin60 units.))
= 6(2((60π*4*4/360)-(½*4*4sin60)))
= 6(2(⅓(8π)-4√(3)))
= 12(⅓(8π)-4√(3))...
Let a be the radius of the inscribed circle.
Let the circle's centre be X.
Calculating a.
(3-a)² = √(3)²+a²
9-6a+a² = 3+a²
6 = 6a
a = 1 unit.
a is the radius of the inscribed circle (XT = XN = XE...
Online Edumath Educators and Learners are Super Smart and Amazingly, Very Clever.
Communicate us to mentor/teach/educate your child/children Mathematics online at affordable tuition, helping them be...
Let R = 1 unit.
sin22.5 = r/(1-r)
r = sin22.5-rsin22.5
r+rsin22.5 = sin22.5
r(1+sin22.5) = sin22.5
r = sin22.5/(1+sin22.5) units.
r = 0.2767686539 units.
It implies;
R/r
= 1/(sin22.5/(1+sin22.5))...
Let BC be 1 unit.
a = 180-60-20-40-10
a = 50°
a is angle ACB.
Observing Sine Rule.
(b/sin50) = (1/sin(60+20))
b = 0.7778619134 units.
b is AB.
c = 180-40-60
c = 80°
c is angle ADB....
Notice.
Radius of the ascribed circle is 7 cm.
Calculating a, radius of the big inscribed circle.
(7-a)² = 2a²
49-14a+a² = 2a²
a²+14a-49 = 0
Resolving the above quadratic equation via completing...
Calculating area green.
tana = 4/8
a = atan(½)°
b = 2a
b = 2atan(½)°
tanc = 8/4
c = atan(2)°
d = 2c
d = 2atan(2)°
It implies, area green is,
2(area sector with radius 4 units and angle 2atan(2...
Calculating yellow area.
It is;
2(area quarter circle with radius 4 units - area triangle with base and height 4 units respectively - area quarter circle with radius 3 units + area triangle with b...
Notice.
The ascribed plane shape is a regular pentagon with side length 10 units.
Let a be the radius of the bigger inscribed circle.
Radius of the smaller pink inscribed circle is r.
c = ⅕(180(5-...
