Let BD = a
Notice!
BC = CD = BD = a
AC = 15 cm.
Half BD = ½(a) cm.
It implies;
sin60 = b/a
½√(3) = b/a
b = ½√(3)a cm.
Therefore;
½(a)+b = AC
½(a)+½√(3)a = 15
a+√(3)a = 30
a(1+√(3)) = 30
a = 10...
Let AC be 1 unit.
a = 180-12-15
a = 153°
a is angle ADC.
Observing Sine Rule.
(1/sin153) = (b/sin12)
b = 0.4579648493 units.
b is CD.
c = 180-12-18-15-33
c = 180-78
c = 102°
c is an...
Observing similar plane shape side lengths ratio.
3 - 5
a - (4-a)
Cross Multiply.
3(4-a) = 5a
12-3a = 5a
8a = 12
a = 3/2 units.
a is the base of the right-angled triangle with height 3 uni...
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...
Area green is;
(Area triangle with height 4 units and base 8√(3) units - Area sector with radius 8 units and angle 30°) + (Area sector with radius 8 units and angle 60° - Area triangle with heig...
Let x³ be a.
Therefore;
a²-a-2 = 0
Resolving the above quadratic equation via completing the square approach.
(a-(1/2))² = 2+(1/2)²
(a-(1/2))² = (9/4)
a-(1/2) = ±(3/2)
a = ½±(3/2)
It implies;
a =...
a = ¼(7*7)π-½(7*7)
a = ¼(49π)-½(49)
a = ¼(49π-98) square units.
b = 2a
b = 2*¼(49π-98)
b = ½(49π-98) square units.
It implies;
Area of red region is;
Half area square with side 14 units - b
= ½...
Let the radius of the small yellow inscribed circle be a.
Let the radius of the big green inscribed circle be b.
Notice!
PQ = a+b
Calculating radius of the ascribed semi circle.
c² = 12²+16²
c =...
Let the side of the three inscribed congruent regular hexagon be 1 unit.
It implies, the side length of the square is;
a = 6 units.
b² = 2-2cos120
b = √(3) units.
c = 1+(½)
c = ½(3) units.
c = 1...
Radius of the semi circle r is;
r = ½(10)
r = 5 cm.
a = 1+5
a = 6 cm.
b = 5-1
b = 4 cm.
a² = b²+c²
c = √(6²-4²)
c = √(20)
c = 2√(5) cm.
cosd = 4/6
d = acos(2/3)
d = 48.189685104...
