Let AB be 1 unit.
Therefore;
BC = 2 units.
It implies, the regular hexagon side length is 3 units.
a = 120-108
a = 12°
b² = 2²+3²-2*2*3cos12
b = 1.1234895599 units.
(1.1234895599/sin12) = (2/s...
Sir Mike Ambrose is the author of the question.
a = 4/2tan(180/9)
a = 2/tan20 units.
b = 4/2tan(180/6)
b = 2/tan30 units.
Therefore;
Red Length is;
a+b
= (2/tan20)+(2/tan30)
= 2(tan...
Sir Mike Ambrose is the author of the question.
Let r be the radius of the three congruent circles.
Calculating r.
(9+r)² = (9-r)²+(9-3r)²
81+18r+r² = 81-18r+r²+81-54r+9r²
36r = 81-54r+9r²
9r²-90r...
a = 6-2
a = 4 units.
b² = 6²-4²
b = 2√(5) units.
c = 6-b
c = (6-2√(5)) units.
d = asin(⅔)°
tan(asin⅔) = e/(6-2√(5))
e = ⅕(12√(5)-20) units.
f = 4+e
f = ⅕(12√(5)) units.
g² = 6²+(⅕(12√(5)))²
g...
Let AB be 1 unit.
a² = 2-2cos108
a = 1.61803398875 units.
a is BE.
(1.61803398875/sin156) = (b/sin6)
b = 0.41582338164 units.
b is BP.
(1.61803398875/sin156) = (c/sin18)
c = 1.22929666779 units....
sin60 = a/20
a = 10√(3) cm.
a+b = 20
b = (20-10√(3)) cm.
c = 20-2b
c = 20-2(20-10√(3))
c = 20-40+20√(3)
c = (20√(3)-20) cm.
c = 14.64101615138 cm.
Therefore;
Shaded Area Red is;
½(14.641016151...
Notice;
Radius r of the circle is 6 units.
Area Red is;
6(area sector with radius 6 units and angle 6°-area triangle with base 6 units and height 6sin60 units)
= 6(60π*36/360 - ½*36sin60)
= 6(6π-...
Let the the side length of the inscribed hexagon side 2 units.
a² = 2-2cos120
a = √(3) units.
a = 1.73205080757 units.
a is the side length of the inscribed square.
Therefore;
Area Square is;
a²...
Our certified, experienced, resilient, productive and committed Math Educators will be glad to be at your service.
We make teaching and learning Mathematics fun for learners, we expose the simplicit...
Let AB be 2 units.
a = atan(½)°
b² = 2²+1²
b = √(5) units.
Let r be the radius of the inscribed circle.
Calculating r.
sin(atan½) = r/c
c = r/sin(atan½) units.
It implies;
1+r+(r/sin(atan½))...
