Let O be the centre of the circle.
Since AE = EB, therefore E is the midpoint of length AB.
AB = 10cm, therefore AE = EB = 10/2 = 5cm.
r will be;
Triangle EBC Area+Triangle EOC Area+Triangle AOE...
Sir Mike Ambrose is the author of the question.
Let the square side be 1 unit.
Calculating Area Yellow.
Length CE is;
(CE/sin105)=(1/sin60)
CE = sin105/sin60
CE = (√(6)+√(2))/2√(3) units.
Area Ye...
Area yellow is;
Area circle with radius 4 cm - 3(area triangle with two side 4 cm each and angle 120°).
= 16π - ¼(3*16√(3))
= 16π-12√(3)
= 4(4π-3√(3)) cm²
Sir Mike Ambrose is the author of the question.
Let the square side be 2 units.
Area Square is;
2²
= 4 square units.
Calculating r, radius of the inscribed circle.
√(2)r+½√(2)r+½√(10)r = ½√(2)*√(2...
Let x be 2 unit.
Therefore;
Area square is;
8²
= 64 square unit.
Calculating area rectangle.
The length is;
√(8²+6²) + 2cos(atan(4/3))
= 10+(6/5)
= 56/5 unit.
The width is;
2sin(atan(4/3))
= 8...
Sir Mike Ambrose is the author of the question.
Let the side of the inscribed square be x.
Calculating x.
(10√(29)-√(29)x)/(5sin(45+atan½)) = (√(2)x)/(sin(90-atan½-atan(2/5))
It implies;
3√(5)x...
Sir Mike Ambrose is the author of the question.
Let the side length of the inscribed equilateral triangle ABD be x.
Calculating x.
(12√(2))² = (½√(3)x)² + ((12√(2))-½(12√(2)-x))²
It implies;...
Sir Mike Ambrose is the author of the question.
Let the square side be x.
Calculating x.
25 = (x²-9)+2x²-2√(2)x*√(x²-9)cos(45+y)
Where y is an angle.
Notice;
cosy = √(x²-9)/x and siny = 3/x.
It...
100 = 49+64-2*7*8cosA
112cosA = 113-100
A = acos(13/112)
A = 83.3345727438°
(7/cosB) = (10/cos83.3345727438)
B = 44.0486256741°
(8/cosC) = (10/cos83.3345727438)
C = 52.6168015821°
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...
