Radius, r of one of the five congruent, inscribed circles is;
8r²+√(3)r-1=0
r = (√(35)-√(3))/16 m.
Therefore;
Total area of the five congruent, inscribed yellow circles is;
5πr²
= 5π( (√(35)-√(3...
Please, browse through the link below.
Screenshot slide video of our teaching and learning activities with our super smart and amazingly, very clever learners.
It is in memory of Late Sound Sultan (...
Online Edumath Educators and Learners are Super Smart and Amazingly, Very Clever.
Communicate us to mentor/teach your child/children Mathematics online at affordable tuition, helping them become a l...
Notice;
The semi circle has inscribed quadrilateral which is a cyclic quadrilateral when the semi circle is made a complete circle considering it's diameter.
Calculating Alpha.
Let it be a.
It i...
cos30 = r/(1-r)
Where r is the radius of each of the three congruent inscribed circles.
It implies;
2r = √(3) - √(3)r
r = √(3)/(2+√(3))
r = (2√(3)-3)/(1)
r = (2√(3)-3) unit.
r = 0.46410161514 uni...
Area blue is;
Area square with side 10 cm - Area triangle with height 5 cm and base 10 cm - Area triangle with height 10 cm and base (4.71404520791sin45) cm - Area triangle with height 5 cm and bas...
Calculating r, radius of the Semi Circle.
½(π)r² = 338π
r² = 676
r = 26 cm.
a = ½(48)
a = 24 cm.
Calculating x.
It implies;
Observing Pythagoras Rule.
26² = 24²+x²
x² = 676-576
x² = 100
x = 10...
a = ⅐*180(7-2)
a = ⅐(180*5)
a = ⅐(900)°
a is the single interior angle of the regular heptagon.
Calculating x and y.
b = 360-2*⅐(900)
b = (360/1)-(1800/7)
b = ⅐(2520-1800)
b = ⅐(720)°
c = ½(b)
c...
a = 10+94
a = 104°
a is angle BAC.
b = 24+34
b = 58°
b is angle ABC.
Let AB = 1 unit.
c = 180-58-104
c = 180-162
c = 18°
c is angle ACB.
It implies;
(1/sin18) = (d/sin104)
d =...
a² = 6²+8²
a = √(100)
a = 10 cm.
a is the hypotenuse of the ascribed right-angled triangle.
Let b be the hypotenuse of the blue inscribed right -angled triangle.
It implies, observing similar tria...
