Screenshot of today's teaching and learning activities with some of our super smart and clever learners;
Oghosa (year 7), Adesuwa (year 5), Derick (year 6), Ogheneruemu (year 4) and Oghenezino (yea...
Radius of the inscribed circle, r is;
r = 2 cm.
Therefore;
Shaded Region is;
area sector with radius 2 cm and angle 120°- area triangle with two equal lengths 2 cm and the angle they for...
Screenshot of some of our numerous teaching and learning activities with our very talented, smart and clever learners.
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Let the side length of the square be 2 units.
Let the radius of the small inscribed circle be x.
Let the radius of the big inscribed circle be y.
Calculating y.
(2-y)² = y²+y²
2-y = √(2)y
2 = √(...
Calculating shaded area considering semi circle PQ.
Diameter of semi circle PQ will be 10cm, therefore radius is 5cm.
Shaded area will be;
Area semi circle PQ of radius 5cm and angle 180° -...
Let angle CAD = a
Let angle BAD = b
Let AD = c
tana = 3/c ----- (1).
tanb = 2/c ----- (2).
Adding (1) and (2).
Therefore;
tana+tanb = 5/c ----- (3).
Notice;
tan(a+b) = tana+tanb/(1-tanatanb)...
Screenshot of today's teaching and learning activities with some of our very talented, smart and clever learners;
Prince (year 4) and Oghenezino (year 4).
Educator: Mr. Oyoma
We are smart and clever...
Let the radius of one the the two congruent inscribed circles be a.
Therefore;
a = 2 cm.
tanb = (2/4)
b = atan(½)°
c = (180-2atan(½))°
c = 126.86989764584°
Shaded Region is;
Half area rectangl...
AC = √(12²+12²)
AC = 12√(2) cm.
AE = √(12²+6²)
AE = 6√(5) cm.
Angle BAE = atan(½)°
Angle CAE = 45-atan(1/2)
Angle CAE = 18.43494882292°
cos18.43494882292 = (AF)/12√(2)
AF = 16.099689438 cm.
EF...
Let AB = a.
Let BE = b.
It implies;
2a+b = 392
b = 392-2a
BC = ½(b-70)
BC = ½(392-2a-70)
BC = ½(322-2a)
Therefore;
BC = DE = 161-a
c = ½(70)+161-a
c = 196-a
Where c is half BE.
d² = a²-(196-a)²...
