Calculating AQ using similar plane shapes (triangle) rule.
12 = 6
6 = a cross multiply.
12a = 36
a = 3 cm.
Where a is AQ.
PQ = √(6²+3²)
PQ =√(45)
PQ = 3√(5) cm.
BQ = 12-3
BQ = 9 cm.
Calculati...
Screenshot of today's teaching and learning activities with some of our amazing, smart and clever learners;
Ogheneruona (year 10), Oghenevwarhe (year 8), Manasseh (year 8) and Oghenemano (year 8).
E...
a = atan(2)°
b = 180-2a
b = (180-2atan(2))°
Where b is angle FBP.
Notice;
CQ = 5 cm.
Calculating FP using sine rule.
Let FP be c.
(c/sin(180-2atan(2))) = (5/sin(atan(2)))
c = 4.472135955 cm....
Let the AB be 4 cm.
Therefore, radius of the inscribed semi circle is;
radius = 2 cm.
Let the radius of the inscribed circle be r.
Calculating r.
(2+r)² = (4-r)²+(2-r)²
4+4r+r² = 16-8r+r²+4-4r+r²...
Let the inscribed blue quarter circle radius be r.
Calculating r.
Considering similar triangles rule.
r = 4
8 = r (cross multiply).
r² = 32
r = 4√(2) cm.
Therefore;
Area blue (area quarter...
Calculating the side length of square ABCD.
2x² = 8²
x² = 32
x = 4√(2) units.
Where x is AB.
Let BG = y
Semi circle radius r is;
r = ½(4√(2)+y) units.
Calculating y.
sin45 = ½(4√(2)+y)/...
Screenshot of yesterday's teaching and learning activities with some of our high flyers, smart and clever learners;
Arithmetical Oghenevwarhe and Mathematical Ogheneruona, we are super proud of you....
a² = 10²+10²-2*10*10cos150
a = 19.31851652578 cm.
Where a is AF which is equal AE.
(19.31851652578/sin150) = (10/sinb)
b = 15°
Where b is angle BAF.
tan15 = c/10
c = 2.67949192431 cm.
It implies;...
Let the small inscribed square side be a.
Let b be the side of the big inscribed square.
It implies;
b = 10-a
Therefore;
Calculating a.
tan(90-75) = a/(10-a)
a(1+tan15) = 10tan15
a = (10tan15)...
1. “Pure mathematics is, in its way, the poetry of logical ideas.” —Albert Einstein
2. “Without mathematics, there is nothing you can do. Everything around you is mathematics. Everything around yo...
