Mathematics Question and Solution
Notice.
Radius of the semi circle is 5 cm.
Let the width of the blue rectangle be a.
Therefore;
Blue rectangle length is 2a.
b² = a²+a²
b = √(2a²)
b = √(2)a
c = ½(b)
c = ½(√(2)a)
d²+c² = 5²
d² = 25-(½(√(2)a))²
d² = 25-½(a²)
d = √(25-½(a²))
e = d-b
e = √(25-½(a²))-½(√(2)a)
r² = b²+e²
Where r is the radius of the semi circle (5 cm).
5² = (√(2)a)²+(√(25-½(a²))-½(√(2)a))²
25 = 2a²+25-½(a²)-a√(50-a²)+½(a)²
0 = 2a²-a√(50-a²)
a√(50-a²) = 2a²
√(50-a²) = 2a
50-a² = 4a²
50 = 5a²
a² = 10
a = √(10) cm.
Again, a is the width of the blue rectangle.
It's length is;
2a
= 2√(10) cm.
It implies;
Area Blue Rectangle is;
Length*width
= 2√(10)*√(10)
= 2*10
= 20 cm²
