Mathematics Question and Solution
Let r be the radius of the ascribed semi circle.
Therefore;
½(r) is the radius of the inscribed circle.
(2r)² = 2a²
a² = 2r²
a = √(2)r cm.
b = ½(a)
b = ½(√(2)r) cm.
c² = a²+b²
c² = (√(2)r)²+(½(√(2)r))²
c² = 2r²+½(r²)
c = √(½(5r²))
c = √(2.5)r cm.
d = 2+c
d = (2+√(2.5)r) cm.
b² = e²+2²
(½(√(2)r)²-2² = e²
e² = ½(r)²-4
e = √(0.5r²-4) cm.
It implies;
Calculating r.
(2r)² = d²+e²
4r² = (2+√(2.5)r)²+√(0.5r²-4)²
4r² = 4+4√(2.5)r+2.5r²+0.5r²-4
r² = 4√(2.5)r
r = 0
Or
r = 4√(2.5) units.
r = 6.3245553203 cm.
It implies;
d = (2+√(2.5)r)
And r = 6.3245553203 cm.
d = 2+√(2.5)*6.3245553203
d = 12 cm.
e = √(0.5r²-4)
Again, r = 6.3245553203 cm.
e = √(0.5*6.3245553203²-4)
e = 4 cm.
f² = e²+d²
f² = 4²+12²
f = √(16+144)
f = √(160)
f = 4√(10) cm.
f is the diameter of the ascribed semi circle.
c = √(2.5)r
And again, r = 6.3245553203 units.
c = √(2.5)*6.3245553203
c = 10 cm.
tang = e/d
tang = 4/12
tang = (⅓)
g = atan(⅓)°
It implies;
Area of the inscribed shaded region (Area pink triangle) is;
Area triangle with height and base 4√(10) cm and 10sin(atan(⅓)) cm respectively.
= 0.5*4√(10)*10sin(atan(⅓))
= 20√(10)*(1/√(10))
= 20 cm²
