Mathematics Question and Solution
a² = 3²+4²
a = √(25)
a = 5 units.
2b² = 5²
b² = (25/2)
b = √(25/2)
b = ½(5√(2)) units.
b is the radius of the semi circle.
c = 90+45
c = 135°
(½(5√(2))/sin135) = (3/sind)
d = 36.8698976462°
e = 180-135-36.8698976462
e = 8.1301023538°
(f/sin8.1301023538) = (0.5*5√(2)/sin135)
f = 0.7071067812 units.
g² = 2r²
g = √(2)r units.
r is the radius of the inscribed circle.
Calculating r.
(0.5(5√(2))-r)² = (√(2)r)²+0.7071067812²
12.5-5√(2)r+r² = 2r²+0.5
r²+5√(2)r-12 = 0
Resolving the quadratic equation.
Calculating r, radius of the inscribed circle via completing the square approach.
(r+(5√(2)/2))² = 12+(5√(2)/2)²
(r+(5√(2)/2))² = 12+(50/4)
(r+(5√(2)/2))² = ¼(48+50)
(r+(5√(2)/2))² = ¼(98)
r = -(5√(2)/2)±√(98/4)
r = -(5√(2)/2)±(7√(2)/2)
r = ½(-5√(2)±7√(2))
It implies;
r ≠ ½(-5√(2)-7√(2))
r = ½(-5√(2)+7√(2))
r = ½(2√(2))
r = √(2) units.
Again, r is the radius of the inscribed circle.
Area inscribed circle is;
πr²
= π*√(2)²
= 2π square units.
