Mathematics Question and Solution
a ~ 6
6 ~ 8
Cross Multiply.
8a = 6*6
2a = 9
a = ½(9) units.
b = a+8
b = ½(9)+8
b = ½(25) units.
b is the diameter of the ascribed half circle.
c = ½(b)
c = ¼(25) units.
c is the radius of the ascribed half circle.
d = c-a
d = ¼(25)-½(9)
d = ¼(7) units.
Let e be the radius of the inscribed circle.
f = e-d
f = e-¼(7)
f = ¼(4e-7) units.
g = c-e
g = ¼(25)-e
g = ¼(25-4e) units.
Calculating e.
g² = e²+f²
(¼(25-4e))² = e²+(¼(4e-7))²
(625-200e+16e²)/16 = e²+(16e²-56e+49)/16
625-200e+16e² = 16e²+16e²-56e+49
16e²+144e-576 = 0
2e²+18e-72 = 0
e²+9e-36 = 0
e²+12e-3e-36 = 0
e(e+12)-3(e+12) = 0
(e+12)(e-3) = 0
Therefore;
e = 3 units.
Again, e is the radius of the inscribed circle.
Area inscribed circle is;
πe²
= π*3²
= 9π square units.
