Mathematics Question and Solution
Calculating x.
Let a be the height of the inscribed isosceles triangle.
b = ½(7)+7
b = 10.5 units.
c² = 10.5²+a²
c = √(a²+110.25) units.
Let d be one of the two equal inscribed angles.
cosd = 10.5/√(a²+110.25) --- (1).
e² = 3.5²+a²
e = √(a²+12.25) units.
f = (10-√(a²+12.25)) units.
It implies;
(10-√(a²+12.25))² = √(a²+110.25)²+14²-2*14*√(a²+110.25)cosd
100-20√(a²+12.25)+a²+12.25 = a²+110.25+196-28√(a²+110.25)cosd
100-20√(a²+12.25) = 294-28√(a²+110.25)cosd
50-10√(a²+12.25) = 147-14√(a²+110.25)cosd
14√(a²+110.25)cosd = 97+10√(a²+12.25) --- (2).
Substituting (1) in (2).
14√(a²+110.25)*(10.5/√(a²+110.25)) = 97+10√(a²+12.25)
14*10.5 = 97+10√(a²+12.25)
50 = 10√(a²+12.25)
25 = a²+12.25
a² = 12.75
a = √(12.75) units.
a = 3.57071421427 units.
Again, a is the height of the inscribed isosceles triangle.
tang = a/(0.5*7)
g = atan(3.57071421427/3.5)
g = 45.5729959992°
Therefore, the required length, x is;
x² = 10²+14²-2*10*14cos45.5729959992
x = 10 units.
