Mathematics Question and Solution
Considering the quadrilateral cyclic.
a² = 2*10²-2*10²cosx
a² = 200-200cosx --- (1).
a² = 9²+7²-2*9*7cosx
a² = 130+126cosx --- (2).
Equating (1) and (2).
a² = a²
200-200cosx = 130+126cosx
200-130 = 126cosx+200cosx
326cosx = 70
x = acos(70/326)
x = 77.6006688181°
y = 180-x
y = 102.3993311819°
Therefore, area cyclic quadrilateral is;
Area triangle with height 10 units and base 10sin77.6006688181 units + Area triangle with height 9 units and base 7sin102.3993311819 units.
= ½*10*10sin77.6006688181+½*9*7sin102.3993311819
= 24√(11) square units.
= 79.5989949685 square units.
≈ 79.60 square units to 2 decimal places.
