Mathematics Question and Solution
Let BD = a
Notice!
BC = CD = BD = a
AC = 15 cm.
Half BD = ½(a) cm.
It implies;
sin60 = b/a
½√(3) = b/a
b = ½√(3)a cm.
Therefore;
½(a)+b = AC
½(a)+½√(3)a = 15
a+√(3)a = 30
a(1+√(3)) = 30
a = 10.9807621135 cm.
½(a) is;
½(10.9807621135)
= 5.4903810568 cm.
It implies;
Area of the kite is;
½(AC)*(BD)
= ½*10.9807621135*15
= 82.3557158513 cm²
Therefore, half the area of the kite is;
½(82.3557158513)
= 41.1778579256 cm²
Calculating radius, r of the inscribed orange circle using half the kite's area.
It implies;
½(CD)r+½(AD)r = 41.1778579256
Calculating AD.
(AD)² = 2*5.4903810568
AD = 7.7645713531 cm.
Therefore;
½(10.9807621135)r+½(7.7645713531)r = 41.1778579256
10.9807621135r+7.7645713531r = 82.3557158513
18.7453334666r = 82.3557158513
r = 82.3557158513/18.7453334666
r = 4.3933982822 cm.
r = 4.4 cm to 1 decimal place.
Again, r is the radius of the inscribed orange circle.
