Mathematics Question and Solution
Notice.
R-r = 10 cm.
It implies;
R = (10+r) cm.
R is the side length of the ascribed square.
Calculating length AB.
a = R-r
a = 10 cm.
b = ½(R) units.
It implies;
R² = 10²+(½(R))²
R²-¼(R²) = 100
¼(3R²) = 100
3R² = 400
R = √(400/3)
R = ⅓(20√(3)) cm.
R = 11.5470053838 cm.
And R-r = 10 cm.
Therefore;
⅓(20√(3))-10 = r
r = ⅓(20√(3)-30) cm.
r = 1.5470053838 cm.
Recall.
b = ½(R) units.
b = ½(11.5470053838)
b = 5.7735026919 cm.
tanc = 10/5.7735026919
c = 60°
d = 90-c
d = 30°
tan(0.5*30) = e/11.5470053838
e = 11.5470053838tan(15)
e = 3.09401076759 cm.
f = R-e
f = 11.5470053838-3.09401076759
f = 8.45299461621 cm.
Therefore, the required length, length AB is;
cos30 = f/(AB)
AB = f/cos30
AB = 8.45299461621/cos30
AB = 9.76067743425 cm.
