Mathematics Question and Solution
Let AB be 4 units.
BC = 6 units.
Let a be the side length of the inscribed square.
Calculating a.
b² = 4²-a²
b = √(16-a²)
It implies;
a = √(16-a²)
6 = 4
Cross Multiply.
4a = 6√(16-a²)
2a = 3√(16-a²)
4a² = 9(16-a²)
4a² = 144-9a²
13a² = 144
a = 12√(13)/13 units.
a = 3.32820117735 units.
c² = 4²-(12/√(13))²
c² = 16-(144/13)
c = √(64/13)
c = 8√(13)/13 units.
It implies, area red is;
0.5*(8√(13)/13)(12√(13)/13)
= 48/13 square units.
sind = (8√(13)/13)/4
d = asin(2√(13)/13)
e = (90-asin(2√(13)/13))°
f = (2asin(2√(13)/13))°
Area Green is;
0.5*3*3sin(2asin(2√(13)/13))
= 54/13 square units.
g = 20√(13)/13 units.
h = (90-asin(2√(13)/13)°
Therefore;
j² = (20√(13)/13)²+7²-2*7*(20√(13)/13)cos(90-asin(2√(13)/13)
j = 3√(53/13) units.
j = 3√(689)/13 units.
((3√(53/13))/sin(90-asin(2√(13)/13)) = (7/sink)
k = (asin(7√(53)/53))°
l = (90-asin(7√(53)/53))°
tan(90-asin(7√(53)/53)) = m/(12√(13)/3)
m = 24√(13)/91 units.
Area Blue is;
(12√(13)/13)²-0.5(12√(13)/13)(24√(13)/91)
= (144/13)-(144/91)
= (1008-144)/91
= 864/91 square units.
It implies;
Area Red : Area Green : Area Blue is;
(48/13) : (54/13) : (864/91)
= 336 : 378 : 864
= 168 : 189 : 432
= 56 : 63 : 144
