Mathematics Question and Solution
Notice.
The ascribed right-angled triangle is not drawn to scale.
Let a be the side length of the inscribed square.
Calculating Area Inscribed Square.
tanb = 3√(3)/(9√(3))
b = atan(⅓)°
tanb = a/c
tan(atan(⅓)) = a/c
c = 3a units.
d = 9√(3)-c-a
d = (9√(3)-4a) units.
e = (3√(3)-a) units.
Calculating a.
(3√(3))² = d²+e²
27 = (9√(3)-4a)²+(3√(3)-a)²
27 = 243-72√(3)a+16a²+27-6√(3)a+a²
17a²-78√(3)a+243 = 0
a²-(78√(3)/(2*17))a = -243/17
(a-(39√(3)/17))² = (-243/17)+(-39√(3)/17))²
(a-(39√(3)/17))² = (432/289)
a = (39√(3)/17)±√(432/289)
a = (39√(3)/17)±(12√(3)/17)
It implies;
a ≠ (39√(3)/17)+(12√(3)/17)
a = (39√(3)/17)-(12√(3)/17)
a = (27√(3)/17) units
Again, a is the side length of the inscribed square.
Therefore, area inscribed square is;
a²
= (27√(3)/17)²
= 2187/289 square units.
= 7.56747404844 square units in decimal.
