01 Area of a right triangle of known median bisecting the hypotenuse

By cosine law:

$a^2 = 3^2 + x^2 - 2(3x) \cos 60^\circ$

$a^2 = 9 + x^2 - 3x$
 

$b^2 = 3^2 + x^2 - 2(3x) \cos 120^\circ$

$b^2 = 9 + x^2 + 3x$

 

 

By Pythagorean Theorem:

$(2x)^2 = a^2 + b^2$

$4x^2 = (9 + x2 - 3x) + (9 + x2 + 3x)$

$2x^2 = 18$

$x = 3 ~ \text{cm}$

 

Area of the triangle
$A = \frac{1}{2}(3x) \sin 60^\circ + \frac{1}{2}(3x) \sin 120^\circ$

$A = \frac{1}{2}(3)(3) \sin 60^\circ + \frac{1}{2}(3)(3) \sin 120^\circ$

$A = 7.79 ~ \text{cm}^2$   ←   answer