Çözüldü Right Triangle's Circumcircle - Trigonometry

https://i.ibb.co/QcZ1CVb/triangle.png
https://images.collegedunia.com/pub...52771489Math Level 1 SAT Practice Test 13.pdf
(The last question with no answer or solution)

Solution Without Using Pythagorean Theorem
∡QSR) = arctan[ 12 / (12·√3) ] = arctan(1 / √3) = 30°
Hypotenuse |QS| is the diameter of the QRS right triangle's circumcircle, so sin(30°) = 12 / |QS| and |QS| = 24 units
Radius of the circumcircle: r = |QS| / 2 = 24 / 2 = 12 units
Area of the circumcircle = π·(12^2) = 144·π unit^2.

Note: In the question, the phrase "QR is perpendicular to RS" is unnecessary because, in the literal notation, the middle letter of any right triangle (here it is R in the QRS right triangle) also indicates the angle at that vertex is 90°, thus it has to be that |QR| ⊥ |RS| by the standard definition.