Question

Let \(ABC\) be an acute-angled triangle with circumradius \(R\). Then \[ \sqrt{a^2b^2-4R^2} + \sqrt{b^2c^2-4R^2} + \sqrt{c^2a^2-4R^2} = \] is equal to:

• A. \(a+b+c\)
• B. \(a^2+b^2+c^2\)
• C. \(\dfrac{a^2+b^2+c^2}{2}\)
• D. \(2(a^2+b^2+c^2)\)

Hint:

For triangle problems involving the circumradius \(R\), remember the important identities \[ a=2R\sin A,\quad b=2R\sin B,\quad c=2R\sin C \] and \[ ab\cos C=\frac{a^2+b^2-c^2}{2}. \] These frequently convert complicated radical expressions into simple algebraic forms.

Answer 1

The Correct Option is C

Step 1: Use the relation between sides and circumradius.
For any triangle, \[ a=2R\sin A,\qquad b=2R\sin B,\qquad c=2R\sin C \] Since the triangle is acute-angled, \[ A,B,C\lt 90^\circ \] and all trigonometric quantities are positive.

Step 2: Simplify the first radical.
Consider \[ \sqrt{a^2b^2-4R^2} \] Using \[ c^2=a^2+b^2-2ab\cos C \] and the identity \[ ab=2Rc, \] we get \[ a^2b^2-4R^2c^2 = a^2b^2-a^2b^2\sin^2C \] \[ = a^2b^2\cos^2C \] Hence, \[ \sqrt{a^2b^2-4R^2c^2} = ab\cos C \] Using the cosine rule, \[ ab\cos C = \frac{a^2+b^2-c^2}{2} \] Therefore, \[ \sqrt{a^2b^2-4R^2c^2} = \frac{a^2+b^2-c^2}{2} \] Similarly, \[ \sqrt{b^2c^2-4R^2a^2} = \frac{b^2+c^2-a^2}{2} \] and \[ \sqrt{c^2a^2-4R^2b^2} = \frac{c^2+a^2-b^2}{2} \]

Step 3: Add the three expressions.
Adding, \[ \frac{a^2+b^2-c^2}{2} + \frac{b^2+c^2-a^2}{2} + \frac{c^2+a^2-b^2}{2} \] \[ = \frac{ (a^2+b^2-c^2) + (b^2+c^2-a^2) + (c^2+a^2-b^2) }{2} \] \[ = \frac{a^2+b^2+c^2}{2} \]

Step 4: Final conclusion.
Hence, \[ \boxed{\frac{a^2+b^2+c^2}{2}} \] Therefore, the correct option is \[ \boxed{\left(3\right)\ \frac{a^2+b^2+c^2}{2}} \]