Question:

In a triangle \(ABC\), the lengths of the sides \(AB\) and \(AC\) equal \(17.5\) cm and \(9\) cm respectively. Let \(D\) be a point on the line segment \(BC\) such that \(AD\) is perpendicular to \(BC\). If \(AD = 3\) cm, then what is the radius (in cm) of the circle circumscribing the triangle \(ABC\)?

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In the right triangles \(ABD\) and \(ACD\), find the sines of the angles at \(B\) and \(C\), then use the law of sines \(a / \sin A = 2R\).
Updated On: Nov 4, 2026
  • 17.05
  • 27.85
  • 26.25
  • 32.25
  • 22.45
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The Correct Option is C

Solution and Explanation

Step 1: Find the sine of angle C:
In the right triangle \(ADC\) (right angle at \(D\)), \(\sin C = \dfrac{AD}{AC} = \dfrac{3}{9} = \dfrac{1}{3}\).

Step 2: Use the law of sines on side AB:
In triangle \(ABC\), \(\dfrac{AB}{\sin C} = 2R\), where \(R\) is the circumradius. So \(2R = \dfrac{17.5}{1/3} = 52.5\).

Step 3: Find R:
\(R = \dfrac{52.5}{2} = 26.25\) cm.

Step 4: Check using angle B:
In the right triangle \(ABD\), \(\sin B = \dfrac{3}{17.5} = \dfrac{6}{35}\). Then \(2R = \dfrac{AC}{\sin B} = \dfrac{9 \times 35}{6} = 52.5\), so \(R = 26.25\) cm again.

Final Answer:
The correct option is (3): \(26.25\).
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