Multiple choice

ABC is a right-angled triangle with angle B = 90°. A circle with BC as diameter cuts the hypotenuse AC at Point P. AB = 15 cm and BC = 8 cm. What is the distance from the point P to vertex A? समकोण त्रिभुज ABC का कोण B = 90° है। एक वृत्त जिसका व्यास BC, AC को बिन्दु P पर काटता है। AB = 15 सेमी और BC = 8 सेमी. तो शीर्ष A से P की दूरी क्या है?

  1. 10 cm/सेमी

  2. 12 cm/सेमी

  3. 13.2 cm/सेमी

  4. 15 cm/सेमी
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C Correct answer
Explanation

In right triangle ABC with angle B=90°, AB=15, BC=8. AC = sqrt(15² + 8²) = 17 cm. Circle with BC as diameter has center at midpoint of BC. Using intersecting chords theorem or power of point: AP × PC = BP². But better: using property that angle BPC = 90° (angle in semicircle), triangles are similar. Answer is 13.2 cm using section formula.

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