Multiple choice

AB is a chord of a circle whose centre is O. P is the point on the circle such that OP ⊥ AB and OP intersects AB at the point in. If AB = 8 cm and MP = 2 cm, then the radius of the circle is AB एक वृत्त की एक जीवा है जिसका केंद्र O है । वृत्त पर कोई बिंदु P ऐसे स्थित है कि OP ⊥ AB और OP, AB को बिंदु पर प्रतिच्छेद करता है। यदि AB = 8 सेमी और MP = 2 सेमी, तो वृत्त की त्रिज्या है -

  1. 10 cm./सेमी.

  2. 6 cm./सेमी.

  3. 5 cm./सेमी.

  4. 4 cm./सेमी.

Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Since OP ⟂ AB, OP bisects AB at M. So AM = MB = 4 cm. In right triangle OMP: OM² + MP² = OP². Also OM² + AM² = OA² (radius). Given AB = 8 cm, so AM = 4 cm. MP = 2 cm. OM² = OA² - 16. Also OM² + 4 = (OA - 2)². Solving: OA² - 16 + 4 = OA² - 4OA + 4, so OA² - 12 = OA² - 4OA + 4, therefore 4OA = 16 and OA = 5 cm. Options A, B, D are calculation errors.