AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is :
a. 17 cm
b. 15 cm
c. 4 cm
d. 8 cm
Solution:
It is given that
AD = 34 cm
AB = 30 cm
Construct OL perpendicular to AB
AL = LB = 1/2 AB = 1/2 (30) = 15 cm
In triangle OLA
Using pythagoras theorem
OA² = OL² + AL²
Substituting the values
17² = OL² + 15²
By further calculation
OL² = 289 - 225 = 64
So we get
OL = 8 cm
Therefore, the distance of AB from the centre of the circle is 8 cm.
✦ Try This: AD is a diameter of a circle and AB is a chord. If AD = 24 cm, AB = 20 cm, the distance of AB from the centre of the circle is :
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 10
NCERT Exemplar Class 9 Maths Exercise 10.1 Problem 1
AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is : a. 17 cm, b. 15 cm, c. 4 cm, d. 8 cm
Summary:
AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is 8 cm
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