Please enter the following details
Free Daily Targets & Mocks
Study Material & Formulas PDFs
1 on 1 Mentorship
Question 120
‘O’ is the centre of the circle, AB is a chord of the circle, OM┴ AB. If AB = 20 cm, OM = 2√11 cm, then radius of the circle is
Solution
OM is perpendicular bisector of AB = 20 cm
=> BM = 10 cm and OM = 2$$\sqrt{11}$$
In $$\triangle$$OBM
radius, OB = $$\sqrt{(OM)^2 + (BM)^2}$$
=> OB = $$\sqrt{100 + 44} = \sqrt{144}$$
=> OB = 12 cm
Create a FREE account and get:
- Free SSC Study Material - 18000 Questions
- 230+ SSC previous papers with solutions PDF
- 100+ SSC Online Tests for Free
