Question

The radius of a circle is 8cm and length of one of its chord is 12cm. Find the distance of the chord from the centre

Answer 1

Answers Best Answer:  1. required distance = √(8² - 6²) = 5.29 cm 

2.length of chord = 2*[√(10² - 5²)] = 17.32 

3. √(R² - 5.5²) - √(R² - 2.5²) = 3 or 
√(R² - 5.5²) = 3 + √(R² - 2.5²) squaring weget 
(R² - 5.5²) = 9 + (R² - 2.5²) +6*√(R² - 2.5²) or 
6*√(R² - 2.5²) = -9 -5.5² + 2.5² = -33 or 
36(R² - 2.5²) = 1089 or 
R² = 1089+225 = 1314 or R = 36.25 cm 

4. Draw diagram. Join center O to B and D too. O is the mid point of AC. It can be proved that the triangles AOB and CODare congruent hence AB = CD 

5. Draw diagram. join AC. angle CAB = 90 degreeas OD is parallel to AC, AC = 2OD 
Diagram shows thatCD > OC and cannot be proved to be equal to 2OD. 

6. required distance = √(10² - 6²) + √(8² - 6²) = 13.29 cm 

7 AB = 6 cm. As triangle AOB will be equilateral 

8.R = √(3² + 4²) = 5 required distance = √(5² - 4²) = 3 cm