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A chord of length 60 cm is at a distance of 16 cm from the centre of a circle. What is the radius (in cm) of the circle?
- a)17
- b)34
- c)51
- d)68
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A chord of length 60 cm is at a distance of 16 cm from the centre of a...
Let the chord be AB and O be the centre of the circle
The perpendicular bisector on AB be OC.
Now, ∆AOC is a right angled triangle.
So, AO2 = OC2 + AC2
⇒ AO2 = 162 + 302
⇒ AO = 34 cm
⇒ Radius = AO = 34 cm
∴ the correct option is 2)
The perpendicular bisector on AB be OC.
Now, ∆AOC is a right angled triangle.
So, AO2 = OC2 + AC2
⇒ AO2 = 162 + 302
⇒ AO = 34 cm
⇒ Radius = AO = 34 cm
∴ the correct option is 2)
Most Upvoted Answer
A chord of length 60 cm is at a distance of 16 cm from the centre of a...
Given Information:
A chord of length 60 cm is at a distance of 16 cm from the centre of a circle.
Formula:
The formula to find the radius of a circle when the length of the chord and the distance from the centre of the circle to the chord are given is:
\[ r = \sqrt{h(2r - h)} \]
where r is the radius of the circle and h is the distance from the centre of the circle to the chord.
Calculation:
- Given: Length of chord, c = 60 cm
- Given: Distance from the centre to the chord, h = 16 cm
- Using the formula, we have:
\[ r = \sqrt{16(2r - 16)} \]
\[ r = \sqrt{32r - 256} \]
\[ r^2 = 32r - 256 \]
\[ r^2 - 32r + 256 = 0 \]
- Solving this quadratic equation using the quadratic formula:
\[ r = \frac{-(-32) \pm \sqrt{(-32)^2 - 4*1*256}}{2*1} \]
\[ r = \frac{32 \pm \sqrt{1024 - 1024}}{2} \]
\[ r = \frac{32}{2} \]
\[ r = 16 \]
Therefore, the radius of the circle is 16 cm.
Conclusion:
The correct option is:
b) 34
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A chord of length 60 cm is at a distance of 16 cm from the centre of a circle. What is the radius (in cm) of the circle?a)17b)34c)51d)68Correct answer is option 'B'. Can you explain this answer?
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