When the circumference and area of a circle are numerically equal, then the diameter is numerically equal to?
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Answered by
3
Area = Circumference
![2\pi \: r = \pi {r}^{2} 2\pi \: r = \pi {r}^{2}](https://tex.z-dn.net/?f=2%5Cpi+%5C%3A+r+%3D+%5Cpi+%7Br%7D%5E%7B2%7D+)
![2r \: = \frac{\pi \: r {}^{2} }{\pi} 2r \: = \frac{\pi \: r {}^{2} }{\pi}](https://tex.z-dn.net/?f=2r+%5C%3A++%3D++%5Cfrac%7B%5Cpi+%5C%3A+r+%7B%7D%5E%7B2%7D+%7D%7B%5Cpi%7D+)
![2r \: = \: {r}^{2} 2r \: = \: {r}^{2}](https://tex.z-dn.net/?f=2r+%5C%3A++%3D++%5C%3A++%7Br%7D%5E%7B2%7D+)
![2r= (r) \times (r) 2r= (r) \times (r)](https://tex.z-dn.net/?f=2r%3D+%28r%29+%5Ctimes+%28r%29)
![\frac{2r}{r } = r \frac{2r}{r } = r](https://tex.z-dn.net/?f=+%5Cfrac%7B2r%7D%7Br+%7D++%3D+r)
![2 = r \: \: \: or \: \: r = 2 2 = r \: \: \: or \: \: r = 2](https://tex.z-dn.net/?f=2+%3D+r+%5C%3A++%5C%3A++%5C%3A+or+%5C%3A++%5C%3A+r+%3D+2)
![but \: \: d = 2r \\ so \: \: \: d = 2 \times 2 \\ \: \: \: \: \: \: \: d = 4 but \: \: d = 2r \\ so \: \: \: d = 2 \times 2 \\ \: \: \: \: \: \: \: d = 4](https://tex.z-dn.net/?f=but+%5C%3A+%5C%3A+d+%3D+2r+%5C%5C+so+%5C%3A++%5C%3A++%5C%3A+d+%3D+2+%5Ctimes+2+%5C%5C++%5C%3A++%5C%3A++%5C%3A+%5C%3A++%5C%3A++%5C%3A++%5C%3A+++d+%3D+4)
Therefore the diameter of the circle = 4
Therefore the diameter of the circle = 4
Answered by
23
Step-by-step explanation:
Circumference of circle ≈ 2πr
Area of circle ≈ πr²
Given that 2πr ≈ πr ²
→ R ≈2
So, radius of circle =2.
Diameter = Double of radius
Diameter = 2 × 2
Diameter = 4
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