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There are two concentric circles whose areas are in the ratio of 9 : 16 and the difference between their diameters is 4 cm. What is the area of the outer circle?
- a)32 cm2
- b)64cm2
- c)36 cm2
- d)48 cm2
Correct answer is option 'B'. Can you explain this answer?
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There are two concentric circles whose areas are in the ratio of 9 : 1...
Given data:
The ratio of the areas of two concentric circles = 9:16
Let the radii of the outer and inner circles be R and r respectively.
The difference between their diameters = 4 cm
Therefore, we have: R - r = 2 cm (since the difference between the diameters is the difference between the radii multiplied by 2)
Calculation:
We know that the area of a circle is proportional to the square of its radius.
Let A1 and A2 be the areas of the inner and outer circles respectively.
Therefore, we have:
A2/A1 = 16/9 (given)
A1 = πr²
A2 = πR²
R - r = 2 cm (given)
We can rewrite this as R = r + 2
Substituting the value of R in terms of r in the equation for A2, we get:
A2 = π(R)² = π(r + 2)² = πr² + 4πr + 4π
Substituting the value of A1 in terms of r in the equation for A2, we get:
A2 = (16/9)A1
πr² + 4πr + 4π = (16/9)πr²
9πr² + 36πr + 36π = 16πr²
7πr² - 36πr - 36π = 0
π(7r² - 36r - 36) = 0
7r² - 36r - 36 = 0
Solving this quadratic equation using the quadratic formula, we get:
r = (36 ± √(36² + 4×7×36)) / (2×7) = (36 ± 30) / 14
We take the positive value of r, which is:
r = 6 cm
Substituting this value of r in the equation for A2, we get:
A2 = πr² + 4πr + 4π
= π×6² + 4π×6 + 4π
= 36π + 24π + 4π
= 64π
Therefore, the area of the outer circle is 64π, which is approximately equal to 201.06 cm².
Answer: (b) 64cm²
The ratio of the areas of two concentric circles = 9:16
Let the radii of the outer and inner circles be R and r respectively.
The difference between their diameters = 4 cm
Therefore, we have: R - r = 2 cm (since the difference between the diameters is the difference between the radii multiplied by 2)
Calculation:
We know that the area of a circle is proportional to the square of its radius.
Let A1 and A2 be the areas of the inner and outer circles respectively.
Therefore, we have:
A2/A1 = 16/9 (given)
A1 = πr²
A2 = πR²
R - r = 2 cm (given)
We can rewrite this as R = r + 2
Substituting the value of R in terms of r in the equation for A2, we get:
A2 = π(R)² = π(r + 2)² = πr² + 4πr + 4π
Substituting the value of A1 in terms of r in the equation for A2, we get:
A2 = (16/9)A1
πr² + 4πr + 4π = (16/9)πr²
9πr² + 36πr + 36π = 16πr²
7πr² - 36πr - 36π = 0
π(7r² - 36r - 36) = 0
7r² - 36r - 36 = 0
Solving this quadratic equation using the quadratic formula, we get:
r = (36 ± √(36² + 4×7×36)) / (2×7) = (36 ± 30) / 14
We take the positive value of r, which is:
r = 6 cm
Substituting this value of r in the equation for A2, we get:
A2 = πr² + 4πr + 4π
= π×6² + 4π×6 + 4π
= 36π + 24π + 4π
= 64π
Therefore, the area of the outer circle is 64π, which is approximately equal to 201.06 cm².
Answer: (b) 64cm²
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Community Answer
There are two concentric circles whose areas are in the ratio of 9 : 1...
I think there is some mistake in the answer or question because radius of two circles are 6 cm and 8 cm and the area of outer circle is 201.14
cm......
cm......
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There are two concentric circles whose areas are in the ratio of 9 : 16 and the difference between their diameters is 4 cm. What is the area of the outer circle?a)32 cm2b)64cm2c)36cm2d)48cm2Correct answer is option 'B'. Can you explain this answer?
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