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The volume and heigh of a right circular...

The volume and heigh of a right circular cone are 1232 ` cm^(3)` and 24 cm respectively , the area of its curved surface ( in ` cm^(2)`) is :

A

1100

B

225

C

616

D

550

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The correct Answer is:
To find the curved surface area of a right circular cone given its volume and height, we can follow these steps: ### Step 1: Write down the formulas The volume \( V \) of a cone is given by the formula: \[ V = \frac{1}{3} \pi r^2 h \] where \( r \) is the radius and \( h \) is the height of the cone. The curved surface area \( A \) of a cone is given by the formula: \[ A = \pi r L \] where \( L \) is the slant height of the cone. ### Step 2: Substitute the known values into the volume formula Given: - Volume \( V = 1232 \, \text{cm}^3 \) - Height \( h = 24 \, \text{cm} \) Substituting into the volume formula: \[ 1232 = \frac{1}{3} \pi r^2 (24) \] ### Step 3: Simplify the equation First, multiply both sides by 3 to eliminate the fraction: \[ 1232 \times 3 = \pi r^2 (24) \] \[ 3696 = \pi r^2 (24) \] Next, divide both sides by \( 24 \): \[ \frac{3696}{24} = \pi r^2 \] Calculating the left side: \[ 154 = \pi r^2 \] ### Step 4: Solve for \( r^2 \) Now, divide both sides by \( \pi \): \[ r^2 = \frac{154}{\pi} \] Using \( \pi \approx \frac{22}{7} \): \[ r^2 = \frac{154 \times 7}{22} = \frac{1078}{22} = 49 \] ### Step 5: Find the radius \( r \) Taking the square root of both sides: \[ r = \sqrt{49} = 7 \, \text{cm} \] ### Step 6: Calculate the slant height \( L \) Using the Pythagorean theorem: \[ L = \sqrt{r^2 + h^2} \] Substituting the values: \[ L = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \, \text{cm} \] ### Step 7: Calculate the curved surface area \( A \) Now, substitute \( r \) and \( L \) into the curved surface area formula: \[ A = \pi r L \] Substituting the values: \[ A = \frac{22}{7} \times 7 \times 25 \] The \( 7 \) cancels out: \[ A = 22 \times 25 = 550 \, \text{cm}^2 \] ### Final Answer The area of the curved surface of the cone is: \[ \boxed{550 \, \text{cm}^2} \]
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