The volume of a cone with height equal to radius, and slant height 5 cm is :
This question asks us to find the volume of a cone given specific information about its dimensions: the height is equal to the radius, and the slant height is 5 cm.
To solve this, we need to use the formulas related to the dimensions of a cone:
We are given:
First, let's use the relationship \( l^2 = r^2 + h^2 \) to find the value of the radius (and height) since \( h = r \).
Substitute the given values into the formula:
\( 5^2 = r^2 + r^2 \)
\( 25 = 2r^2 \)
Now, solve for \( r^2 \):
\( r^2 = \frac{25}{2} \)
We can find \( r \) by taking the square root:
\( r = \sqrt{\frac{25}{2}} = \frac{\sqrt{25}}{\sqrt{2}} = \frac{5}{\sqrt{2}} \)
Since \( h = r \), the height is also \( h = \frac{5}{\sqrt{2}} \) cm.
Next, we need to calculate the volume of the cone using the formula \( V = \frac{1}{3} \pi r^2 h \).
Substitute the values we found for \( r^2 \) and \( h \) into the volume formula:
\( V = \frac{1}{3} \pi \left(\frac{25}{2}\right) \left(\frac{5}{\sqrt{2}}\right) \)
Now, simplify the expression:
\( V = \frac{1}{3} \pi \frac{25 \times 5}{2 \times \sqrt{2}} \)
\( V = \frac{1}{3} \pi \frac{125}{2\sqrt{2}} \)
\( V = \frac{125 \pi}{3 \times 2\sqrt{2}} \)
\( V = \frac{125 \pi}{6\sqrt{2}} \)
The volume of the cone is \( \frac{125 \pi}{6\sqrt{2}} \) cubic centimeters.
| Dimension | Value |
|---|---|
| Slant Height (l) | 5 cm |
| Radius (r) | \( \frac{5}{\sqrt{2}} \) cm |
| Height (h) | \( \frac{5}{\sqrt{2}} \) cm (since h=r) |
| Volume (V) | \( \frac{125 \pi}{6\sqrt{2}} \) cm\(^3\) |
Comparing our result with the given options, we find that the calculated volume matches one of the options.
| Concept | Formula | Description |
|---|---|---|
| Volume of a Cone | \( V = \frac{1}{3} \pi r^2 h \) | Requires radius (r) and height (h). |
| Slant Height Relationship | \( l^2 = r^2 + h^2 \) | Relates slant height (l), radius (r), and height (h). Useful when one dimension is missing or there's a relationship between them. |
| Surface Area (Base) | \( A_{base} = \pi r^2 \) | Area of the circular base. |
| Surface Area (Lateral) | \( A_{lateral} = \pi r l \) | Area of the curved surface. |
| Surface Area (Total) | \( A_{total} = \pi r^2 + \pi r l \) | Sum of base area and lateral area. |
A cone is a three-dimensional geometric shape that tapers smoothly from a flat base (usually circular) to a point called the apex or vertex. The base of a cone is a circle, and the apex is on the line perpendicular to the center of the base.
Understanding these basic concepts and formulas is crucial for solving problems involving the volume and surface area of cones.
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