Cone Section Area Calculation: 8m Radius, 16m Height

Hey guys! Today, we're diving deep into the fascinating world of geometry, specifically tackling a problem involving cones. If you're into math, especially the kind that involves shapes and calculations, you're going to love this. We've got a circular cone with a radius of 8 meters and a height of 16 meters. The core question is: what's the area of the cross-section created when a plane slices through it, parallel to the base, and specifically at a distance of 10 meters from the vertex? This isn't just about crunching numbers; it's about understanding how geometric properties scale and interact. We'll break down the concepts, show you the steps, and figure out which of the options – (A) 25π m², (B) 24π m², (C) 18π m², or (D) 10π m² – is the correct answer. Get ready to flex those brain muscles because we're about to embark on a cool mathematical journey!

Understanding the Geometry of Cones

Alright, let's start by getting our heads around the basics of the cone we're dealing with. We’re talking about a right circular cone, which is the standard cone shape you probably picture – a circular base and a vertex directly above the center of that base. Our cone has a radius (r) of 8 meters and a height (h) of 16 meters. Now, imagine a plane cutting through this cone. The problem states this plane is parallel to the base. This is a super important detail, guys. When a plane cuts a cone parallel to its base, the cross-section it creates is always another circle, similar to the base. The second key piece of information is the distance of this plane from the vertex. The vertex is that pointy tip of the cone. The plane is located 10 meters from the vertex. This distance is measured along the height of the cone. So, we have a smaller cone formed above the cutting plane, and this smaller cone is similar to the original, larger cone. Understanding similarity is crucial here because it allows us to use ratios to find the dimensions of the cross-sectional circle.

The Power of Similar Triangles

The concept of similar triangles is our best friend when solving problems like this. If you were to slice the cone vertically through its vertex and center of the base, you'd see an isosceles triangle. The height of the cone would be one leg of a right-angled triangle, the radius of the base would be the other leg, and the slant height would be the hypotenuse. When our plane cuts the cone parallel to the base, it creates a smaller, similar right-angled triangle within the larger one. Let's denote the height of the original cone as $H$ and its radius as $R$. We are given $R = 8$ m and $H = 16$ m. The cutting plane is at a distance $h'$ from the vertex. We are given $h' = 10$ m. Let the radius of the circular cross-section created by this plane be $r'$. Because the cutting plane is parallel to the base, the triangle formed by the vertex, the center of the cross-section, and a point on the edge of the cross-section is similar to the triangle formed by the vertex, the center of the base, and a point on the edge of the base. This similarity means the ratio of corresponding sides is equal. Specifically, the ratio of the height of the smaller cone (from the vertex to the cutting plane) to the height of the original cone is equal to the ratio of the radius of the cross-section to the radius of the original cone's base. Mathematically, this is expressed as: $r'/R = h'/H$. This relationship is the cornerstone of our calculation. It elegantly connects the dimensions of the original cone to the dimensions of the cross-section without needing to know the slant height or any complex calculus.

Calculating the Radius of the Cross-Section

Now that we've established the relationship using similar triangles, let's plug in the numbers and find the radius of our cross-sectional circle. We have the following values: Original cone's radius $R = 8$ m, original cone's height $H = 16$ m, and the distance from the vertex to the cutting plane $h' = 10$ m. We want to find the radius of the cross-section, $r'$. Using the similarity ratio we just discussed, we have:

$$\frac{r'}{R} = \frac{h'}{H}$$

Let's substitute our known values into this equation:

$$\frac{r'}{8} = \frac{10}{16}$$

To solve for $r'$, we can multiply both sides of the equation by 8:

$$r' = 8 \times \frac{10}{16}$$

Now, we simplify the fraction $\frac{10}{16}$. Both 10 and 16 are divisible by 2, so $\frac{10}{16} = \frac{5}{8}$.

$$r' = 8 \times \frac{5}{8}$$

See how the 8s cancel out? This makes the calculation super straightforward:

$$r' = 5 \text{ m}$$

So, the radius of the circular cross-section created by the plane is 5 meters. This is a pretty neat result, and it shows how predictable geometric scaling can be. We didn't need any advanced formulas, just the fundamental principle of similar figures. The radius of the smaller circle is directly proportional to its height from the vertex, relative to the main cone's dimensions. This value, $r' = 5$ m, is what we need to calculate the area of the cross-section.

The Area Formula for a Circle

We've found the radius of the circular cross-section, which is $r' = 5$ m. The question asks for the area of this cross-section. Since the cross-section is a circle, we use the standard formula for the area of a circle, which is $A = \pi r^2$. In our case, the radius is $r'$, so the area of the cross-section, let's call it $A'$, will be:

$$A' = \pi (r')^2$$

Substitute the value of $r'$ we just calculated:

$$A' = \pi (5)^2$$

Calculate the square of 5:

$$A' = \pi \times 25$$

So, the area of the cross-section is 25π square meters.

Final Answer and Verification

We have successfully calculated the area of the cross-section made by the plane. The radius of the original cone is 8m, and its height is 16m. The plane cuts parallel to the base at a distance of 10m from the vertex. Using the properties of similar triangles, we found that the ratio of the radius of the cross-section ($r'$) to the radius of the base ($R$) is equal to the ratio of the distance from the vertex to the plane ($h'$) to the total height of the cone ($H$). This gave us the equation $\frac{r'}{8} = \frac{10}{16}$, which simplified to $r' = 5$ m.

Then, we used the formula for the area of a circle, $A = \pi r^2$, to find the area of the cross-section. With $r' = 5$ m, the area $A'$ is $\pi (5)^2 = 25\pi$ square meters.

Let's double-check our work. The original cone has a radius of 8m and a height of 16m. The cutting plane is at 10m from the vertex. This means the plane is more than halfway up the cone's height (10m out of 16m). If it were exactly halfway (8m from the vertex), the radius of the cross-section would be half the base radius, i.e., 4m, and the area would be $16\pi$ m². Since 10m is further up, the radius should be larger than 4m, and the area larger than $16\pi$ m². Our result of 5m radius and $25\pi$ m² area fits this expectation. The ratio of heights is 10/16 = 5/8. So the ratio of radii should also be 5/8. $r' = (5/8) * 8 = 5$m. The area is $\pi * 5^2 = 25\pi$ m².

Comparing our calculated area with the given options:

(A) ab = 25 π m² (B) ab = 24 π m² (C) ab = 18 π m² (D) ab = 10 π m²

Our result, 25π m², matches option (A). So, the correct answer is (A). It's awesome how these geometric principles work out so neatly, right? Keep practicing, and you'll master these problems in no time!