Sure, let's solve the problem step-by-step:
1. Finding the slant height:
Given:
- Diameter of the circular base, [tex]\(d = 10\)[/tex] cm
- Total surface area of the cone, [tex]\(S = 90\pi\)[/tex] cm²
First, calculate the radius of the base:
[tex]\[ r = \frac{d}{2} = \frac{10}{2} = 5 \text{ cm} \][/tex]
The total surface area [tex]\(S\)[/tex] of a cone is given by:
[tex]\[ S = \pi r (r + l) \][/tex]
where [tex]\(r\)[/tex] is the radius and [tex]\(l\)[/tex] is the slant height.
Substituting the given values:
[tex]\[ 90\pi = \pi \cdot 5 \cdot (5 + l) \][/tex]
Cancel out [tex]\(\pi\)[/tex] from both sides:
[tex]\[ 90 = 5 \cdot (5 + l) \][/tex]
Solving for [tex]\(l\)[/tex]:
[tex]\[ 90 = 5 \cdot (5 + l) \][/tex]
[tex]\[ 90 = 25 + 5l \][/tex]
[tex]\[ 90 - 25 = 5l \][/tex]
[tex]\[ 65 = 5l \][/tex]
[tex]\[ l = \frac{65}{5} = 13 \text{ cm} \][/tex]
So, the slant height [tex]\(l\)[/tex] is 13 cm.
2. Finding the vertical height:
We will use the Pythagorean theorem to find the vertical height [tex]\(h\)[/tex]. For a cone, the relationship between the radius [tex]\(r\)[/tex], slant height [tex]\(l\)[/tex], and vertical height [tex]\(h\)[/tex] is:
[tex]\[ l^2 = r^2 + h^2 \][/tex]
Substitute the known values:
[tex]\[ 13^2 = 5^2 + h^2 \][/tex]
[tex]\[ 169 = 25 + h^2 \][/tex]
Solving for [tex]\(h^2\)[/tex]:
[tex]\[ h^2 = 169 - 25 \][/tex]
[tex]\[ h^2 = 144 \][/tex]
Taking the square root of both sides:
[tex]\[ h = \sqrt{144} = 12 \text{ cm} \][/tex]
So, the vertical height [tex]\(h\)[/tex] is 12 cm.
3. Finding the volume:
The volume [tex]\(V\)[/tex] of a cone is given by:
[tex]\[ V = \frac{1}{3} \pi r^2 h \][/tex]
Substitute the known values:
[tex]\[ V = \frac{1}{3} \pi (5)^2 (12) \][/tex]
[tex]\[ V = \frac{1}{3} \pi (25) (12) \][/tex]
[tex]\[ V = \frac{1}{3} \pi \cdot 300 \][/tex]
[tex]\[ V = 100\pi \][/tex]
So, the volume [tex]\(V\)[/tex] is:
[tex]\[ V = 100\pi \text{ cm}^3 \][/tex]
Therefore, the solutions are:
(i) The slant height [tex]\(l\)[/tex] is 13 cm.
(ii) The vertical height [tex]\(h\)[/tex] is 12 cm.
(iii) The volume [tex]\(V\)[/tex] is [tex]\(100\pi\)[/tex] cm³, which is approximately [tex]\(314.16\)[/tex] cm³ (since [tex]\(\pi \approx 3.14159\)[/tex]).
