A pyramid has an equilateral triangle as its base, of which each side is 8 cm. Its slant edge is 24 cm. The whole surface area of the pyramid (in cm2) is:
Finding the Whole Surface Area of a Pyramid with an Equilateral Base
The problem asks us to find the total surface area of a pyramid that has an equilateral triangle as its base. We are given the side length of the equilateral base and the length of the pyramid's slant edge.
The whole surface area of any pyramid is the sum of the area of its base and the area of all its lateral faces.
Whole Surface Area = Area of Base + Area of Lateral Faces
Calculating the Area of the Base Triangle
The base is an equilateral triangle with a side length of 8 cm.
The formula for the area of an equilateral triangle with side 'a' is \(\frac{\sqrt{3}}{4} a^2\).
Here, the base side 'a' = 8 cm.
Area of Base = \(\frac{\sqrt{3}}{4} \times (8\, \text{cm})^2\)
Area of Base = \(\frac{\sqrt{3}}{4} \times 64\, \text{cm}^2\)
Area of Base = \(16\sqrt{3}\, \text{cm}^2\)
Calculating the Area of the Lateral Faces
A pyramid with an equilateral triangle base has three lateral faces. Each lateral face is an isosceles triangle.
- The base of each isosceles lateral triangle is a side of the equilateral base, which is 8 cm.
- The two equal sides of each isosceles lateral triangle are the slant edges of the pyramid, which are 24 cm.
To find the area of an isosceles triangle, we need its height (which is the slant height of the pyramid face).
Let 's' be the slant edge length (24 cm) and 'a' be the base side length of the pyramid (8 cm).
Consider one lateral face. It's an isosceles triangle with sides 24 cm, 24 cm, and 8 cm. The height of this triangle (\(h_s\)), dropped from the apex to the base (8 cm side), bisects the base.
Using the Pythagorean theorem in the right-angled triangle formed by the height (\(h_s\)), half of the base (\(\frac{a}{2}\)), and the slant edge (s):
\(h_s^2 + \left(\frac{a}{2}\right)^2 = s^2\)
\(h_s^2 + \left(\frac{8}{2}\right)^2 = 24^2\)
\(h_s^2 + 4^2 = 24^2\)
\(h_s^2 + 16 = 576\)
\(h_s^2 = 576 - 16\)
\(h_s^2 = 560\)
\(h_s = \sqrt{560}\)
We can simplify \(\sqrt{560}\). \(560 = 16 \times 35\).
\(h_s = \sqrt{16 \times 35} = \sqrt{16} \times \sqrt{35} = 4\sqrt{35}\) cm.
Now, we can find the area of one lateral face (isosceles triangle):
Area of one lateral face = \(\frac{1}{2} \times \text{base} \times \text{height}\)
Area of one lateral face = \(\frac{1}{2} \times 8\, \text{cm} \times 4\sqrt{35}\, \text{cm}\)
Area of one lateral face = \(4 \times 4\sqrt{35}\, \text{cm}^2\)
Area of one lateral face = \(16\sqrt{35}\, \text{cm}^2\)
Since there are 3 lateral faces, the total area of the lateral faces is:
Total Area of Lateral Faces = 3 \(\times\) Area of one lateral face
Total Area of Lateral Faces = 3 \(\times 16\sqrt{35}\, \text{cm}^2\)
Total Area of Lateral Faces = \(48\sqrt{35}\, \text{cm}^2\)
Calculating the Whole Surface Area
Now, we add the area of the base and the total area of the lateral faces:
Whole Surface Area = Area of Base + Total Area of Lateral Faces
Whole Surface Area = \(16\sqrt{3}\, \text{cm}^2 + 48\sqrt{35}\, \text{cm}^2\)
Whole Surface Area = \((16\sqrt{3} + 48\sqrt{35})\, \text{cm}^2\)
Comparing this result with the given options, we find that it matches option 4.
| Component | Calculation | Area (\(\text{cm}^2\)) |
|---|---|---|
| Base Area (Equilateral Triangle) | \(\frac{\sqrt{3}}{4} \times 8^2\) | \(16\sqrt{3}\) |
| Slant Height (\(h_s\)) of Lateral Face | \(\sqrt{24^2 - 4^2} = \sqrt{576 - 16} = \sqrt{560} = 4\sqrt{35}\) | - |
| Area of One Lateral Face (Isosceles Triangle) | \(\frac{1}{2} \times 8 \times 4\sqrt{35}\) | \(16\sqrt{35}\) |
| Total Lateral Surface Area (3 faces) | \(3 \times 16\sqrt{35}\) | \(48\sqrt{35}\) |
| Whole Surface Area | Base Area + Total Lateral Area | \((16\sqrt{3} + 48\sqrt{35})\) |
Summary of the Pyramid Surface Area Calculation
- Identify the base shape and dimensions.
- Calculate the area of the base. For an equilateral triangle of side 'a', Area = \(\frac{\sqrt{3}}{4} a^2\).
- Identify the shape of the lateral faces (isosceles triangles).
- Determine the base and equal sides of the lateral faces.
- Calculate the height (slant height) of a lateral face using the Pythagorean theorem with the slant edge and half the base side of the pyramid.
- Calculate the area of one lateral face. For an isosceles triangle, Area = \(\frac{1}{2} \times \text{base} \times \text{height}\).
- Multiply the area of one lateral face by the number of lateral faces (3 for a triangular base) to get the total lateral surface area.
- Add the base area and the total lateral surface area to find the whole surface area.
Revision Table: Pyramid Surface Area
| Concept | Formula/Method | Application in this problem |
|---|---|---|
| Area of Equilateral Triangle | \(\frac{\sqrt{3}}{4} a^2\) | Base Area = \(\frac{\sqrt{3}}{4} \times 8^2 = 16\sqrt{3}\) |
| Pythagorean Theorem | \(a^2 + b^2 = c^2\) | Finding slant height: \(h_s^2 + 4^2 = 24^2\) |
| Area of Triangle | \(\frac{1}{2} \times \text{base} \times \text{height}\) | Area of one lateral face: \(\frac{1}{2} \times 8 \times 4\sqrt{35}\) |
| Whole Surface Area of Pyramid | Base Area + Total Lateral Area | \(16\sqrt{3} + 3 \times 16\sqrt{35}\) |
Additional Information on Pyramid Geometry
A pyramid is a polyhedron formed by connecting a polygonal base and a point, called the apex. Each base edge and apex form a triangle, called a lateral face. The type of pyramid is determined by the shape of its base.
- Regular Pyramid: A pyramid where the base is a regular polygon and the apex is directly above the center of the base. In a regular pyramid, all lateral faces are congruent isosceles triangles.
- Slant Height: The height of a lateral face is called the slant height of that face. For a regular pyramid, all slant heights are equal.
- Slant Edge: The edge connecting the apex to a vertex of the base. For a regular pyramid, all slant edges are equal.
- Height of the Pyramid: The perpendicular distance from the apex to the plane of the base. This is different from the slant height.
In this specific problem, the base is an equilateral triangle, which is a regular polygon. However, the problem only gives the slant edge (distance from apex to base vertex), not necessarily the height or slant height directly. We used the slant edge to find the slant height of the lateral triangular faces, which is crucial for calculating their area.
Important Questions from Solid Figures
A cylindrical tube, open at both ends, is made of a metal sheet which is 0.5 cm thick. Its outer radius is 4 cm and length is 2 m. How much metal (in cm 3) has been used in making the tube?
The volume of a right circular cone is 308 cm 3 and the radius of its base is 7 cm. What is the curved surface area (in cm 2) of the cone? (Take π = \(\frac{22}{7} \) )
The slant height and radius of a right circular cone are in the ratio 29 ∶ 20. If its volume is 4838.4 π cm 3, then its radius is:
Six cubes, each of edge 2 cm, are joined end to end. What is the total surface area of the resulting cuboid in cm 2?
A solid cube of side 8 cm is dropped into a rectangular container of length 16 cm, breadth 8 cm and height 15 cm which is partly filled with water. If the cube is completely submerged, then the rise of water level (in cm) is:
