The base of a right pyramid is an equilateral triangle with side 8 cm, and its height is 30 \(\sqrt{3}\) cm. The volume (in cm 3)
480 cm3
The problem asks us to find the volume of a right pyramid. We are given that the base is an equilateral triangle with a side length of 8 cm, and the height of the pyramid is $30 \sqrt{3}$ cm.
To find the volume of a pyramid, we use the formula:
$$ \text{Volume} = \frac{1}{3} \times \text{Base Area} \times \text{Height} $$
First, we need to calculate the area of the equilateral triangular base. The formula for the area of an equilateral triangle with side length 'a' is:
$$ \text{Area} = \frac{\sqrt{3}}{4} a^2 $$
Given the side of the equilateral triangle base is 8 cm, the base area is:
$$ \text{Base Area} = \frac{\sqrt{3}}{4} (8 \text{ cm})^2 = \frac{\sqrt{3}}{4} \times 64 \text{ cm}^2 $$
Simplifying the base area calculation:
$$ \text{Base Area} = 16\sqrt{3} \text{ cm}^2 $$
Now, we use the volume formula with the calculated base area and the given height:
Given Height = $30\sqrt{3}$ cm
$$ \text{Volume} = \frac{1}{3} \times (16\sqrt{3} \text{ cm}^2) \times (30\sqrt{3} \text{ cm}) $$
Let's perform the multiplication:
$$ \text{Volume} = \frac{1}{3} \times 16 \times 30 \times \sqrt{3} \times \sqrt{3} \text{ cm}^3 $$
We know that $\sqrt{3} \times \sqrt{3} = 3$. Substituting this into the equation:
$$ \text{Volume} = \frac{1}{3} \times 16 \times 30 \times 3 \text{ cm}^3 $$
We can cancel out the $\frac{1}{3}$ and the 3:
$$ \text{Volume} = 16 \times 30 \text{ cm}^3 $$
Finally, calculate the product:
$$ \text{Volume} = 480 \text{ cm}^3 $$
Therefore, the volume of the right pyramid is 480 cm3.
| Concept | Formula |
|---|---|
| Area of Equilateral Triangle (side 'a') | $\frac{\sqrt{3}}{4} a^2$ |
| Volume of a Pyramid | $\frac{1}{3} \times \text{Base Area} \times \text{Height}$ |
A right pyramid is a pyramid where the apex is directly above the centroid of the base. In the case of an equilateral triangle, the centroid is the point where the medians intersect, and it is also the center of the circumscribed and inscribed circles.
The curved surface area and the volume of a cylindrical object are 88 cm 2and 132 cm 3, respectively. The height (in cm) of the cylindrical object is:
(Take π = \(\frac{{22}}{7}\) )
The circumference of the base of a cylindrical vessel is 264 cm and its height is 50 cm. The capacity (in litres) of the vessel is:
(Take π = \(\frac{22}{7}\) )
What will be the total cost (in Rs.) of polishing the curved surface of a wooden cylinder at rate of Rs. 50 per m 2, if its diameter is 70 cm and height is 6 m? (Take π = \(\frac{22}{7}\) )
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?
The volume of a cone is 73920 cm3. If the height of the cone is 160 cm, then find the diameter of its base.
The volume of a cone with height equal to radius, and slant height 5 cm is :
How many metres of 2-m-wide cloth will be required to make a conical tent with a diameter of the base as 14 m and slant height as 9 m? (ignore wastage)
A cone and a hemisphere have equal bases and volumes. What is the ratio of the height of the cone to the radius of the hemisphere?
A metallic solid cuboid of dimensions 36 cm × 18 cm × 12 cm is melted and recast in the form of cubes of side 6 cm. Find the number of cubes so formed.
If the surface area of a sphere is 64 π cm 2, then the volume of the sphere is:
Find the surface area of a sphere of diameter 21 cm. (Use π = \(\frac{{22}}{7}\) )
A cube is 7 cm of an edge and another cube is 14 cm on an edge. The ratios of their surface areas are