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.
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