A right square pyramid has been truncated so that its height is 2/3 of the original height of 4.5 m. Find: A. The length x; B. The volume of the truncated pyramid; C. The TSA of the pyramid.

Here are the solutions to your questions.

QUESTION 4

Given:
Original height of the pyramid, $H = 4.5$ m.
Height of the truncated pyramid (frustum), $h_f = \frac{2}{3} \times 4.5 \text{ m} = 3$ m.
Side length of the base of the original pyramid, $L = 6$ m.
Side length of the top square of the frustum, $x$.

First, determine the height of the small pyramid that was removed from the top.
Height of small pyramid, $h_s = H - h_f = 4.5 \text{ m} - 3 \text{ m} = 1.5$ m.

A. The length x

Step 1: Use similar triangles formed by the cross-section of the pyramids. The ratio of heights is equal to the ratio of corresponding base lengths.

$$\frac{h_s}{H} = \frac{x}{L}$$ $$\frac{1.5 \text{ m}}{4.5 \text{ m}} = \frac{x}{6 \text{ m}}$$ $$\frac{1}{3} = \frac{x}{6 \text{ m}}$$

Step 2: Solve for $x$.

$$x = \frac{6 \text{ m}}{3}$$ $$x = 2 \text{ m}$$

The length $x$ is $\boxed{\text{2 m}}$.

B. The volume of the truncated pyramid
The volume of the truncated pyramid (frustum) is the volume of the original pyramid minus the volume of the small pyramid removed from the top. The formula for the volume of a pyramid is $V = \frac{1}{3} \times \text{base area} \times \text{height}$.

Step 1: Calculate the volume of the original pyramid ($V_H$).

$$V_H = \frac{1}{3} \times L^2 \times H$$ $$V_H = \frac{1}{3} \times (6 \text{ m})^2 \times 4.5 \text{ m}$$ $$V_H = \frac{1}{3} \times 36 \text{ m}^2 \times 4.5 \text{ m}$$ $$V_H = 12 \text{ m}^2 \times 4.5 \text{ m} = 54 \text{ m}^3$$

Step 2: Calculate the volume of the small pyramid ($V_s$).

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