Question

The top of a triangular-based pyramid(tetrahedron) is cut off. The cut is madeparallel to the base. If the vertical height ofthe top is 6 cm, calculate:a the height of the truncated piece,b the volume of the small pyramid,c the volume of the original pyramid.​

Answer 1

Given:

A pyramid.

The surface area of a pyramid length 7.2 cm.

To find:

We need to calculate

           a) The height of the truncated piece

           b) Volume of small pyramid

           c) Volume of original pyramid

Solution:

The cut created the highest of the larger pyramid that includes a aspect length of twelve cm is parallel to the bottom. Doing this provides a smaller pyramid of aspect length three cm. the peak of the frustum is 6 cm.

The volume of the pyramid is capable to $V=\left( \frac{1}{3} \right)B.h$ where B is that the space of bottom and h is that the vertical height of the pyramid.

If the peak of the smaller pyramid is $x$, $\[\frac{x}{3}=\frac{6+x}{12}\]$

$& \Rightarrow 12x=18+3x \\ & \\ & \Rightarrow 12x-3x=18 \\ & \\ & \Rightarrow 9x=18 \\ & \\ & \Rightarrow x=2$

Height of the original pyramid is equal to 8cm.

The volume of the initial pyramid is

$& =\left( \frac{1}{3} \right)\times 8\times 144 \\ \\ & =384$

The volume of the frustum that is cut off is

$& =\left( \frac{1}{3} \right)\times 2\times 9 \\ \\ & =6$

The volume of the truncated frustum is $\[384-6=378\]$.

Therefore, the required volume the truncated frustum is $\[378c{{m}^{3}}\]$.

Answer 2

Answer:

x/x+6=3/12

4x=x+6

3x=6

x=2

total heigt= 6+2=8

Step-by-step explanation: