Pyramid Minus Pyramid

The volume of a frustum of a regular pyramid is given by the formula:

$V=\dfrac{\left(A_1+A_2+\sqrt{A_1 \times A_2}\right)(h)}{3}$ where: $A_1$ and $A_2$ are area of the bases and $h$ = height

$A_1=5 \times 5=25; A_2=7 \times 7=49$

Substituting, we get

$V=\dfrac{\left(25+49+\sqrt{25 \times 49}\right)(3)}{3}=74+35=$ $\boxed{109}$

The volume of the truncated pyramid = the volume of pyramid with $7\times 7$ base - the volume of pyramid with $5\times 5$ base.

The volume of pyramid is given by $V = \dfrac {b^2h}3$ , where $b$ is the side length of the base square and $h$ , the height. For this pyramid, the relation between $b$ and $h$ is given by $\dfrac hb = \dfrac 32$ $\implies h = \dfrac {3b}2$ . Therefore, $V= \dfrac {b^3}2$ and the volume of the truncated pyramid is $V_{truncated} = \dfrac {7^3}2 - \dfrac {5^3}2 = \boxed{109}$ .

Suppose $x$ be the height of the removed pyramid of square length of $5$ .

Then the volume of the original pyramid = $(\dfrac{x+3}{3})(7^2)$ .

On the other hand, the volume of the removed pyramid = $(\dfrac{x}{3})(5^2)$

Thus, the solution volume = $49 - \dfrac{x}{3}(49-25) = 49 + 8x$ .

Then due to triangle similarity, $\dfrac{x}{x+3} = \dfrac{5}{7}$ .

$7x = 5x + 15$ .

$x = 7.5$

Therefore, the solution volume = $49 + 60 = \boxed{109}$ .

Note : the formula for the cut pyramid's volume $V$ with upper square base length $a$ , bigger square base length $b$ , and height $h$ between the planes can be generalized as:

$V = \dfrac{h}{3}(a^2 + ab + b^2)$

The solid is in the shape of a Frustum of a Regular Pyramid and the formula for the volume is

$V=$ $\dfrac{h}{3}($ $A_1 + A_2 + \sqrt{A_1\times A_2})$ where $A_1$ and $A_2$ are area of the bases

Solving for $A_1$ we have,

$A_1=7\times7=49$

Solving for $A_2$ , we have

$A_2=5\times5=25$

Substituting, we obtain

$V=$ $\dfrac{h}{3}($ $49 + 25 + \sqrt{49\times25}) =$ $\boxed{109}$