A geometry problem by A Former Brilliant Member

The density of the whole pyramid is $ρ=\dfrac{W}{V} = \dfrac{900}{V}$ . The density of the first cut is $ρ=\dfrac{W_1}{V_1}$ . But the two cuts are equal in weight so $2W_1=W=900$ . From here, $W_1=450$ . Equating $ρ=ρ$ , we get

$\dfrac{900}{V}=\dfrac{450}{V_1}$ $\implies$ $\dfrac{V_1}{V}=\dfrac{1}{2}$

We know that the volume of similar solids $(V_1,V)$ have the same ratio as the cubes of any two corresponding lines. So

$\dfrac{V_1}{V} = \dfrac{x^3}{10^3}$

Equating $\dfrac{V_1}{V} = \dfrac{V_1}{V}$ , we have

$\dfrac{1}{2} = \dfrac{x^3}{10^3}$

$x = 7.93700526$

Finally,

$y = 10-x =$ $\boxed{\color{#20A900}2.06~feet}$