The volume of both pyramids.

The volume of the square pyramid is \(\displaystyle{V}=\frac{{1}}{{3}}{a}^{{2}}{h}\) where a is a one side length of the square of base of the pyramid.

suppose assume that a=10, h=10.

The volume of the pyramid is

\(\displaystyle{V}=\frac{{1}}{{3}}{10}^{{2}}{10}\)

\(\displaystyle=\frac{{1}}{{3}}{\left({1000}\right)}\)

=333.333

suppose height increase 60% and reduce the length of an arm of a square is reduced by 35%.

\(\displaystyle{h}_{{1}}={10}+\frac{{60}}{{100}}{\left({1}\right)}\)

=16

\(\displaystyle{a}_{{1}}={10}-\frac{{35}}{{100}}{\left({10}\right)}\)

=6.5

Volume of new pyramid is computed as follows,

\(\displaystyle{V}_{{1}}=\frac{{1}}{{3}}{\left({6.5}\right)}^{{216}}=\frac{{1}}{{3}}{\left({42.25}\right)}{16}\)

=225.333

Here, the new volume decreases, the difference of volume is,

\(\displaystyle{V}-{V}_{{1}}={333.333}-{225.333}\)

=108

Thus, the new volume decreases 108 units.

Compute the size of pyramid decreases

\(\displaystyle\frac{{{V}-{V}_{{1}}}}{{V}}\times{100}=\frac{{108}}{{333.333}}\times{100}\)

\(\displaystyle={0.3240003}\times{100}\)

=32.4

Therefore, the size of new pyramid decreases 32.4%.

The volume of the square pyramid is \(\displaystyle{V}=\frac{{1}}{{3}}{a}^{{2}}{h}\) where a is a one side length of the square of base of the pyramid.

suppose assume that a=10, h=10.

The volume of the pyramid is

\(\displaystyle{V}=\frac{{1}}{{3}}{10}^{{2}}{10}\)

\(\displaystyle=\frac{{1}}{{3}}{\left({1000}\right)}\)

=333.333

suppose height increase 60% and reduce the length of an arm of a square is reduced by 35%.

\(\displaystyle{h}_{{1}}={10}+\frac{{60}}{{100}}{\left({1}\right)}\)

=16

\(\displaystyle{a}_{{1}}={10}-\frac{{35}}{{100}}{\left({10}\right)}\)

=6.5

Volume of new pyramid is computed as follows,

\(\displaystyle{V}_{{1}}=\frac{{1}}{{3}}{\left({6.5}\right)}^{{216}}=\frac{{1}}{{3}}{\left({42.25}\right)}{16}\)

=225.333

Here, the new volume decreases, the difference of volume is,

\(\displaystyle{V}-{V}_{{1}}={333.333}-{225.333}\)

=108

Thus, the new volume decreases 108 units.

Compute the size of pyramid decreases

\(\displaystyle\frac{{{V}-{V}_{{1}}}}{{V}}\times{100}=\frac{{108}}{{333.333}}\times{100}\)

\(\displaystyle={0.3240003}\times{100}\)

=32.4

Therefore, the size of new pyramid decreases 32.4%.