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(C) Let the side of the square base of the pyramid be $x$ and the height of the pyramid be $y$.The volume of the pyramid is $V = \frac{1}{3} x^2 y$.Wh

Vedclass · Accepted Answer

$60 < p < 65$

Vedclass · Answer

(C) Let the side of the square base of the pyramid be $x$ and the height of the pyramid be $y$.The volume of the pyramid is $V = \frac{1}{3} x^2 y$.When the side length $x$ is increased by $p \%$,the new side length is $x' = x \left(1 + \frac{p}{100}\right) = x \left(\frac{100+p}{100}\right)$.When the height $y$ is decreased by $p \%$,the new height is $y' = y \left(1 - \frac{p}{100}\right) = y \left(\frac{100-p}{100}\right)$.Since the volume remains the same,$V = \frac{1}{3} (x')^2 y' = \frac{1}{3} x^2 y$.Substituting the new values:$\frac{1}{3} x^2 y = \frac{1}{3} \left[ x \left(\frac{100+p}{100}\right) \right]^2 \left[ y \left(\frac{100-p}{100}\right) \right]$$1 = \left(\frac{100+p}{100}\right)^2 \left(\frac{100-p}{100}\right)$$100^3 = (100+p)^2 (100-p)$$1000000 = (10000 + 200p + p^2)(100 - p)$$1000000 = 1000000 - 10000p + 20000p - 200p^2 + 100p^2 - p^3$$p^3 + 100p^2 - 10000p = 0$Since $p > 0$,we can divide by $p$:$p^2 + 100p - 10000 = 0$Using the quadratic formula $p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$p = \frac{-100 \pm \sqrt{100^2 - 4(1)(-10000)}}{2} = \frac{-100 \pm \sqrt{10000 + 40000}}{2} = \frac{-100 \pm \sqrt{50000}}{2}$$p = \frac{-100 \pm 100\sqrt{5}}{2} = -50 \pm 50\sqrt{5}$Since $p > 0$,$p = 50(\sqrt{5} - 1) \approx 50(2.236 - 1) = 50(1.236) = 61.8$.Thus,$60 < p < 65$.

Suppose the height of a pyramid with a square base is decreased by $p \%$ and the lengths of the sides of its square base are increased by $p \%$ (where $p > 0$). If the volume remains the same,then:

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