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the lateral surface area of cone a is exactly $\frac{1}{2}$ the lateral…

Question

the lateral surface area of cone a is exactly $\frac{1}{2}$ the lateral surface area of cylinder b.
a. true
b. false

Explanation:

Step1: Recall lateral - surface - area formulas

The lateral - surface area of a cone \(S_{cone}=\pi rl\), where \(l = \sqrt{h^{2}+r^{2}}\) (slant height), and the lateral - surface area of a cylinder \(S_{cylinder}=2\pi rh\).

Step2: Compare the two areas

Given the cone and cylinder have the same height \(h\) and radius \(r\). The slant height of the cone \(l=\sqrt{h^{2}+r^{2}}\gt h\) (by the Pythagorean theorem). So, \(S_{cone}=\pi rl=\pi r\sqrt{h^{2}+r^{2}}\) and \(S_{cylinder}=2\pi rh\).
If \(S_{cone}=\frac{1}{2}S_{cylinder}\), then \(\pi r\sqrt{h^{2}+r^{2}}=\frac{1}{2}\times(2\pi rh)\), which simplifies to \(\sqrt{h^{2}+r^{2}} = h\). But \(\sqrt{h^{2}+r^{2}}\gt h\) for \(r\gt0\).

Answer:

B. False