Question

the lateral area of a cone is 574 pi cm2. The radius is 19.6 cm. What is the slant height to the nearest tenth of a centimeter?

Answer 1

Answer:

29.3 cm.

Step-by-step explanation:

We have been given that the lateral area of a cone is 574 pi cm2. The radius is 19.6 cm. We are asked to find the slant height of the cone.

We will use lateral area of cone formula to solve our given problem.

$\text{Lateral area of cone}=\pi rl$, where,

r = Radius of cone,

l = Slant height of cone.

Upon substituting our given values in above formula we will get,

$574\pi\text{ cm}^2=\pi*19.6\text{ cm}*l$

$574\pi\text{ cm}^2=\pi*19.6\text{ cm}*l$

Dividing both sides by pi we will get,

$\frac{574\pi\text{ cm}^2}{\pi}=\frac{\pi*19.6\text{ cm}*l}{\pi}$

$574\text{ cm}^2=19.6\text{ cm}*l$

Dividing both sides by 19.6 cm we will get,

$\frac{574\text{ cm}^2}{19.6\text{ cm}}=\frac{19.6\text{ cm}*l}{19.6\text{ cm}}$

$\frac{574\text{ cm}}{19.6}=l$

$29.2857142857\text{ cm}=l$

$l\approx 29.3\text{ cm}$

Therefore, the slant height to the nearest tenth of a centimeter is 29.3 cm.

Answer 2

$\bf \textit{lateral area of a cone}\\\\ LA=\pi r\sqrt{r^2+h^2}~~ \begin{cases} r=radius\\ \sqrt{r^2+h^2}=slant~height\\ -----------\\ LA=574\pi \\ r=19.6 \end{cases} \\\\\\ 574\pi =\pi (19.6)\sqrt{r^2+h^2}\implies \cfrac{574\pi }{19.6\pi }=\sqrt{r^2+h^2} \\\\\\ \cfrac{574}{19.6}=\sqrt{r^2+h^2}$