Surface Area of a Cone

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The surface area of a cone is equal to the curved surface area plus the area of the base: $\pi r^2 + \pi L r,$ where $r$ denotes the radius of the base of the cone, and $L$ denotes the slant height of the cone. The curved surface area is also called the lateral area.

In the cone below, $h$ is the height, $L$ is the slant height, $c$ is the circumference of the base, and $r$ is the radius of the base. How does it look when we unroll it?

If we unroll it, the shape is as follows: It is a sector of a circle with radius $L$ and arc length $c$. So the curved surface area of the cone is the area of the sector above. The area of a sector given the arc length $c$ and radius $L$ is given by $A=\dfrac{1}{2}cL$. Now applying this to the cone, we have $A=\frac{1}{2}cL,$ where $L$ is the slant height and $c$ is the circumference of the base. After some manipulations, $A=\pi Lr,$ as given in the definition.

A cone was formed by rolling a thin sheet of metal in the form of a sector of a circle with diameter $72\text{ cm}$ and central angle $150^\circ$. Find the curved surface area of the cone formed.

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The curved surface area is the area of the sector of a circle, so we have

$$A=\dfrac{\theta}{360}(\text{area of the circle})=\dfrac{150}{360}(\pi)\big(36^2\big)=540 \pi.\ _\square$$

A pile of sand is in the shape of a right circular cone. It has a height of $3$ feet and a base diameter of $8$ feet. Find its surface area in square feet.

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After some manipulations, we can see that a 3-4-5 right triangle is formed. Therefore, the slant height is $5$ feet. So,

$$A=\pi r^2+\pi L r=\pi \big(4^2\big) +\pi(5)(4)=36 \pi.\ _\square$$

What is the lateral area of a cone that has a height of $8$ and a radius of $5?$

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The first step is to compute the slant height $L.$ Applying Pythagorean theorem, we have

$$L=\sqrt{8^2+5^2}=\sqrt{64+25}=\sqrt{89}.$$

Applying the formula, we have

$$A= \pi Lr=\pi \big(\sqrt{89}\big)(5)=5 \pi \sqrt{89}.\ _\square$$