# Answer to Question #7700 in Geometry for Mallory

Question #7700

An ice cream parlor uses waffle comes in the shape of right circular cones. The height of a chocolate dipped cone is 5 and the diameter is 3.

Some of the waffle cones are dipped, opening side down, into a 2-inch-deep bowl of chocolate. To the nearest tenth of a square inch, what is the area of the part of the waffle cone that is covered in chocolate?

Name all the geometric figures that are formed by horizontal or vertical cross sections of the cone. Explain where these cross sections are found.

Some of the waffle cones are dipped, opening side down, into a 2-inch-deep bowl of chocolate. To the nearest tenth of a square inch, what is the area of the part of the waffle cone that is covered in chocolate?

Name all the geometric figures that are formed by horizontal or vertical cross sections of the cone. Explain where these cross sections are found.

Expert's answer

An ice cream parlor uses waffle comes in the shape of right circular cones. The height of a chocolate dipped cone is 5 and the diameter is 3.

Some of the waffle cones are dipped, opening side down, into a 2-inch-deep bowl of chocolate. To the nearest tenth of a square inch, what is the area of the part of the waffle cone that is covered in chocolate?

The total area of a waffle is S = πRL, where R = 3/2 = 1.5 (hall of diameter, i.e. radius) and L = √(5²+3²) = √34 (length of the generatrix). So, S = 1.5√34π.

The area of dipped part of a waffle is s = πrl, here r/R = (5-2)/5 =& l/L (from the similarity of triangles). So, r = 3/5·R = 3/5·3/2 = 9/10, l = 3/5·L = 3√34/5 and s = π·9/10·3√34/5 = 27√34π/50.

Finally, area of dipped waffle is S - s = 1.5√34π - 27√34π/50 ≈ 17.6 in².

Some of the waffle cones are dipped, opening side down, into a 2-inch-deep bowl of chocolate. To the nearest tenth of a square inch, what is the area of the part of the waffle cone that is covered in chocolate?

The total area of a waffle is S = πRL, where R = 3/2 = 1.5 (hall of diameter, i.e. radius) and L = √(5²+3²) = √34 (length of the generatrix). So, S = 1.5√34π.

The area of dipped part of a waffle is s = πrl, here r/R = (5-2)/5 =& l/L (from the similarity of triangles). So, r = 3/5·R = 3/5·3/2 = 9/10, l = 3/5·L = 3√34/5 and s = π·9/10·3√34/5 = 27√34π/50.

Finally, area of dipped waffle is S - s = 1.5√34π - 27√34π/50 ≈ 17.6 in².

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