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Problem 833

Prove: The lateral area of a regular pyramid is equal to one-half the product of its slant height and the perimeter of I its base.

Expert verified

The lateral area of a regular pyramid can be found by dividing the pyramid into 'n' isosceles triangles, where 'n' is the number of sides of the base. The area of one isosceles triangle is calculated as Area = \( (1/2) * a * l \), where 'a' is the side length and 'l' is the slant height. The total lateral area is the sum of the areas of all isosceles triangles, which can be written as \( A_l = n * (1/2) * a * l \). Replacing \( a * n \) with the perimeter of the base 'P', we get \( A_l = (1/2) * l * P \), which proves that the lateral area of a regular pyramid is equal to one-half the product of its slant height and the perimeter of its base.

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Chapter 48

The base of a right prism is a regular hexagon with area \(24 \sqrt{3}\). If the lateral faces of the prism are squares, what is the lateral area?

Chapter 48

Show that the lateral area of a prism is equal to the product of the perimeter of a right section and the length of a lateral edge.

Chapter 48

A sphere has radius \(7 .\) What is the area enclosed by a spherical triangle whose angles have measures $\mathrm{a}=100^{\circ}, \mathrm{b}=120^{\circ}, \mathrm{c}=140^{\circ} ?\( [Take \)\left.\pi=(22 / 7)\right]$.

Chapter 48

Find the surface area of a regular tetrahedron when each edge is of length a) \(1 ;\) b) 2 .

Chapter 48

Find the lateral area and the total area of a right circular cone in which the radius measures 14 in. and the slant height measures 20 in. [use $\pi=(22 / 7)]$.

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