Question

identify the solid that is represented by the net. then find its surface area. (all lengths are in feet.)

Explanation:

Step1: Identify the solid

The net represents a triangular - pyramid. It has a triangular base and 3 triangular faces meeting at a vertex.

Step2: Calculate the area of the base

The base is a triangle with base $b = 6$ feet and height $h = 5.2$ feet. The area of a triangle is $A_{base}=\frac{1}{2}\times b\times h$. So, $A_{base}=\frac{1}{2}\times6\times5.2 = 15.6$ square feet.

Step3: Calculate the area of the lateral faces

There are 3 lateral faces. Each lateral face has a base equal to a side of the base - triangle and height equal to 5 feet.
The three lateral - face areas:
For the first lateral face with base 6 feet: $A_1=\frac{1}{2}\times6\times5 = 15$ square feet.
For the second and third lateral faces (assuming the base - triangle is isosceles or we can calculate each separately), if we consider the other sides of the base - triangle, and using the same formula for the area of a triangle $A=\frac{1}{2}\times base\times height$, and since the height of each lateral face is 5 feet. Let's assume the base - triangle has sides such that the sum of the areas of the three lateral faces: $A_{lateral}=3\times\frac{1}{2}\times base\times5$. Since one base is 6 feet and assuming the other two are equal (from the symmetry of the net), we have $A_{lateral}=3\times\frac{1}{2}\times6\times5=45$ square feet.

Step4: Calculate the surface area

The surface area $A$ of the triangular - pyramid is the sum of the base area and the lateral - face areas. So, $A = A_{base}+A_{lateral}=15.6 + 45=60.6$ square feet.

Answer:

The solid is a triangular pyramid and its surface area is 60.6 square feet.