Aregular pentagonal prism has a height of 14 in. and a base edge length of 10 in. identify its lateral area and surface area.
can anyone me solve this?

The lateral faces of the prism are 5 rectangles with dimensions 10 by 14.

So the lateral area = 5 * 10 * 14 =  700 (in squared)

Consider one of the bases of the prism.

As shown in the figure, the pentagon is made up of 5 isosceles triangles.

The vertex of each of these 5 triangles is 360°/5=72°.

Draw the altitude, which forms 2 right triangles with angle measures 36°-54°-90°.
 
Let he hypotenuse be s and the altitude be h in, as shown in the figure.

x*cos54°=5

x=5/(cos54°)=5/(0.588)=8.5

then, 

h=x*sin54°=8.5*(0.8)=6.8 (in)

Thus, the area of one of the 5 congruent isosceles triangles is 

(in squared)

The area of the pentagonal base is 34*5 = 170 (in squared)

Thus the surface area = lateral area + 2 base area
           = 700+2*170 = 700+340 = 1040 (in squared)

 

lateral area: 700 in squared

surface area: 1040 in squared

Google it that might work