McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
2. Surface Areas of Prisms and Cylinders
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Exercise 10 Page 818

Use the formula for the surface area of a prism.

Lateral Area: About 126 square meters
Surface Area: About 151.9 square meters

Practice makes perfect

Let's analyze the given solid. It is helpful to first to rotate the solid so that the hexagonal base is horizontal.

We assume that the section of the rhombus formed by the four segments in the base is a square and that the prism is a right prism.

We are asked to find its lateral and surface areas and round the answers to the nearest tenth if necessary.


Lateral Area

First, let's solve for the lateral area. It is equal to L= P h, where P is the perimeter of the base and h is the height of the prism. This tells us that h= 9 meters and P=2* 3+4* 2= 14 meters.
L= P h
L=( 14)( 9)
L= 126
Therefore, the lateral area is L= 126 square meters.

Surface Area

Now, let's solve for the surface area of the prism. It is equal to S= L+2B, where B is the area of the base. Let's analyze the base of the hexagon.

The base area is the sum of the area of square A_(□) (side length of 3 meters) and twice the area of △ ABC. By using the formula for the area of a square, we find that A_(□)=3^2=9 square meters. Now, let's find the area of △ ABC, A_(△), by the following formula. A_(△)=1/2AC* DB AC=3 meters. Since △ ABC is an isosceles triangle, BD is its altitude and it divides AC into two equal parts. This tells us that AD= 32=1.5. Now, let's use the Pythagorean Theorem for the right △ ADB.
AD^2+DB^2=AB^2
1.5^2+DB^2= 2^2
Solve for DB
2.25+DB^2=4
DB^2=1.75
sqrt(DB^2)=sqrt(1.75)
DB=sqrt(1.75)
DB=1.322875...
DB≈ 1.32
Now, let's find the area of △ ABC, A_(△).
A_(△)=1/2AC* DB
A_(△)=1/2( 3)( 1.32)
Simplify right-hand side
A_(△)=1/2* 3.96
A_(△)=3.96/2
A_(△)=1.98
Now, let's calculate the area of the base B.
B=A_(□)+2A_(△)
B= 9+2( 1.98)
Simplify right-hand side
B=9+3.96
B=12.96
We find that the area of the base is about B=12.96 square meters. The lateral area is L= 126 square meters. Finally, let's find the surface area of the solid, S. Remember that we should round the answer to the nearest tenth.
S= L+2B
Substitute values and evaluate
S= 126+2(12.96)
S=126+25.92
S=151.92
S≈ 151.9
The surface area of the given solid is about 151.9 square meters.