McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
4. Volumes of Prisms and Cylinders
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Exercise 52 Page 838

Use the formulas for the lateral and surface area.

Lateral Area: 212.1 ft^2
Surface Area: 255.4 ft^2

Practice makes perfect

The given solid is a pyramid.

To calculate the lateral area of a pyramid, we can use the known formula where P is the perimeter of the base and l is the slant height. L=1/2Pl The base of the pyramid is an equilateral triangle with the side length of 10ft. Let's calculate its perimeter!
P = 3s
P = 3( 10)
P = 30

Let's now find the slant height.

To find the slant height, we can use the Pythagorean Theorem. When doing this, notice that the slant height l is one of the arms of a right triangle. One of the edges of the pyramid is the hypotenuse and half of one of the edges of the base is the other leg.

Let's use these values to solve for l.
a^2+b^2=c^2
( 5)^2+( l)^2=( 15)^2
Solve for l
25+l^2=225
l^2=200
l = sqrt(200)
l = sqrt(100* 2)
l = 10sqrt(2)
Since a negative lateral height doesn't make sense, we only considered positive values of l. We can now calculate the lateral area of the pyramid by substituting P = 30 and l = 10sqrt(2) into the formula.
L = 1/2 P l
L = 1/2 ( 30)( 10sqrt(2))
L = 150sqrt(2)
L = 212.123034...
L = 212.1
We have found that the lateral area is 212.1ft^2. To find the surface area, all we have to do is add the base area B to the lateral area L. The base is an equilateral triangle. Let's find its height h using the Pythagorean Theorem.
We can see that the legs have measures a=5 and b=h, while the hypotenuse has measure c = 10. Let's substitute those values into the Pythagorean Theorem and solve for h.
a^2+b^2=c^2
( 5)^2+( h)^2=( 10)^2
Solve for h
25+h^2=100
h^2 = 75
h = sqrt(75)
h = sqrt(25*3)
h = sqrt(25)*sqrt(3)
h=5sqrt(3)
A negative height makes no sense, so we can consider only positive values of h. We can now calculate the area of the base using the formula for the area of a triangle. B = 1/2bh Let's substitute b=10 and h=5sqrt(3) into the formula and calculate B.
B = 1/2bh
B = 1/2( 10)( 5sqrt(3))
B = 25sqrt(3)
B = 43.301270...
B = 43.3
Now we can calculate the surface area of the pyramid.
S=L+B
S=212.1+43.3
S=255.4
The surface area of the pyramid is 255.4 ft^2.