McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
5. Volumes of Pyramids and Cones
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Exercise 25 Page 845

Use the formula for the volume of a pyramid.

About 32.13 cubic feet

Practice makes perfect

A greenhouse is a regular octagonal pyramid with a height of h=5 feet. The base has side lengths of 2 feet.

We are asked to find the volume of the greenhouse. To do this, first, we must find the area of its base. Then, we will use the formula for the volume of a pyramid.

Area of Base

Let's analyze the base of the greenhouse.

The base is a regular octagon. To find its area B we will use the formula for the area of a regular polygon. B= 12aP The variable a represents the apothem and P represents the perimeter of the polygon. Each base edge is 2 feet. This tells us that P=8* 2=16 feet. To find a, let's divide the base into eight congruent isosceles triangles.

Notice the following.

  • The angle ∠ ACB is a central angle, so m∠ ACB= 360^(∘)8=45^(∘).
  • Since CD is an apothem, it is an altitude of isosceles Δ ABC. This tells us that CD bisects ∠ ACB. Therefore, m∠ ACD= 12m∠ ACB=22.5^(∘).
  • The altitude CD bisects AB, too. Therefore AD= 12AB=1 foot.
To find a=CD, let's use the trigonometric ratios for right △ ADC.
cot(m∠ ACD)=CD/AD
cot 22.5^(∘)=CD/1
Solve for CD
cot 22.5^(∘)=CD
CD=cot 22.5^(∘)
CD=2.414213...
CD≈ 2.41
Therefore, the apothem is about a≈ 2.41 feet. Now, we can substitute the values into the formula for the area of the base.
B=1/2aP
Substitute values and evaluate
B=1/2( 2.41)( 16)
B=1/2* 38.56
B=38.56/2
B=19.28
The area of the base is about 19.28 square feet.

Volume of Greenhouse

The area of the base of the greenhouse is B=19.28 square feet, and its height is h=5 feet. Now, let's use the formula for the volume of a pyramid to find the volume of the greenhouse.
V=1/3Bh
Substitute values and evaluate
V=1/3(19.28)(5)
V=1/3* 96.4
V=96.4/3
V≈ 32.1333...
V≈ 32.13
Finally, the volume of the greenhouse is about 32.13 cubic feet.