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Big Ideas Math Geometry, 2014

BI

Big Ideas Math Geometry, 2014 View details

6. Volumes of Pyramids

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Exercise 23 Page 640

Find the area of the base of the pyramid.

About 9.22 cubic feet

Practice makes perfect

We are given the following regular pentagonal pyramid.

We are asked to find the volume of the given pyramid and round the answer to the nearest hundredth. We will use the formula for the volume of a pyramid. V=B h/3 Here, the variable B is the area of the base, and the variable h is the height of the prism. First, we will find the area of the base. Next, the height.

Base

To find 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 3 feet. This tells us that P=5* 3=15 feet. To find a, let's divide the base into five congruent isosceles triangles.

Notice the following.

The angle ∠ AFB is a central angle, so m∠ AFB= 360^(∘)5=72^(∘).
Since FG is an apothem, it is an altitude of isosceles Δ AFB. This tells us that FG bisects ∠ AFB. Therefore, m∠ AFG= 12m∠ AFB=36^(∘).
By the Triangle Sum Theorem, the sum of the angle measures of right Δ GAF is 180^(∘).

m∠ GAF+m∠ AFG+m∠ FGA=180^(∘)

m∠ AFG= 36^(∘), m∠ FGA= 90^(∘)

m∠ GAF+ 36^(∘)+ 90^(∘)=180^(∘)

▼

Solve for m∠ GAF

Add terms

m∠ GAF+126^(∘)=180^(∘)

LHS-126^(∘)=RHS-126^(∘)

m∠ GAF=54^(∘)

To find a, let's use trigonometric ratios for right Δ GAF.

tan(m∠ GAF)=FG/AG

m∠ GAF= 54^(∘), AG= 1.5

tan 54^(∘)=FG/1.5

▼

Solve for FG

LHS * 1.5=RHS* 1.5

1.5tan 54^(∘)=FG

Rearrange equation

FG=1.5tan 54^(∘)

Use a calculator

FG=2.06457...

Round to 2 decimal place(s)

FG≈ 2.06

Therefore, the apothem is about 2.06 feet. Now, we can substitute the values into the formula for the area of the base.

B=1/2aP

▼

Substitute values and evaluate

a= 2.06, P= 15

B=1/2( 2.06)( 15)

Multiply

B=1/2(30.9)

MoveRightFacToNum

a/c* b = a* b/c

B=1(30.9)/2

1* a=a

B=30.9/2

Calculate quotient

B=15.45

The area of the base is about 15.45 square feet.

Height

To find the height, first, let's find BF.

To find BF, let's use trigonometric ratios for right Δ BFG.

sin(m∠ GBF)=a/BF

m∠ GBF= 54^(∘), a= 2.06

sin 54^(∘)=2.06/BF

▼

Solve for BF

LHS * BF=RHS* BF

BFsin 54^(∘)=2.06

.LHS /sin 54^(∘).=.RHS /sin 54^(∘).

BF=2.06/sin 54^(∘)

Multiply

BF=2.546300...

Round to 2 decimal place(s)

BF≈ 2.55

Now, let's look at the pyramid.

To find the height of the pyramid, h=SF, let's use trigonometric ratios for right Δ SFB.

tan(m∠ FBS)=SF/BF

m∠ FBS= 35^(∘), BF= 2.55

tan 35^(∘)=SF/2.55

▼

Solve for SF

LHS * 2.55=RHS* 2.55

2.55tan 35^(∘)=SF

Rearrange equation

SG=2.55tan 35^(∘)

Multiply

SG=1.785529...

Round to 2 decimal place(s)

SF≈ 1.79

Therefore, the height of the pyramid, h=SF, is about 1.79 feet.

Volume

Finally, we can substitute 15.45 for B, and 1.79 for h into the formula for the volume of the given solid.

V=B h/3

▼

Substitute values and evaluate

B= 15.45, h= 1.79

V=15.45( 1.79)/3

Multiply

V=27.6555/3

Calculate quotient

V=9.2185

Round to 2 decimal place(s)

V≈ 9.22

Therefore, the volume of the given solid is about 9.22 cubic feet.

Subchapter links

Communicate Your Answer p.635

Monitoring Progress p.637-638

Exercises p.639-640