Exercise 8 Page 875

A tea bag can be modeled by the following square pyramid with a base edge of s= 4 centimeters and a height of l= 5 centimeters.

Since AB is half the side length, AB=2 centimeters. Now, let's use the Pythagorean Theorem for right △ ABC to find the height of the pyramid h.

AB^2+AC^2=BC^2

SubstituteIII

Substitute AB= 2, AC= h, BC= 5

2^2+ h^2= 5^2

▼

Solve for h

CalcPow

Calculate power

4+h^2=25

SubEqn

LHS-4=RHS-4

h^2=21

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt(h^2)=sqrt(21)

SqrtPowToNumber

sqrt(a^2)=a

h=sqrt(21)

Now, let's use the formula for the volume of a pyramid to find the volume of the tea bag, \textcolor{darkorange}{V_\text{tea bag}}. \begin{gathered} \textcolor{darkorange}{V_\text{tea bag}}=\dfrac{1}{3}\textcolor{darkviolet}{B}h \end{gathered} The variable B represents the area of the base. Since the base is a square with a side s=4, its area is B=4^2=16 square centimeters. Now, let's substitute the values into the formula for \textcolor{darkorange}{V_\text{tea bag}}.

\textcolor{darkorange}{V_\text{tea bag}}=\dfrac{1}{3}\textcolor{darkviolet}{B}h

▼

Substitute values and evaluate

SubstituteII

B= 16, h= sqrt(21)

\textcolor{darkorange}{V_\text{tea bag}}=\dfrac{1}{3}(\textcolor{darkviolet}{16})(\sqrt{21})

Multiply

Multiply

\textcolor{darkorange}{V_\text{tea bag}}=\dfrac{1}{3}\cdot 16\sqrt{21}

MoveRightFacToNumOne

1/b* a = a/b

\textcolor{darkorange}{V_\text{tea bag}}=\dfrac{16\sqrt{21}}{3}

UseCalc

Use a calculator

\textcolor{darkorange}{V_\text{tea bag}}=24.440403\ldots

RoundDec

Round to 1 decimal place(s)

\textcolor{darkorange}{V_\text{tea bag}}\approx \textcolor{darkorange}{24.4}

Therefore, the volume of the tea bag is about 24.4 square centimeters.