# Tuesday, April 11

Assignment: 10.4 on p. 532: 1-11. Topic: Density and problem solving with volume.

Topic: Density and problem solving with volume. Students should be able to use equations to find the missing lengths using the volume. For example in a triangular pyramid with volume 180, height of the base triangle = 12, and altitude of the pyramid = 6, we could first notice that the volume of a pyramid is

$\dpi{300}\inline V = \dfrac{1}{3}B \cdot h$

And then that $\dpi{300}\inline B$ is the area of the triangular base, so $\dpi{300}\inline B = \dfrac{1}{2}bh_2$. So we are essentially solving

$\dpi{300}\inline V = \dfrac{1}{3} \cdot \dfrac{1}{2} \cdot b \cdot h_2 \cdot h$

And we know all the values of variables except for $\dpi{300}\inline b$ so we can re-arrange by the commutative property of multiplication

$\dpi{300}\inline V = \dfrac{1}{3} \cdot \dfrac{1}{2} \cdot h_2 \cdot h \cdot {\color{red}b}$

With the known information we have

$\dpi{300}\inline 180 = \dfrac{1}{3} \cdot \dfrac{1}{2} \cdot 12 \cdot 6 \cdot {\color{red}b}$

So

$\dpi{300}\inline 180 = \dfrac{1}{6} \cdot 12 \cdot 6 \cdot {\color{red} b}$

Then

$\dpi{300}\inline 180 = 12{\color{red} b}$

And finally

$\dpi{300}\inline 15 = {\color{red} b}$