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Dylan is a skater who enjoys exploring the math behind the ramps, obstacles, and shadows at a local skate park. One sunny day, he took a break under a tall tree and noticed his shadow next to the tree's shadow. This sparked his curiosity about the tree's height, and he decided to use its shadow to calculate it.

Solution

Begin by labeling the vertices for each of the triangles in the diagram.

In this diagram, $\overline{AD}$ corresponds to $\overline{AE},$ and $\overline{BD}$ corresponds to $\overline{CE}.$ As the triangles $ABD$ and $ACE$ are similar, the ratio between their corresponding sides will be the same.

$$\begin{array}{c}\frac{AD}{AE}=\frac{BD}{CE}\end{array}$$

Note that $AE$ equals the sum of $AD$ and $DE.$

$$\begin{array}{c}\frac{AD}{AD+DE}=\frac{BD}{CE}\end{array}$$

In this case, $AD,$ $BD,$ $AD,$ and $DE$ are known. Substitute these values into the proportion and solve for $CE$ to find the height of the tree.

$\frac{AD}{AD+DE}=\frac{BD}{CE}$

SubstituteValues

Substitute values

$\frac{13}{13+66}=\frac{6.5}{CE}$

AddTerms

Add terms

$\frac{13}{79}=\frac{6.5}{CE}$

MultEqn

$\text{LHS}\cdot CE=\text{RHS}\cdot CE$

$\frac{13}{79}\cdot CE=6.5$

MultEqn

$\text{LHS}\cdot 79=\text{RHS}\cdot 79$

$13\cdot CE=513.5$

DivEqn

$\text{LHS}/13=\text{RHS}/13$

$CE=39.5$

Dylan determined that the height of the tree is $39.5$ feet.

Dylan is enjoying his time skating at the park. He drops-in on a triangular-shaped ramp. He begins his descent from a height of $15$ feet, which is also the height of the ramp. After he glides down the slope for $20$ feet, he realizes that he is now only $70\%$ as high as when he started.

Solution

The problem states that the height of a ramp is $15$ feet. Dylan rides down $20$ feet from the top of the ramp and lands at a height that is $70\%$ of his initial height. Dylan's current height can be found by multiplying his initial height by $0.70.$

$$\begin{array}{c}\text{Dylan’s Current Height}\\ 0.70\cdot 15=10.5\text{ ft}\end{array}$$

This means that his current height is $10.5$ feet. Use the base of the ramp, its height, the distance that Dylan covered, Dylan's current height, and the slope of the ramp to draw a diagram of the situation.

Based on the diagram, it can be seen that both triangles $ABC$ and $DBE$ have a right angle each and share a common angle at vertex $B.$ Therefore, these triangles are similar because of the Angle-Angle Similarity Theorem.

$$\begin{array}{c}△ABC\sim △DBE\end{array}$$

Consequently, $\overline{DE}$ corresponds to $\overline{AC}$ and $\overline{BE}$ corresponds to $\overline{BC}.$

$$\begin{array}{ccc}\overline{DE}& \leftrightarrow & \overline{AC}\\ \overline{BE}& \leftrightarrow & \overline{BC}\end{array}$$

Additionally, $BC$ equals the sum of $CE$ and $BE.$ Since the ratios between the lengths of corresponding sides of the triangles are equal, the following equation holds true.

$$\begin{array}{c}\frac{DE}{AC}=\frac{BE}{BC}\\ \Downarrow \\ \frac{DE}{AC}=\frac{BE}{CE+BE}\end{array}$$

In this equation, only $BE$ is unknown. The known values can be substituting in this proportion to find the value of $BE.$

$\frac{DE}{AC}=\frac{BE}{CE+BE}$

SubstituteValues

Substitute values

$\frac{10.5}{15}=\frac{BE}{20+BE}$

CalcQuot

Calculate quotient

$0.7=\frac{BE}{20+BE}$

MultEqn

$\text{LHS}\cdot \left(20+BE\right)=\text{RHS}\cdot \left(20+BE\right)$

$0.7\left(20+BE\right)=BE$

Distr

Distribute $0.7$

$14+0.7\cdot BE=BE$

SubEqn

$\text{LHS}-0.7BE=\text{RHS}-0.7BE$

$14=0.3\cdot BE$

RearrangeEqn

Rearrange equation

$0.3\cdot BE=14$

DivEqn

$\text{LHS}/0.3=\text{RHS}/0.3$

$BE=\frac{14}{0.3}$

UseCalc

Use a calculator

$BE=46.6\overline{6}$

After solving for $BE,$ the length of $\overline{BE}$ can be added to $CE$ to determine the length of the slope.

$$\begin{array}{c}\text{The Slope’s Length}\\ 20+46.6\overline{6}\approx 66.67\text{ feet}\end{array}$$

Dyland found that the length of the slope is approximately $66.67$ feet.

Later that day, Dylan returned home and realized that he still forgot about his math homework. He needs to finish it so that he can have free time on Saturday to go skateboarding again. His homework requires him to compare a pair of slope triangles that lie on the same line.

Consider that the side of each square in the grid is $1$ unit. Help Dylan complete his homework.

Solution

a Consider that $△ABC$ and $△BDE$ are similar as they are slope triangles on the same line.

$$\begin{array}{c}△ABC\sim △BDE\end{array}$$

Due to this similarity, the ratio between corresponding sides is equal. In this case, $AC$ corresponds to $BE$ and $BC$ corresponds to $DE.$ This information can be used to write the following proportion.

$$\begin{array}{c}\frac{AC}{BE}=\frac{BC}{DE}\end{array}$$

Since it is asked to compare the rise to the run for each of the similar slope triangles, this proportion will be rewritten to have the ratio of the rise to the run of $△ABC$ on one side and the ratio of the rise to the run of $△BDE$ on the other side.

$\frac{AC}{BE}=\frac{BC}{DE}$

MultEqn

$\text{LHS}\cdot BE=\text{RHS}\cdot BE$

$AC=\frac{BC}{DE}\cdot BE$

MultEqn

$\text{LHS}\cdot DE=\text{RHS}\cdot DE$

$AC\cdot DE=BC\cdot BE$

DivEqn

$\text{LHS}/BC=\text{RHS}/BC$

$\frac{AC\cdot DE}{BC}=BE$

DivEqn

$\text{LHS}/DE=\text{RHS}/DE$

$\frac{AC}{BC}=\frac{BE}{DE}$

This indicates that the ratios of the rise to the run are equal in both slope triangles.

b In the previous part, the proportion that compares the rise to the run for each of the similar slope triangles was written.

$$\begin{array}{c}\frac{AC}{BC}=\frac{BE}{DE}\end{array}$$

The given diagram will be used to find the numerical value of this proportion. Count the number of units it takes to get from point $A$ to point $C$ to find the length of $\overline{AC}.$ Similarly, count the number of units it takes to get from point $B$ to point $C$ to find the length of $\overline{BC}.$

This means that $AC=3$ and $BC=2.$ Apply a similar reasoning to find the lengths of $\overline{BE}$ and $\overline{DE}.$

Once of the lengths in the proportion are known, substitute them into the proportion to find its numerical value.

$$\begin{array}{c}\frac{AC}{BC}=\frac{BE}{DE}\\ \Downarrow \\ \frac{3}{2}=\frac{6}{4}\end{array}$$

At Dylan's skate park, there is a cool shelter that covers the entire skateboard track, providing shade for everyone during their skate sessions. Seeing the roof line, Dylan thinks it is the perfect chance to apply his new school knowledge about slope triangles on the same line.

With the help of a friend, Dylan measured the rise and run at various points on one of the slopes of the roof and sketched a diagram.

Now, he is curious whether the slope of the roof line remains consistent over the different areas. Help Dylan figure out the roof's slope and see if it remains the same everywhere.