Exercise 14 Page 134

We want to know how long it will take us to swim across the pond from point D to point E. Let's take a look at the given picture.

We will begin by finding the length of side DE. Next, using the information from the exercise, we will calculate the time it takes for us to swim across the pond. First, notice that ∠ B is congruent to ∠ E. Since they are both right angles and their measures are 90^(∘). ∠ D = ∠ E = 90^(∘)

From the picture, we can see that ∠ ACB and ∠ DCE are vertical angles. We know that vertical angles are congruent. Let's mark this on our graph.

We found that two angles in △ ABC are congruent to two angles in △ CDE. Because of this, the third angles are also congruent, which means that triangles are similar. Now we can use the fact that corresponding sides in similar triangles have equivalent ratios. Let's take a look at what this means in our case. AB/BC=DE/CE [0.3em] ⇕ [0.3em] 105/140=d/80 Let's solve the equation for d.

105/140=d/80

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Solve for d

MultEqn

LHS * 140=RHS* 140

105/140* 140=d/80* 140

FracMultDenomToNumber

a/140* 140 = a

105=d/80* 140

MoveRightFacToNum

a/c* b = a* b/c

105=140d/80

CalcQuot

Calculate quotient

105=1.75d

DivEqn

.LHS /1.75.=.RHS /1.75.

105/1.75=1.75d/1.75

CalcQuot

Calculate quotient

60=d

RearrangeEqn

Rearrange equation

d=60

We found that distance from point D to point E is 60 meters. From the exercise, we know that we can swim at 3.6 kilometers per hour, or 3600 meters per 60 minutes. Let's use this information to create a proportion and find out how long it will take us to swim across the pond. We will mark the time needed to swim across the pond as y. 3600/60=60/y Let's solve this equation for y!

3600/60=60/y

▼

Solve for y

ReduceFrac

a/b=.a /60./.b /60.

60=60/y

MultEqn

LHS * y=RHS* y

60y=60

DivEqn

.LHS /60.=.RHS /60.

y=1

It will take us 1 minute to swim across the pond.