Today, we learned to solve some proportion-related problems.

Similar Triangles

Similar triangles is one of the more basic ones. You have to solve only one simple proportion to figure it out.

If you have similar triangles, the ratios of the corresponding sides are the same. For example, if \(\triangle ABC \sim \triangle XYZ\) (the order does matter, it tells you which sides are corresponding), \(\frac{AB}{AC} = \frac{XY}{XZ}\).

Basically, if you divide any side of one triangle by another side of the same triangle, that fraction is the same for the corresponding sides of the other triangle.

Then, you just create a proportion with the side you’re solving for in it, cross-multiply, and solve.

Other Simple Proportions

These types of proportions are even simpler than the similar triangle proportions.

Basically, they give you an equation to solve that looks like this: \(\frac{a}{b} = \frac{x}{c}\).

To solve this, nothing special is needed: just cross-multiply.

Speed Comparison Problems

Another harder type of problem involves the \(distance = rate * time\) equation.

It goes like this: Some things do the rate in the same time period. One of them has a distance of \(a\), and the other one has a distance of \(b\). One of them has a rate of \(r\), and the other one has a rate \(x\) higher. What is the rate of both of them?

We have two unknowns here: what we’re trying to find, \(x\), and \(t\). To simplify this, we could cancel out \(t\) because it appears in both equations.

To do this, find \(t\) in terms of the other variables. This gives: \(\frac{a}{r} = t\) and \(\frac{b}{r+x} = t\). Then, the equation two equations can be equated, and now, you have a “Other Simple Proportions” to solve.

Working Together Time Problems?

Say you have two people working together. One of them can finish a job in \(a\) hours, and the other, \(b\) hours. How fast could they finish a job together?

We could add the two times together, but that doesn’t make sense: their time should become lower. We could take the average of their times, but that also doesn’t make sense if we consider the case where they can do the job in the same amount of time.

Basically, time doesn’t add up well in these types of problems. Looking at the distance-rate-time equation again, we see something that does add up well: rate.

For example, if you have a 60 mph car and you add a rocket booster to it that can add 40 mph, you will have a 100-mph rocket car (not counting inefficiencies like friction), not a 50 mph car or a 20 mph car.

In the distance-rate-time equation, we can assume that the distance is \(1\), for \(100%\) completed. That means that person 1’s speed would be \(d/t = 1/a\), and the same for person 2.

Then, we can add the speeds together for their combined speed. If we equate this to \(d/t = 1/t\), we get the equation: \(\frac{1}{a} + \frac{1}{b} = \frac{1}{t}\). Solving for t, we take the reciprocal of everything. Then, we see that the time is the reciprocal of the sums of the reciprocals of the individual times for both of them. This is kind of like the harmonic mean, and the harmonic mean does come up sometimes in some distance-speed-time problems, so that makes sense.

Basically, time doesn’t add up well, so we add up speed instead. After adding up the speed, we can find the time for them together.

Conclusion

If you were wondering why there haven’t been any new posts for a long time, I’ve been kind of busy, and there hasn’t been much intuition to cover. If you were wondering, we learned to solve rational functions, which is basically just multiplying by some number to cancel out the denominators and solving the remaining polynomial.