q's comes from module 2 week 4 challenge Q#20

[victorioussheep]

here is a question that relates to calculating a triangle's area when the question has already given you the three sides. My question is why choose 6K, 8K, and 10K? Why not use the original three numbers - 3K,4K, and 5K?

[quacker88]

Hey @victorioussheep !

So the area of the triangle is $K$, and we know this is equal to $\frac12 \cdot \text{(base)} \cdot \text{(height)}$. Keep in mind that altitude and height are the same thing, and the problem actually gives the values for the altitudes!

Since one of the altitudes is $\frac13$, we can say that

$K=\frac12 \cdot \text{(base)} \cdot \frac13$
$3K=\frac12 \cdot \text{(base)}$
$6K=\text{(base)}$

So, the length of the base that has altitude $\frac13$ is equal to $6K$.

We can actually do the same thing for the other two bases!

Base with altitude $\frac14$:
$K=\frac12 \cdot \text{(base)} \cdot \frac14$
$8K=\text{(base)}$

Base with altitude $\frac15$:
$K=\frac12 \cdot \text{(base)} \cdot \frac15$
$10K=\text{(base)}$

That means the three bases (which are the sides) of the triangle are $6K, 8K,$ and $10K$! Hope this cleared up the confusion

[victorioussheep]

@quacker88
okkkk..... so they basically use 6K, 8K, and 10K because they are the whole numbers so it's convenient to use?

[quacker88]

@victorioussheep not quite -- the numbers were actually calculated.

Remember, $K$ is the variable we set for the area. The goal of the problem is to find $K$. But here's the thing, we need the side lengths of the triangle to find the area, right? So, the calculations I did earlier were to find the side lengths of the triangle.

Remember, $\textbf{(area)} = \frac12 \cdot \textbf{(base)} \cdot \textbf{(height)}$, right? (that rhymes haha)

Well, we set the area as $K$ earlier, and the problem says that the length of one of the altitudes (the height) is equal to $\frac13$. Here's a (very not to scale) sketch of what we have so far:

Well, we know the area and we know the height, so let's solve for the base!

$\text{(area)} = \frac12 \cdot \text{(base)} \cdot \text{(height)}$
$K=\frac12 \cdot \text{(base)} \cdot \frac13$
$6K=\text{(base)}$

So the length of the base is $6K$. Remember, we don't know exactly what $K$ is yet (it's what we're solving for!) but we do know that the base is equal to $6K$.

You can do this process for all three sides and you'll get that the side lengths are $6K, 8K,$ and $10K$! Does this make a little more sense now?

[victorioussheep]

@quacker88
oh yes! it definitely makes more sense now! I really like how you use visuals in between your explanation! It makes me understand the concept a lot quicker, and thanks again!

[quacker88]

@victorioussheep Glad that it helped!