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

  • M0★ M1★ M2★ M3★ M4

    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?

  • MOD

    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

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

    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 \(\frac14\):
    \(K=\frac12 \cdot \text{(base)} \cdot \frac14\)

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

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

  • M0★ M1★ M2★ M3★ M4

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

  • MOD

    @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!

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

    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?

  • M0★ M1★ M2★ M3★ M4

    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!

  • MOD

    @victorioussheep Glad that it helped! 😄