@quacker88 Thank you for your response! It helps a lot! 😁

@CoherentMango Thanks for asking! 🙂 This is true whenever you have two triangles that share a baseline and a common vertex that meets the altitude. I've colored the baseline \(\textcolor{red}{\text{ red }} \) below. The common vertex is labeled \( A.\)

The obtuse \(\textcolor{purple}{\text{ purple }} \) triangle has an altitude which is perpendicular to the baseline, emanating from point \(A,\) shown as the dotted \(\textcolor{purple}{\text{ purple }} \) line. Similarly, the green acute triangle has its altitude perpendicular to the baseline and emanating from the point \(A,\) shown as a dotted \(\textcolor{green}{\text{ green } } \) line.

M2W3D13-ch-part-3-why-two-triangles-same-height.png

Turned around, it might be easier to see that they share a common baseline and altitude!

🙂

@TSS-Graviser Hi again!

You have a good point! If you're just in the middle of a calculation, and you get a fraction with a radical in the denominator, but it's only an intermediate step, you don't really have to take out the square root from the bottom. It's just for you to look at, right?

$$ \frac{2}{\sqrt{5} - 1} $$

cartoon-po-happier-30-percent.jpg
"I would probably leave it like that." -- Prof. Loh
 

However, if your final answer has a square root in the bottom of the fraction, I'd recommend simplifying to remove it. This is called "rationalizing the denominator." It's perhaps easier for teachers to check students' answers if they all are in the same format.

Like, imagine if you were a teacher, and your students gave you answers that looked like:

$$ \frac{1}{2+1}, \text{ } \frac{3-2}{4-1}, \text{ } \frac{0.5}{1.5}, \text{ and } \frac{5}{15}$$

These are all equal to \(\frac{1}{3},\) but it wouldn't be a lot of fun to check that!

The reason might be just due to convention, but there are some discussions on other websites explaining that, to name a reason, it's easier to add fractions when their denominators are integers. For example, it's easier to add

$$ \frac{\sqrt{3}}{3} + \frac{\sqrt{6}-\sqrt{3}}{3} $$

compared with adding

$$ \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{6}}. $$

cartoon-po-happier-30-percent.jpg

"That's ugly!" says Prof. Loh