Here's the question you clicked on:
rebeccaskell94
In △PQR, what is the length of line segment QR? http://learn.flvs.net/webdav/assessment_images/educator_geometry_v14/pool_Geom_3641_0404_02/image0064e2eda71.gif 28 28radical 2 56radical 3 56radical 2 *I think the answer is A?*
this is a "special right triangle" 45-45-90 The idea is to remember that the hypotenuse (longest side, opposite the 90 deg angle) is sqrt(2) times bigger than either leg (which are equal) h = a*sqrt(2) or (divide by sqrt(2) ) a= h/sqrt(2)
or (because people do not sqrt(2) in the denominator) a= sqrt(2)*h/(sqrt(2)*sqrt(2)) (multiply top and bottom by sqrt(2)) you do this to get rid of the sqrt(2) in the bottom a= sqrt(2)*h/2
Whoa. That was a lot.
too much to get the answer?
I'm not sure. I'm looking at it like @ . @
the side is h/sqrt(2)= 56/sqrt(2) but none of your answers are in that form. so we have to "rationalize" to get an answer that matches.
@Parthkohli this one :D If you are given a leg of a 45-45-90 triangle, you just put 2√ in front of it to get the hypotenuse ^_^ • If the leg is 8, then the hypotenuse is 82√ • If the leg is 1281283283283249483483, then the hypotenuse is 12812832832832494834832√
The short-cut that I gave you doesn't work here too much, FYI. I just told you in simple words to remove/put sqrt2. What that short-cut actually says is that you have to multiply sqrt2 to get the hypotenuse and divide sqrt2 from the hypotenuse to get the leg.
can we just use trigo. here? \[\cos \theta =\frac{adjacent}{hyp}\]
So, you divide \(\sqrt2\) from the hypotenuse to get the leg.
Or, the Pythagorean. \( \color{Black}{\Rightarrow a^2 + a^2 = 56^2 }\) \( \color{Black}{\Rightarrow 2a^2 = 3136}\)
If you ever get confused with that short-cut, just come back to Pythagorean ;)
Ah! okay um wait...that just goes back to 28?
\( \color{Black}{\Rightarrow a^2 = 1568}\) You find \(\sqrt{1568}\).
Okay √1568 = 39.5 and 28√2 = 39.5 as well :D