## rebeccaskell94 Group Title 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?* 2 years ago 2 years ago

1. phi Group Title

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)

2. phi Group Title

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.

4. phi Group Title

too much to get the answer?

I'm not sure. I'm looking at it like @ . @

6. phi Group Title

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√

8. ParthKohli Group Title

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.

9. kaiz122 Group Title

can we just use trigo. here? $\cos \theta =\frac{adjacent}{hyp}$

10. ParthKohli Group Title

So, you divide $$\sqrt2$$ from the hypotenuse to get the leg.

11. kaiz122 Group Title

12. ParthKohli Group Title

Or, the Pythagorean. $$\color{Black}{\Rightarrow a^2 + a^2 = 56^2 }$$ $$\color{Black}{\Rightarrow 2a^2 = 3136}$$

13. ParthKohli Group Title

If you ever get confused with that short-cut, just come back to Pythagorean ;)

Ah! okay um wait...that just goes back to 28?

15. ParthKohli Group Title

$$\color{Black}{\Rightarrow a^2 = 1568}$$ You find $$\sqrt{1568}$$.

16. ParthKohli Group Title

Okay √1568 = 39.5 and 28√2 = 39.5 as well :D

18. ParthKohli Group Title

Yep :)