Question 1188478
<font face="Times New Roman" size="+2">


I presume you meant *[tex \Large \sqrt{288}] when you wrote 288-------√.


The appropriate plain text notation for the square root of an argument is sqrt{argument}.  So you would have been more understandable if you had written sqrt{288}.  Please use standard plain text notation in the future so that we don't have to guess what you mean.


The measures of the three sides of a right triangle are related by the Pythagorean Theorem thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ c^2\ =\ a^2\ +\ b^2]


where *[tex \Large c] is the measure of the hypotenuse and *[tex \Large a] and *[tex \Large b] are the measures of the other two legs.


You are given the relationship between the hypotenuse and one of the legs.  We can call the leg side *[tex \Large a] because the labels on the legs are arbitrary.


So given that the measure of leg a is *[tex \Large a], the measure of the hypotenuse is *[tex \Large 3a].


Rearranging Pythagoras' formula, we get:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ c^2\,-\,a^2\ =\ b^2]


Since we are given *[tex \Large b\ =\ \sqrt{288}], then *[tex \Large b^2\ =\ 288]


Furthermore, since we know that *[tex \Large c\ =\ 3a], we can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (3a)^2\ -\ a^2\ =\ 288]


Solve for *[tex \Large a] and then calculate *[tex \Large 3a]


																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
*[illustration darwinfish.jpg]

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
I > Ø
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}
</font>