March 18th, 2015, 03:16 PM  #1 
Newbie Joined: Mar 2015 From: fl Posts: 3 Thanks: 1  Right Triangle Driving me mad
So I have a problem driving me crazy. There is a right triangle. Side B is given, it is 13.5. Side A and C are not given, nor are any angles besides the right angle. The only other information I am given is that side C is three times side A. So, if you multiplied side A by three, you would get side C, the hypotenuse. What formula do I use to solve this?? I drew a picture to help people understand. http://puu.sh/gG2Z3/ef4f7df2a7.jpg 
March 18th, 2015, 03:21 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,184 Thanks: 481 Math Focus: Calculus/ODEs 
Use the Pythagorean theorem as follows: $\displaystyle A^2+B^2=(3A)^2$ $\displaystyle A^2+B^2=9A^2$ $\displaystyle 8A^2=B^2$ $\displaystyle A^2=\frac{B^2}{8}$ $\displaystyle A=\frac{B}{2\sqrt{2}}$ Since you know $B$, you now know $A$ and $C=3A$. 
March 18th, 2015, 03:28 PM  #3 
Newbie Joined: Mar 2015 From: fl Posts: 3 Thanks: 1 
I don't understand it completely, could you explain it a bit more please? Where did the 8 come from in the third line, would I substitute 8 for the 13.5? 
March 18th, 2015, 03:35 PM  #4 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,184 Thanks: 481 Math Focus: Calculus/ODEs 
After squaring $3A$ in the right side, we have: $\displaystyle A^2+B^2=9A^2$ Now, subtract $A^2$ from both sides to get: $\displaystyle A^2A^2+B^2=9A^2A^2$ $\displaystyle B^2=8A^2$ Divide both sides by $8$: $\displaystyle \frac{B^2}{8}=\frac{8A^2}{8}$ $\displaystyle \frac{B^2}{8}=\frac{\cancel{8}A^2}{\cancel{8}}$ $\displaystyle \frac{B^2}{8}=A^2$ If $a=b$ then $b=a$ so reverse sides to put the variable for which we are solving on the left: $\displaystyle A^2=\frac{B^2}{8}$ Take the square root of both sides, and discard the negative root since all quantities represent measures of the sides of a triangle: $\displaystyle \sqrt{A^2}=\sqrt{\frac{B^2}{8}}$ $\displaystyle A=\frac{\sqrt{B^2}}{\sqrt{8}}$ $\displaystyle A=\frac{B}{\sqrt{2^2\cdot2}}$ $\displaystyle A=\frac{B}{\sqrt{2^2}\cdot\sqrt{2}}$ $\displaystyle A=\frac{B}{2\sqrt{2}}$ Now, to find $A$, plug in the value of $B$, and then to get $C$, multiply $A$ by $3$. 
March 18th, 2015, 03:52 PM  #5 
Newbie Joined: Mar 2015 From: fl Posts: 3 Thanks: 1 
Oh ok, I get it! Thank you very much. It turns out the answer they put was wrong, haha. They said it was 4.6 when in fact it was 4.77... Thanks!


Tags 
driving, mad, triangle 
Search tags for this page 
Click on a term to search for related topics.

Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
I have this homework that is driving me crazy  DANDIS  Advanced Statistics  2  December 24th, 2011 01:12 PM 
factoring that is driving me insane  aaronmath  Algebra  3  November 8th, 2011 08:17 PM 
Please help. This has been driving me insane.  thfc1799  Advanced Statistics  1  November 21st, 2010 10:31 PM 
These equations are driving me to insanity!  frankenstein  Algebra  4  August 22nd, 2010 05:52 PM 
Number of possible outcomes when driving  groundwar  Algebra  6  October 27th, 2009 01:08 PM 