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: 12,155 Thanks: 465 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: 12,155 Thanks: 465 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!


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