My Math Forum Right Triangle Driving me mad

 Geometry Geometry Math Forum

 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 $C=3A$. Thanks from SilverOrange
 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: $\displaystyle A^2-A^2+B^2=9A^2-A^2$ $\displaystyle B^2=8A^2$ 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 $3$. Thanks from SilverOrange
 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! Thanks from MarkFL

### right triangle in driving

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post DANDIS Advanced Statistics 2 December 24th, 2011 01:12 PM aaron-math Algebra 3 November 8th, 2011 08:17 PM thfc1799 Advanced Statistics 1 November 21st, 2010 10:31 PM frankenstein Algebra 4 August 22nd, 2010 05:52 PM groundwar Algebra 6 October 27th, 2009 01:08 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top