Algebra(Radical Stuff); I have a right triangle. There are two sides given. The longer leg of the two legs is 4x and the shorter one is x. I have to find the total perimeter of the triangle. My answer choices are A.) 68√x B.) √68x C.) 5x+x+√17 D.) 68x Note: I am home schooled and my parents haven't done algebra in years so I'm practically teaching myself. Any help would be appreciated!
wtf was that lol dude wrote for like 7 minutes and then left? that's cold
Google can't even answer this problem... -_-
well I was looking at it and was kinda afraid because the answer I'm getting doesn't match any of these
Same here! I don't even know what I'm doing, but I can't move on in my lessons until I complete the check at the end... Soooo.... I'm messed up.
I believe it is c though
well we have to find the third side in order to find the perimeter. The third side is the hypotenuse so we use Pythagorean's theorom \[a ^{2}+b ^{2}=c ^{2}\]
so in order to get just c let's rearrange it to our liking \[c=\sqrt{a ^{2}+b ^{2}}\]
Oh! Okay. I gotcha :D Sorta.. haha
now substitute \[c=\sqrt{(4x)^{2}+x ^{2}}\rightarrow \sqrt{16x ^{2}+x ^{2}}\rightarrow \sqrt{17x ^{2}}\]
so now that's your c but perimeter is all the sides added up together so \[4x+x+\sqrt{17x ^{2}}\rightarrow 5x+\sqrt{17x ^{2}}\] I'm sure that can be simplified down further into what you have as the answer c I'm just not positive how and Don't want to give you false information.
Are you sure c is \[5x+x+\sqrt{17}\] and it's not \[5x+x \sqrt{17}\] because that is the only way I know to simplify it further
I am guessing it is choice C (but C has a typo) the answer is 5x + x sqrt(17)
thank you for clarifying that. I probably should have been a little more confident in myself and called it out but.......idk I just figured it was something I had forgotten
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