anonymous
  • anonymous
The last question I have I know I am dumb, but The thing is I'm better at Algebra :/
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Idk why I can't set this up!
1 Attachment
whpalmer4
  • whpalmer4
okay, can you tell me anything about the relationship between the small triangle and the big triangle?
okdutchman7
  • okdutchman7
The big and small triangles are similar.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

okdutchman7
  • okdutchman7
They would be proportional to each other
anonymous
  • anonymous
Their base, and I guess you would call it a leg is congruent
whpalmer4
  • whpalmer4
uh, no, the bases are not congruent — the two segments that make up the base of the larger triangle are congruent...
okdutchman7
  • okdutchman7
Are you a girl? Cause girls are better at algebra and guys are better at geometry.
whpalmer4
  • whpalmer4
I'm a counter-example to that theory :-)
anonymous
  • anonymous
Yes I am a girl but my BFF is way better at geometry but she's a lesson behind so she couldn't help!
okdutchman7
  • okdutchman7
9 times of 10 it's true though : )
anonymous
  • anonymous
I know the theory but I wasn't in class , and my teacher is really hard with her tests
whpalmer4
  • whpalmer4
|dw:1449196971961:dw|
anonymous
  • anonymous
and she didn't help me when I asked
whpalmer4
  • whpalmer4
those two segments labeled \(c\) are congruent, so we know that the ratio \(4:c\) is the same as the ratio of the hypotenuse of the big triangle to the base of the big triangle can you write an equation for that?
okdutchman7
  • okdutchman7
4c = c(5x+2) Try to solve it from there
anonymous
  • anonymous
Okay!
okdutchman7
  • okdutchman7
Sorry that would be 4(2c) = c(5x+2)
anonymous
  • anonymous
Okay so the first part is 8c - 1c = 8=5x+2 subtract two from both sides then you get 7 , so then you would divided by 5 and get 5/7 :/
whpalmer4
  • whpalmer4
no, that's still not the correct equation... please, try to write the equation yourself first, you'll need to know how to do that to solve these in the future.
anonymous
  • anonymous
I thought it would be 5x-2 divided by 1/2= 4 , or somethign along that line
whpalmer4
  • whpalmer4
Here's the ratio of the hypotenuse to the base for the little triangle: \[\frac{4}{c}\] What is the ratio of the hypotenuse of the big triangle to the base of the big triangle? Those two will be equal, so you can set them equal to each other, cross-multiply and solve for \(x\).
whpalmer4
  • whpalmer4
\(c\) will magically disappear without our ever having needed to know its value :-)
anonymous
  • anonymous
Would I cross mulitply like this X/5 *4/c ?
whpalmer4
  • whpalmer4
first you write the ratio I asked you for...
anonymous
  • anonymous
I apologize to my medicine is off I can't think clearly!
whpalmer4
  • whpalmer4
once you're a master at these problems you can start skipping steps, but first you have to become a master :-)
whpalmer4
  • whpalmer4
what is the hypotenuse of the big triangle? what is the base of the big triangle?
okdutchman7
  • okdutchman7
You will need to cross multiply, but rethink what you are cross multiplying
anonymous
  • anonymous
Okay can you explain the ratio to me
anonymous
  • anonymous
5x/2, and 4/c ?
okdutchman7
  • okdutchman7
5x-2 on the big triangle is proportional to 4 on the little triangle 2c on the big triangle is proportional to c on the little triangle
anonymous
  • anonymous
Oh so it would be 5x-2/2c * c/4?
whpalmer4
  • whpalmer4
the hypotenuse of the big triangle is 5x-2 the base of the big triangle is the base of the little triangle plus another segment the same length, or c + c = 2c that gives us: \[\frac{4}{c} = \frac{5x-2}{2c}\]
anonymous
  • anonymous
Okay so now I cross multiply to get 5x-2*4 /2
anonymous
  • anonymous
would I multiply -2 to 4
anonymous
  • anonymous
Guys I'm sorry I;m so confused now.........
anonymous
  • anonymous
And you left :/
anonymous
  • anonymous
I got the hint ....... okay thanks anyways
whpalmer4
  • whpalmer4
\[\frac{4}{c} = \frac{5x-2}{2c}\] Okay, if we cross-multiply, we multiply the numerator of one fraction by the denominator of the other, and vice-versa. \[4*2c = c*(5x-2)\]\[8c = 5cx - 2c\] Can you do the rest?
anonymous
  • anonymous
Yes add 2 c to the 8c which would be 10c=5cx and since they have a common variable then I can divided so x= 2?
whpalmer4
  • whpalmer4
\[8c = 5cx-2c\]\[8c+2c = 5cx-2c+2c\]\[10c = 5cx\]we can divide both sides by \(c\) \[\frac{10c}{c} = \frac{5cx}{c}\]\[10 = 5x\]\[x=2\] Now we do the important part, which is to check our answer! If our hypotenuse on the big triangle is \(5x-2\) and we think \(x=2\), then the hypotenuse is \(5(2)-2 = 10-2=8\). Is that twice the length of the other hypotenuse? Yes, it is, just like our base is twice the length of the other base. Our answer is correct.
anonymous
  • anonymous
Okay so it is definitely 2 thank you so much for putting up with me!
whpalmer4
  • whpalmer4
You're welcome! By the way, I did the ratio as hypotenuse / base, but any way you want to do it is fine, so long as you do it the same way for both triangles.
anonymous
  • anonymous
Okay I'll keep that in mind thank you again!
okdutchman7
  • okdutchman7
@whpalmer4 you did a great job explaining it!
anonymous
  • anonymous
By the way my subjects I am good at is Biology, and English. The others nope!
Hero
  • Hero
whpalmer4
  • whpalmer4
@Hero you made some mistakes but got the right answer, and I think it is interesting to see why that is. Small triangle you wrote: \[a^2+b^2=4\](but it should be \(a^2+b^2=4^2 = 16\)) Large triangle you wrote: \[4(a^2+b^2)=(5x-2)^2\]\[a^2+b^2=\frac{(5x-2)^2}{4}\]then Notice: \(4 = \frac{(5x-2)^2}{4}\) Solve for x \[64 = (5x-2)^2\] of course you can only get there from substituting the correct \(a^2+b^2=16\) earlier: \[(2a)^2+(2b)^2 = (5x-2)^2\]\[4a^2+4b^2=(5x-2)^2\]\[4(a^2+b^2) = (5x-2)^2\]\[4(16) = (5x-2)^2\]\[64 = (5x-2)^2\]\[\sqrt{64} = \sqrt{(5x-2)^2}\]etc. Probably just wrote it down wrong after doing it correctly in your head. That's not the interesting part. This approach using the Pythagorean theorem assumes that we have a right triangle, but the problem does not specify that it is! The similar triangles approach works for any pair of similar triangles where the legs are doubled in length, not just a right triangle. In fact, it is the natural of similar triangles that makes this work — double the length of the two congruent sides and you double the length of the third side, so the problem is always just \[2*4 = 5x-2\]\[x=2\]and even my approach could have been simplified. You can write down the work as you did it for triangles which are not right triangles and even though the math is not correct (Pythagorean theorem will not apply), you will still get the same answer! All hail the power of similar triangles! Here is such a pair of triangles. |dw:1449248409607:dw|

Looking for something else?

Not the answer you are looking for? Search for more explanations.