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Vsevolod [243]
3 years ago
11

When finding out if 2 triangles are congruent, is it possible to only use a side and angle if those are the only 2 provided? Hel

p ASAP please.

Mathematics
1 answer:
PIT_PIT [208]3 years ago
8 0
Short answer: No and yes. 
Yes First.
If the lower right point (looks like an C perhaps or an F) and the middle point and the upper left point lie on the same line and if D the middle point and the upper right point lie n the the same line, then yes. You have angle side angle. The lower right angle is given as equal to the upper left angle. The angles at C are vertically opposite angles and they are equal and you have been given 2 enclosed sides that are equal.

No
If the points I've described are not on the same line, then the triangles are not congruent. If I can, I'll edit this a second time. 

Please Note:
You have done something that looks a bit illegal to me. I wouldn't just trade pictures. It only costs you another 5 points to issue another question. I do think, however, that it is very clever. In answering about 1700 questions, I've never seen this done. That's commendable.
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Answer:

Step-by-step explanation:

If you graph there would be two different regions. The first one would be

y = \sqrt{x} \,\,\,\,, 0\leq x \leq 1 \\

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If you rotate the first region around the "y" axis you get that

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And if you rotate the second region around the "y" axis you get that

{\displaystyle A_2 = 2\pi \int\limits_{1}^{2} x(2-x) dx = \frac{4\pi}{3} = 4.188 }

And the sum would be  2.51+4.188 = 6.698

If you revolve just the outer curve you get

If you rotate the first  region around the x axis you get that

{\displaystyle A_1 =\pi \int\limits_{0}^{1} ( \sqrt{x})^2 dx = \frac{\pi}{2} = 1.5708 }

And if you rotate the second region around the x axis you get that

{\displaystyle A_2 = \pi \int\limits_{1}^{2} (2-x)^2 dx = \frac{\pi}{3} = 1.0472 }

And the sum would be 1.5708+1.0472 = 2.618

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