Hi,
Trying to wrap my mind around whether knowing the hypotenuse of a right triangle and the area of the triangle will allow me to find the value of the two legs. For example, let the hypotenuse be 10 (common right triangle) and let the area be 25 (did not choose 24 on purpose so its not a 6,8,10 right triangle).
I wanna say it's not possible to find the specific value for the 2 legs using what's known: a^2+b^2 = 100, and (1/2)ab = 25. I tried to play with the two equations to prove that there are multiple possibilities for a&b but the equations got really messy so I am wondering if someone can help me on that.
thanks,
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- Yu-MingH64
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- Sage Pearce-Higgins
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Re: Right Triangle, Hypotenuse, Area
Interesting question. You can try to think of it conceptually: if you know the length of the hypotenuse, and that it's a right-angled triangle, then knowing just the length of one more side is enough to give you the complete triangle. As the length of this next side changes, you might be able to see that the area of the triangle changes, suggesting that we have enough information. If that doesn't make sense, or you're having trouble imagining it, then think of the algebra too.
You set up the equations really nicely. However, why pick a nasty number like 25 for the area? Let's keep it at 24 so that we know that one solution will be the 6-8-10 triangle, but we can look for more.
Our two equations:
a^2+b^2 = 100, and (1/2)ab = 24
so, ab = 48, and b = 48/a
If I sub that into the first equation
a^2 + (48/a)^2 = 100
a^2 + 2304/a^2 = 100
Multiply both sides by a^2:
a^4 + 2304 = 100a^2
a^4 - 100a^2 +2304 = 0
This is an odd looking equation: it's called a polynomial. Since there's an a^4, I should expect as many as 4 solutions. I can factorize in the usual way I do for quadratics, but with a^2:
(a^2 - 36)(a^2 - 64) = 0
so a^2 = 36 or 64,
a = 6, -6, 8, -8
Wait, hang on, all those solutions give me the same triangle, just "turned round" as it were. As you suspected, there's only one triangle that fits if I know the area and the hypotenuse and that it's a right triangle.
We've gone beyond the kind of thing that you're expected to do in a GMAT test, but this kind of "deep study" is really useful as a workout, and for giving you a deeper understanding of problems, particularly DS ones. If you haven't done problem 48 from the Diagnostic Test in the 2018 Official Guide, then try it out.
You set up the equations really nicely. However, why pick a nasty number like 25 for the area? Let's keep it at 24 so that we know that one solution will be the 6-8-10 triangle, but we can look for more.
Our two equations:
a^2+b^2 = 100, and (1/2)ab = 24
so, ab = 48, and b = 48/a
If I sub that into the first equation
a^2 + (48/a)^2 = 100
a^2 + 2304/a^2 = 100
Multiply both sides by a^2:
a^4 + 2304 = 100a^2
a^4 - 100a^2 +2304 = 0
This is an odd looking equation: it's called a polynomial. Since there's an a^4, I should expect as many as 4 solutions. I can factorize in the usual way I do for quadratics, but with a^2:
(a^2 - 36)(a^2 - 64) = 0
so a^2 = 36 or 64,
a = 6, -6, 8, -8
Wait, hang on, all those solutions give me the same triangle, just "turned round" as it were. As you suspected, there's only one triangle that fits if I know the area and the hypotenuse and that it's a right triangle.
We've gone beyond the kind of thing that you're expected to do in a GMAT test, but this kind of "deep study" is really useful as a workout, and for giving you a deeper understanding of problems, particularly DS ones. If you haven't done problem 48 from the Diagnostic Test in the 2018 Official Guide, then try it out.
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