A right angle triangle has area 6 cm square. Is it possible to find the perimeter of the triangle?
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You can find the minimum value of the perimeter, unfortunately with the given information, the perimeter can be any value larger than that... Where did you get the question? – Macavity Oct 08 '15 at 12:38
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2No. The perimeter can range anywhere from $4\sqrt3+2\sqrt6$ to $+\infty$ – user137794 Oct 08 '15 at 12:39
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My daughter's Form Two (Grade 8) math question... – ClemKY Oct 08 '15 at 12:41
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2Then perhaps the (implicit?) intention is to have integer sides. Then you have a unique solution. – Macavity Oct 08 '15 at 12:47
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you can't find it but you can know that it is greater than or equal to $\sqrt{24}$, if a,b,c (c hypotenuse) are the edges of the triangle, the area is equal to $\frac{ab}{2}$,so the perimeter equal to:
$$ a+b + \sqrt{(a+b)^2 - 2ab} = a+b+\sqrt{(a+b)^2 - 24}$$
so $a+b \geq \sqrt{24}$, and the perimeter $\geq \sqrt{24}$
D. A.
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In fact one can show that the perimeter is $\ge 2\sqrt3(2+\sqrt2) \approx 11.83 >$ twice your estimate. – Macavity Oct 08 '15 at 12:50
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$$6=0.5ab$$ $$p=a+b+\sqrt{a^2+b^2}$$
these two equtions with three variables( impossible to find it)
E.H.E
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Let's try. Since $\frac{1}{2}bh=6$ then $bh=12$. Let's try to use the information that also $b^2+h^2=c^2$ where $c$ is the length of the hypotenuse. The thing we're looking for is $b+h+c$. We know we can eliminate $c$ by $\sqrt{b+h}$ and we can eliminate $h$ with $12/b$. So the perimeter has expression $b+12/b+\sqrt{b+12/b}$.
Looks like there's no way to actually find its number-value without knowing $b$, or at least one of the other edges.
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