How do you find the dimensions of a right triangle if a right triangle has area 960 and hypotenuse length = 52?

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How do you find the dimensions of a right triangle if a right triangle has area 960 and hypotenuse length = 52?

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Answer 1 · 2018-05-11T10:30:50

When we solve simultaneous equations

$1/2 a b = 960$

$a^2 + b^2 = 52 ^2$

we find no real solutions, so no such right triangle exists.

Explanation:

Call the legs $a$ and $b$ .

$1/2 a b = 960$

$a^2 + b^2 = 52 ^2$

Let's eliminate $b$ first.

$b = 1920/a$

$a^2 + (1920/a)^2 = 52^2$

$a^4 - 52^2 a^2 + 1920^2 = 0$

That's a quadratic equation in $a^2$ with a pretty negative discriminant of

$52^4 - 4(1920)^2 = -7433984$

so no real solutions for $a^2$ and thus no real solutions for $a,b$ .

That's the end, but we can look into it a bit deeper.

The hypotenuse is just too small to support this area. Let's find the general formula for the minimum hypotenuse for a real triangle of a given area $A$ .

$1/2 ab = A$

$b = {2A}/a$

$a^2 + b^2 = c^2$

$c^2 = a^2 + {4A^2}/a^2$

$a^4 - c^2 a^2 + 4A^2 = 0$

The discriminant must be positive or zero:

$c^4 - 16 A^2 ge 0$

$c^4 ge 16 A^2$

Everything is positive, so

$c ge 2 sqrt{A}$

In our case

$c = 2 sqrt{960} = 16 sqrt{15} approx 61.97$ is the minimum hypotenuse than can support this area, presumably corresponding to the isosceles right triangle with $a=b.$

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How do you find the dimensions of a right triangle if a right triangle has area 960 and hypotenuse length = 52?

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