Question

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

Answer 1

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`.

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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.`