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Draws a triangle based on 3 side lengths and randomises asking for hypotenuse or not.

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Find $x$.

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Only round your final answer to 1 decimal place.

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Use this link to find some resources to help you revise how to use pythagoras' theorem.

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Avoid using rounded values in calculations and just round for the final answer.

Pythagoras Theorem states that, in a right angled triangle, with hypotenuse $c$:

\\n

\\\\[a^2 + b^2 = c^2\\\\]

\\n

Let\\'s call the unknown value $x$, therefore we can write:

\\n

$a = \\\\var{sh1}$,  $b =\\\\var{sh2}$ and $c = x$

\\n

So

\\\\[\\\\var{sh1}^2 + \\\\var{sh2}^2 = x^2\\\\]

\\n

and therefore

\\n

\\\\[x^2 = \\\\var{sh1^2} + \\\\var{sh2^2}\\\\]

\\\\[x = \\\\sqrt{\\\\var{sh1^2} + \\\\var{sh2^2}}\\\\]

\\n

\\\\[x = \\\\sqrt{\\\\var{sh1^2+sh2^2}}\\\\]

$x = \\\\var{hyp}$ to 1 d.p.

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Avoid using rounded values in calculations and just round for the final answer.

Pythagoras Theorem states that, in a right angled triangle, with hypotenuse $c$:

\\n

\\\\[a^2 + b^2 = c^2\\\\]

\\n

Let\\'s call the unknown value $x$, therefore we can write:

\\n

$a = x$,  $b =\\\\var{sh2}$ and $c = \\\\var{hyp}$

\\n

So

\\n

\\\\[x^2 + \\\\var{sh2}^2 = \\\\var{hyp}^2\\\\]

\\n

and therefore

\\n

\\\\[x^2 = \\\\var{hyp^2} - \\\\var{sh2^2}\\\\]

\\n

\\\\[x = \\\\sqrt{\\\\var{hyp^2-sh2^2}}\\\\]

$x = \\\\var{sh1}$ to 1 d.p.

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one of two shortest sides for calculations.

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$x=$[[0]] to 1 d.p.

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