How do you draw a right triangle with a hypotenuse of $sqrt72$ units?

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Answer 1 · 2017-06-01T21:34:14

Draw the arms of a $90°$ angle with a length of $6$ units and the hypotenuse will have a length of $sqrt72$

Find two square numbers which add up to $72$

According to Pythagoras' Theorem:

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

One possibility of such a right-angled triangle is an isosceles right -angled triangle with equal sides of $6$ units.

The length of the hypotenuse can be calculated as:

$x^2 = 6^2 + 6^2$

$x^2 = 36+36$

$x^2 =72$

$x = sqrt72$

So, if you draw the arms of a $90°$ angle with a length of $6$ units, the hypotenuse will have a length of $sqrt72$

How do you draw a right triangle with a hypotenuse of sqrt72sqrt72 units?