the diagonal of a rectangle is square root of (w^2 +l^2)
so square root of (14^2 +10^2) = 17.2 inches
since 20" is longer than 17.2 it can't be a rectangle
Rectangle has its adjacent sides perpendicular to each other. The diagonal and the two adjacent sides aren't satisfying Pythagoras theorem.
If ABC is a triangle with AC as the hypotenuse and angle B with 90 degrees then we have:
[tex]|AC|^2 = |AB|^2 + |BC|^2[/tex]
Where |AB| = length of line segment AB.
If a triangle doesn't satisfy the Pythagoras theorem, then none of its angles are of 90 degree measures. And if it satisfies it, then one of its angle is of 90 degrees measure.
For the given case, we know that in a rectangle, its sides are perpendicular to each other.
Let one of its side be of 'a' unit length.
let the other adjacent side be of length 'b' units.
Let the diagonal made by connecting the ends of those two perpendicular sides be of length 'd' units.
Then they will follow Pythagoras theorem where diagonal is hypotenuse and the rectangle is right angled as there is 90 degrees angle between the two adjacent sides considered.
Thus,
[tex]d^2 = a^2 + b^2[/tex]
For the given case, it was told that:
a = 14 inches
b = 10 inches
d = 20 inches,
But we see that
[tex]d^2 = 20^2 = 400\\\\a^2 + b^2 = 14^2 + 10^2 = 296\\\\\text{and, thus}\\\\d^2 \neq a^2 + b^2[/tex]
Thus, there is no right angle between those two adjacent sides, and thus, frame is not rectangular.
Learn more about Pythagoras theorem here:
https://brainly.com/question/12105522
