What is the solution to the equation below? Round your answer to two
1 answer:
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Since a right triangle has special rules to it such that the hypotenuse squared equals the sum of the squares of the other 2 sides. In other words, if the hypotenuse = c, and the 2 smaller sides are a and b, then:
Solving for c (hypotenuse), we get:
Therefore c is the root of the square of the other sides. So by having root18, it's like saying:
Getting rid of the square root, so sides a and b must have their squares total 18:
the only squares < 18 are: 16 (4×4), 9 (3×3), 4 (2×2), and 1 (1×1)
of those above added in any order of two of them, only 16+4
I HOPE THAT IS ALONG THE LINE THAT WAS ASKED FOR!!! :-D
Solving for c (hypotenuse), we get:
Therefore c is the root of the square of the other sides. So by having root18, it's like saying:
Getting rid of the square root, so sides a and b must have their squares total 18:
the only squares < 18 are: 16 (4×4), 9 (3×3), 4 (2×2), and 1 (1×1)
of those above added in any order of two of them, only 16+4
I HOPE THAT IS ALONG THE LINE THAT WAS ASKED FOR!!! :-D
