Find the hypotenuse,

, of the triangle.
triangle23-138314-1438770434573
​ Figure not drawn to scale

A
158\sqrt{158}
158


B
160\sqrt{160}
160


C
121\sqrt{121}
121


D
46\sqrt{46}
46

1 answer

To find the hypotenuse of a right triangle, we can use the Pythagorean theorem, which states:

\[ c = \sqrt{a^2 + b^2} \]

where \(c\) is the length of the hypotenuse and \(a\) and \(b\) are the lengths of the other two sides.

However, since you provided options (A, B, C, D) with expressions containing \(\sqrt{n}\), it's likely that the provided values correspond to \(a^2\) and \(b^2\) in each case.

Let's calculate \(a^2 + b^2\) for each option:

A. \(158\sqrt{158}\)

Assuming \(a^2 + b^2 = 158\sqrt{158}\):

\[ c = \sqrt{158\sqrt{158}} = \sqrt{158 \times 158^{1/2}} = \sqrt{158^{3/2}} = (158^{3/4}) = 158^{0.75} \]

B. \(160\sqrt{160}\)

Assuming \(a^2 + b^2 = 160\sqrt{160}\):

\[ c = \sqrt{160\sqrt{160}} = \sqrt{160^{3/2}} = (160^{3/4}) = 160^{0.75} \]

C. \(121\sqrt{121}\)

Assuming \(a^2 + b^2 = 121\sqrt{121}\):

\[ c = \sqrt{121\sqrt{121}} = \sqrt{121^{3/2}} = (121^{3/4}) = 121^{0.75} \]

D. \(46\sqrt{46}\)

Assuming \(a^2 + b^2 = 46\sqrt{46}\):

\[ c = \sqrt{46\sqrt{46}} = \sqrt{46^{3/2}} = (46^{3/4}) = 46^{0.75} \]

These calculations give us expressions depending on the values within the square roots.

Since we don't have concrete values for \(a\) and \(b\), but rather the answers represent the length of the hypotenuse squared, we can directly evaluate them as they stand.

The question asks for the hypotenuse, and from the options you presented, only one can represent a hypotenuse calculated correctly with accurate side lengths according to the Pythagorean theorem.

If you provide the lengths of the triangle sides, I could confirm more precisely the hypotenuse calculation. But from the current data, we can surmise that each provided option reflects the square of the hypotenuse appropriately based on their comparative values. If you're determining which is correct based on triangle dimensions, compare the relative sizes knowing that \(c > a\) and \(c > b\) where \(c\) is always larger.

If additional specific angles and side lengths are available/provided from the triangle, I can further compute to confirm the hypotenuse.