Question

find the exact value of x.
x =
do the side lengths form a pythagorean triple?
\bigcirc yes
\bigcirc no
(note: the image shows a right triangle with hypotenuse 24, one leg 9, and the other leg labeled x.)

Explanation:

Step1: Apply Pythagorean theorem

For a right triangle, the Pythagorean theorem states that \( a^2 + b^2 = c^2 \), where \( c \) is the hypotenuse and \( a, b \) are the legs. Here, hypotenuse \( c = 24 \), one leg \( b = 9 \), and the other leg \( a = x \). So we have \( x^2 + 9^2 = 24^2 \).

Step2: Solve for \( x^2 \)

Calculate \( 9^2 = 81 \) and \( 24^2 = 576 \). Then \( x^2 = 24^2 - 9^2 = 576 - 81 = 495 \).

Step3: Simplify \( x \)

Take the square root of 495. \( 495 = 9\times55 \), so \( \sqrt{495} = \sqrt{9\times55} = 3\sqrt{55} \). Wait, no, wait, maybe I made a mistake. Wait, hypotenuse is the longest side. Wait, maybe I mixed up hypotenuse and leg. Wait, the right angle is between \( x \) and 9, so hypotenuse should be 24? Wait, no, if the triangle has legs \( x \) and 9, and hypotenuse 24, then \( x^2 + 9^2 = 24^2 \). But 9 and 24, let's check: 24 is longer than 9, but is 24 the hypotenuse? Wait, the diagram: the right angle is at the vertex with sides \( x \) and 9, so the hypotenuse is 24. So \( x^2 = 24^2 - 9^2 = 576 - 81 = 495 \), so \( x = \sqrt{495} = 3\sqrt{55} \approx 22.25 \), but 495 factors: 495 = 955 = 9511, so can't be simplified more. Wait, but Pythagorean triple: a Pythagorean triple consists of three positive integers \( a, b, c \) such that \( a^2 + b^2 = c^2 \). Here, 9, \( x \), 24: \( x = \sqrt{495} \) which is not an integer, so the side lengths do not form a Pythagorean triple. Wait, but maybe I mixed up the hypotenuse. Wait, maybe the hypotenuse is \( x \), and the legs are 9 and 24? Let's check that. If hypotenuse is \( x \), then \( 9^2 + 24^2 = x^2 \). Then \( 81 + 576 = 657 \), so \( x = \sqrt{657} = \sqrt{9\times73} = 3\sqrt{73} \), still not integer. Wait, no, the diagram: the right angle is at the bottom, so the two legs are \( x \) (horizontal) and 9 (vertical), and the hypotenuse is 24 (the slant side). So \( x^2 + 9^2 = 24^2 \). But 9, 24, and \( x \): 9 and 24 are not part of a Pythagorean triple. Wait, maybe the problem has a typo? Wait, no, let's recalculate. 24 squared is 576, 9 squared is 81, 576 - 81 is 495. 495 is 955, so square root of 495 is 3*sqrt(55). So \( x = 3\sqrt{55} \), and since \( x \) is not an integer, the side lengths do not form a Pythagorean triple.

Wait, maybe I made a mistake in identifying the hypotenuse. Let's check the diagram again. The right angle is at the vertex where \( x \) and 9 meet, so the hypotenuse is the side opposite the right angle, which is 24. So yes, \( x^2 + 9^2 = 24^2 \). So \( x^2 = 576 - 81 = 495 \), so \( x = \sqrt{495} = 3\sqrt{55} \). And since 9, \( 3\sqrt{55} \), and 24 are not all integers, they do not form a Pythagorean triple.

Answer:

First, for \( x \): \( x = \sqrt{24^2 - 9^2} = \sqrt{576 - 81} = \sqrt{495} = 3\sqrt{55} \).

For the Pythagorean triple question: No, because \( x = 3\sqrt{55} \) is not an integer, so the side lengths (9, \( 3\sqrt{55} \), 24) do not form a Pythagorean triple (which requires all sides to be positive integers).

So, \( x = 3\sqrt{55} \) and the answer to the Pythagorean triple question is No.

But let's check again. Wait, maybe the hypotenuse is \( x \), and the legs are 9 and 24. Then \( x^2 = 9^2 + 24^2 = 81 + 576 = 657 \), \( x = \sqrt{657} = 3\sqrt{73} \), still not integer. So regardless, \( x \) is not integer, so the triple is not Pythagorean.

So final answers: \( x = 3\sqrt{55} \) (or simplified as \( \sqrt{495} \)) and the Pythagorean triple answer is No.

But let's confirm the Pythagorean theorem application. Right triangle, so \( a^2 + b^2 = c^2 \), where \( c \) is hypotenuse (longest side). If 24 is hypotenuse, then \( x \) is leg, so \( x = \sqrt{24^2 - 9^2} = \sqrt{576 - 81} = \sqrt{495} = 3\sqrt{55} \approx 22.25 \), which is less than 24, so that's correct (hypotenuse is longest). So \( x = 3\sqrt{55} \), and since 9, \( 3\sqrt{55} \), 24 are not all integers, they don't form a Pythagorean triple.

So, for the first part, \( x = 3\sqrt{55} \) (or \( \sqrt{495} \)), and for the second part, the answer is No.