Question

1. two out of three. if a right triangle has legs of length 1 and 2, what is the length of the hypotenuse? if it has one leg of length 1 and a hypotenuse of length 3, what is the length of the other leg? 3. hypotenuse hype. if a right triangle has legs of length 1 and x, what is the length of the hypotenuse? 4. assessing area. suppose you know the base of a rectangle has a length of 4 inches and a diagonal has a length of 5 inches. find the area of the rectangle. 5. squares all around. how does the figure below relate to the pythagorean theorem?

Explanation:

2.

First part:

Step1: Recall Pythagorean theorem

For a right - triangle with legs \(a\) and \(b\) and hypotenuse \(c\), \(a^{2}+b^{2}=c^{2}\). Here \(a = 1\), \(b = 2\).
\[c^{2}=1^{2}+2^{2}=1 + 4=5\]

Step2: Solve for \(c\)

\[c=\sqrt{5}\]

Second part:

Step1: Apply Pythagorean theorem

Let one leg \(a = 1\), hypotenuse \(c = 3\), and the other leg be \(b\). Then \(a^{2}+b^{2}=c^{2}\), so \(b^{2}=c^{2}-a^{2}\).
\[b^{2}=3^{2}-1^{2}=9 - 1=8\]

Step2: Solve for \(b\)

\[b = 2\sqrt{2}\]

Step1: Use Pythagorean theorem

For a right - triangle with legs \(a = 1\) and \(b=x\), and hypotenuse \(c\), by \(a^{2}+b^{2}=c^{2}\), we have \(c^{2}=1^{2}+x^{2}=1 + x^{2}\).

Step2: Solve for \(c\)

\[c=\sqrt{1 + x^{2}}\]

Step1: Find the height of the rectangle

Let the base of the rectangle \(a = 4\) inches and the diagonal \(c = 5\) inches. In a rectangle, using the Pythagorean theorem for the right - triangle formed by the base, height \(b\), and diagonal. \(b^{2}=c^{2}-a^{2}\).
\[b^{2}=5^{2}-4^{2}=25 - 16 = 9\]
\[b = 3\]

Step2: Calculate the area of the rectangle

The area of a rectangle \(A=ab\), where \(a = 4\) and \(b = 3\).
\[A=4\times3=12\] square inches

Answer:

The length of the hypotenuse in the first case is \(\sqrt{5}\), and the length of the other leg in the second case is \(2\sqrt{2}\)

3.