Question

QRST is a square. PQ= square root 2 RU= 4 What is the length os SU? Round to the nearest tenth​

Answer 1

The length of SU is approximately 3.45 units.

In the given scenario, you've established that ΔPQT is a 45-45-90 triangle, indicating that the angle at P is 45 degrees, the angle at Q is 45 degrees, and the angle at T is 90 degrees.

According to the properties of such triangles, if QT is the hypotenuse, then QT = PQ√2.

Given QT = 2 (as mentioned), you conclude that PQ is 2/√2. However, it appears there might be a mistake in stating that QT is also 2.

If QT is indeed equal to PQ√2, then QT should be approximately 2.83 units.

Moving on to the square RSUV, you correctly note that RS is 2 units since all sides of a square are congruent.

You then use the Pythagorean theorem to find SU, considering the right-angled triangle RSU within the square.

However, there seems to be an error in the calculation.

The correct equation is RS² + SU² = RU², which should be 2² + SU² = 4². The correct calculation is 4 + SU² = 16, leading to SU =
`√(12)`.

Therefore, SU is the square root of 12, which is approximately 3.46 units, not 3.5 units as stated.

The correct values are QT ≈ 2.83 units and SU ≈ 3.45 units.

Answer 2

3.5 units

Explanation:

Because ΔPQT is 45-45-90, QT = PQ√2. Therefore, QT is 2.

RS is also 2 because all sides of a square are congruent.

Knowing this, we can use the pythagorean theorem to find SU.

RS² + SU² = RU² -> 2² + SU² = 4² -> 4 + SU² = 16 -> SU² = 12

So, SU = √12 ≈ 3.5 units