Question

special right triangles
find the value of each variable. write your answer as an integer or in simplest radical form.
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7 home plate, first base, second base, and third base form a perfect square. the distance from each consecutive
base is 90 feet. what is the distance from home plate directly to second base?

Explanation:

1.

Step1: Identify the triangle type

It is a \(45 - 45-90\) triangle. In a \(45 - 45-90\) triangle, the legs are equal, so \(x = 13\).

Step2: Use the hypotenuse formula

The hypotenuse \(y\) of a \(45 - 45-90\) triangle is \(y=\sqrt{2}\times\) leg. So \(y = 13\sqrt{2}\).

2.

Step1: Legs of \(45 - 45-90\) triangle

Since it is a \(45 - 45-90\) triangle, \(x=\frac{50}{\sqrt{2}}=\frac{50\sqrt{2}}{2}=25\sqrt{2}\).

Step2: Hypotenuse

The hypotenuse \(y=\sqrt{2}\times x\), so \(y = 50\).

3.

Step1: Legs of \(45 - 45-90\) triangle

In a \(45 - 45-90\) triangle, legs are equal. Let the leg be \(x\). The hypotenuse is \(7\sqrt{2}\). Using the formula \(hypotenuse=\sqrt{2}\times leg\), we have \(7\sqrt{2}=\sqrt{2}x\), so \(x = 7\). Then \(y = 7\).

4.

Step1: Legs of \(45 - 45-90\) triangle

Let the leg be \(x\). The hypotenuse \(y=\sqrt{2}x\). Given \(y=\sqrt{10}\), then \(x=\frac{\sqrt{10}}{\sqrt{2}}=\sqrt{5}\).

5.

Step1: Hypotenuse of \(45 - 45-90\) triangle

In a \(45 - 45-90\) triangle, if the hypotenuse is \(24\), then the leg \(x=\frac{24}{\sqrt{2}}=\frac{24\sqrt{2}}{2}=12\sqrt{2}\).

6.

Step1: Leg of \(45 - 45-90\) triangle

In a \(45 - 45-90\) triangle, if the hypotenuse is \(6\), then the leg \(x=\frac{6}{\sqrt{2}}=\frac{6\sqrt{2}}{2}=3\sqrt{2}\).

7.

Step1: \(45 - 45-90\) triangle application

The distance between consecutive bases is \(90\) feet. The distance from home plate to second base is the hypotenuse of a \(45 - 45-90\) triangle with leg \(a = 90\). Using \(c=\sqrt{2}a\), we get \(c = 90\sqrt{2}\) feet.

Answer:

1. \(x = 13,y=13\sqrt{2}\)
2. \(x = 25\sqrt{2},y = 50\)
3. \(x = 7,y = 7\)
4. \(x=\sqrt{5},y=\sqrt{10}\)
5. \(x = 12\sqrt{2}\)
6. \(x = 3\sqrt{2}\)
7. \(90\sqrt{2}\) feet