Question

in 21–23, the altitude to the hypotenuse of a right triangle divides the hypotenuse into two segments.

1. if the lengths of the segments are 5 inches and 20 inches, find the length of the altitude.
2. if the length of the altitude is 8 feet and the length of the shorter segment is 2 feet, find the length of the longer segment.
3. if the ratio of the lengths of the segments is 1:9 and the length of the altitude is 6 meters, find the lengths of the two segments.

Explanation:

Problem 21

Step1: Apply geometric mean theorem

In a right - triangle, the altitude \(h\) to the hypotenuse is the geometric mean of the lengths of the two segments \(a\) and \(b\) of the hypotenuse. The formula is \(h=\sqrt{ab}\).

Step2: Substitute values

Given \(a = 5\) inches and \(b=20\) inches. Then \(h=\sqrt{5\times20}\).

$$h=\sqrt{100}$$

$$h = 10$$

Step1: Use geometric mean theorem

The altitude \(h\) to the hypotenuse of a right - triangle satisfies \(h^{2}=a\times b\), where \(a\) is the shorter segment and \(b\) is the longer segment.

Step2: Solve for \(b\)

Given \(h = 8\) feet and \(a=2\) feet. We have \(b=\frac{h^{2}}{a}\).

$$b=\frac{8^{2}}{2}=\frac{64}{2}$$

$$b = 32$$

Step1: Let the lengths of the segments

Let the lengths of the segments be \(x\) and \(9x\) (since the ratio of the lengths of the segments is \(1:9\)).

Step2: Apply geometric mean theorem

Using \(h^{2}=x\times9x\), where \(h = 6\) meters.

$$36=9x^{2}$$

$$x^{2}=4$$

$$x = 2$$

(we take the positive value since length cannot be negative)

Step3: Find the lengths of the segments

If \(x = 2\), then \(9x=18\)

Answer:

The length of the altitude is \(10\) inches.

Problem 22