Question

The length of the hypotenuse of the right triangle is 25 cm, the length of one of the legs is 10 cm. What is the length of the segment, formed by the altitude to the hypotenuse that is adjacent to the other leg?

Answer 1

Answer:

$\large\boxed{AD=21\ and\ CD=2\sqrt{21}}$

Step-by-step explanation:

Look at the picture.

Use the Pythagorean theorem for calculate the length of CA:

$CA^2+CB^2=AB^2$

We have

$CA=x,\ CB=10,\ AB=25$

Substitute:

$x^2+10^2=25^2$

$x^2+100=625$             subtract 100 from both sides

$x^2=525\to x=\sqrt{525}\\\\x=\sqrt{25\cdot21}\\\\x=\sqrt{25}\cdot\sqrt{21}\\\\x=5\sqrt{21}\ cm$

ΔADC and ΔACB are similar. Therefore the corresponding sides are in proportion:

$\dfrac{AD}{AC}=\dfrac{AC}{AB}$

We have:

$AD=y,\ AC=5\sqrt{21},\ AB=25$

Substitute:

$\dfrac{y}{5\sqrt{21}}=\dfrac{5\sqrt{21}}{25}$

$\dfrac{y}{5\sqrt{21}}=\dfrac{\sqrt{21}}{5}$              cross multiply

$5y=5(\sqrt{21})^2$            divide both sides by 5

$y=21\ cm$

If you mean the length of segment CD, then use the proportion:

$\dfrac{CD}{AD}=\dfrac{CB}{CA}$

Substitute:

$\dfrac{CD}{21}=\dfrac{10}{5\sqrt{21}}\\\\\dfrac{CD}{21}=\dfrac{2}{\sqrt{21}}\qquad\text{cross multiply}\\\\CD\sqrt{21}=42\qquad\text{multiply both sides by}\ \sqrt{21}\\\\21CD=42\sqrt{21}\qquad\text{divide both sides by 21}\\\\CD=2\sqrt{21}\ cm$