Can you find the length of ON using Pythagoras and similarity?

[alextrainer]

In attached file, I understand 50 is the base; no idea how to use the 24 height to calculate length of ON - must have to do with property of right triangles?

View attachment 6367

 

[HallsofIvy]

These are, of course, right triangles so you can use the Pythagorean theorem To determine the length of the hypotenuse of the small triangle. Then, since the angles of the large and small triangles are the same, they are similar triangles. Corresponding parts of the two right triangles are proportional.

 

[alextrainer]

These are, of course, right triangles so you can use the Pythagorean theorem To determine the length of the hypotenuse of the small triangle. Then, since the angles of the large and small triangles are the same, they are similar triangles. Corresponding parts of the two right triangles are proportional.

Thanks for the smaller triangle I get hypotenuse of 46.65. Then each leg and hypotenuse is multipled by a proportion?

So larger triangle would be 50 leg - no idea how to figure height or hypotenuse unless I multiple by 10/40 percent all the known lengths.

 

[MarkFL]

Let:

\(\displaystyle \overline{NO}=x\)

Then, by similarity, we may state:

\(\displaystyle \frac{x}{50}=\frac{24}{40}=\frac{3}{5}\)

Hence:

\(\displaystyle x=50\cdot\frac{3}{5}=10\cdot3=30\)

And then by Pythagoras:

\(\displaystyle \overline{OP}=\sqrt{30^2+50^2}=10\sqrt{3^2+5^2}=10\sqrt{34}\)