First let’s label a few things, lengths of three segments with question marks above, three angles in the big triangle, and the point along the hypotenuse of the big triangle where the interior line meets it at a right angle:

We have three right triangles, one big one, XYZ, and two small ones, XWZ and XYW. We label the segments with question marks a, b, and d. We are going to use the Pythagorean theorem three times! It says, in a right triangle with legs a and b and hypotenuse c, a2 + b2 = c2.

1. In triangle XYZ, a² + b² = (16 +9)² = 25²
2. In triangle XWZ, 9² + d² = a²
3. In triangle XYW, 16² + d² = b²

So we substitute 9² + d² for a² in (1):

4. 9² + d² + b² = 25²

Next we substitute 16² + d² for b² in (4):

5. 9² + d² +16² + d² = 25²

So 2 d² = 25² – 16² – 9² = 625 – 256 – 81 = 288

d² = 144

d = 12

From (2),

a² = 9² + d² = 81 + 144 = 225

a = 15

From (1),

a² + b² = 25²,

so b² = 25² - 15² = 625 – 225 = 400

b = 20

So a = 15, b = 20, and d = 12

So the perpendicular in the figure has length 12 units, and the two legs have lengths 15 and 20 units.

We did it!