Right angle at A, AD ⊥ BC, DC = 9, BC = 25, find AC

Step by step, with the reason for every step · Right triangle with its height to the hypotenuse · Geometry

Step by step

Question: triangle A B C has a right angle at A, and A D is at 90° to B C, with DC = 9, BC = 25: find AC.

The height A D cuts the big triangle into two small ones.

All three triangles have a right angle, and they share their other angles.

So they are similar: the same shape.

Triangles D A C and A B C: D C over A C equals A C over B C.

Cross-multiplied, that gives the rule: AC squared is DC times BC.

AC² = DC × BC

Careful!

It is DC times BC, not plus.

Careful: not AC² = DC + BC ✗

Put the numbers in.

We call AC x.

x² = 9 × 25

9 times 25 is 225.

x² = 225

x is the square root of 225.

A length is positive: 15.

x = √225 = 15

Let's check with Pythagoras in the small triangle A B D: A D squared plus B D squared is 400 the same as A B squared.

So it fits!

triangle ABD: AD² + BD² = 144 + 256 = 400 = AB² = 400

Let's look back.

First: find the similar triangles.

Then: write the rule.

Then: put the numbers in.

Then: take the square root.

Last: check.

Answer

x = √225 = 15

Try this one yourself, then check:

Find AC.

Show the answer

AC = 18