Right angle at A, AD ⊥ BC, DC = 3 cm, BC = 10 cm, 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 = 3 cm, BC = 10 cm: 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 = 3 × 10

3 times 10 is 30.

x² = 30

x is the square root of 30.

A length is positive: 5.5 cm, to the nearest tenth.

x = √30 ≈ 5.5 cm

Let's check with Pythagoras in the small triangle A B D: A D squared plus B D squared is 70, about the same as A B squared.

So it fits!

triangle ABD: AD² + BD² = 21 + 49 = 70 = AB² = 70

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 = √30 ≈ 5.5 cm

Try this one yourself, then check:

Find AC.

Show the answer

AC ≈ 5.7 cm