Right angle at A, AD ⊥ BC, BD = 3, DC = 5, find AD

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 BD = 3, DC = 5: find AD.

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

Cross-multiplied, that gives the rule: AD squared is BD times DC.

AD² = BD × DC

Careful!

It is BD times DC, not plus.

Careful: not AD² = BD + DC ✗

Put the numbers in.

We call AD x.

x² = 3 × 5

3 times 5 is 15.

x² = 15

x is the square root of 15.

A length is positive: 3.9, to the nearest tenth.

x = √15 ≈ 3.9

Let's check with Pythagoras in the small triangle A B D: A D squared plus B D squared is 24, about the same as A B squared.

So it fits!

triangle ABD: AD² + BD² = 15 + 9 = 24 = AB² = 24

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 = √15 ≈ 3.9

Try this one yourself, then check:

Find AD.

Show the answer

AD ≈ 4.9