Right angle at A, AD ⊥ BC, AB = 10, BC = 25, find BD

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 AB = 10, BC = 25: find BD.

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

Cross-multiplied, that gives the rule: AB squared is BD times BC.

AB² = BD × BC

Careful!

It is BD times BC, not plus.

Careful: not AB² = BD + BC ✗

Put the numbers in.

We call BD x.

10² = x × 25

10 squared is 100.

100 = x × 25

To remove times 25, we divide both sides by 25: 4.

x = 10025 = 4

Let's check with Pythagoras in the small triangle A B D: A D squared plus B D squared is 100 the same as A B squared.

So it fits!

triangle ABD: AD² + BD² = 84 + 16 = 100 = AB² = 100

Let's look back.

First: find the similar triangles.

Then: write the rule.

Then: put the numbers in.

Then: remove the 25.

Last: check.

Answer

x = 10025 = 4

Try this one yourself, then check:

Find BD.

Show the answer

BD = 1