In this chapter we will derive the formula to calculate the length of altitude of right triangle.

## Formula for altitude length – Right triangle

The triangle in which **one angle measure 90 degree** is called **right angle triangle.**

Given below is the right triangle ABC with ∠B = 90 degree. The line BM is the altitude of the triangle which intersect hypotenuse AC at right angle.

The altitude intersecting the hypotenuse **divides the triangle into two similar triangle**.

Hence, \mathtt{\triangle ABM\ \sim \ \triangle MBC}

**In similar triangle, the corresponding sides are proportional.**

\mathtt{MB^{2} =\ AB\ \times \ BC}\\\ \\ \mathtt{h^{2} =\ AB\ \times \ BC}\\\ \\ \mathtt{h\ =\ \sqrt{AB.\ BC}}

Hence, the altitude of right triangle is the square root of the product of two sides of the triangle.

I hope you understood the formula, let us solve some problems for further clarity.

## Altitude of right triangle – Solved problems

**Example 01**

Given below is the right triangle ABC. Find the length of altitude BM.

**Solution**

In the above triangle;

AB = 6 cm

BC = 8 cm

The formula for length of altitude is given as;

\mathtt{h\ =\ \sqrt{AB.\ BC}}

Putting the values;

\mathtt{h\ =\ \sqrt{8\times 6}}\\\ \\ \mathtt{h\ =\ \sqrt{48}}\\\ \\ \mathtt{h\ =\ 4\ \sqrt{3} \ cm}

Hence, altitude length is \mathtt{\ 4\ \sqrt{3} \ cm\ } cm.

**Example 02**

Find the length of altitude BP of below right angled triangle.

**Solution**

Given above is right angle triangle ABC where;

AB = 5 cm

AC = 13 cm

To find the length of altitude BP, we need to first find length of BC.

Applying Pythagoras theorem

\mathtt{AC^{2} =\ AB^{2} +\ BC^{2}}\\\ \\ \mathtt{13^{2} =\ 5^{2} +\ BC^{2}}\\\ \\ \mathtt{169\ =\ 25\ +\ BC^{2}}\\\ \\ \mathtt{BC^{2} =\ 169\ -\ 25}\\\ \\ \mathtt{BC^{2} =\ 144}\\\ \\ \mathtt{BC=\ 12}

Now using the altitude formula for right triangle.

\mathtt{h\ =\ \sqrt{AB.\ BC}}\\\ \\ \mathtt{h\ =\ \sqrt{5\times 12}}\\\ \\ \mathtt{h\ =\sqrt{60}}\\\ \\ \mathtt{h\ =\ 2\sqrt{15} \ cm}

Hence, length of altitude is \mathtt{2\sqrt{15} \ cm\ }